RS Aggarwal Class 10 Mathematics Solutions Chapter 15 Probability

 

Question 1. Fill in the blanks:
(i) The probability of an impossible event is ……. .
(ii) The probability of a sure event is …….. .
(iii) For any event E, P(E) + P (not E) = …….. .
(iv) The probability of a possible but not a sure event lies between ……… and …….. .
(v) The sum of probabilities of all the outcomes of an experiment is ……. .
Answer: (i) The probability of an impossible event is zero. (ii) The probability of a sure event is one. (iii) For any event E, P(E) + P(not E) = one. (iv) The probability of a possible but not a sure event lies between zero and one. (v) The sum of probabilities of all the outcomes of an experiment is one.
In simple words: Impossible events never happen - probability zero. Sure events always happen - probability one. Any event and its opposite always add to one. A normal event falls somewhere between zero and one.

Exam Tip: These are fundamental definitions in probability - memorise them. They form the foundation for solving all probability problems.

 

Question 2. A coin is tossed once. What is the probability of getting a tail?
Answer: When a coin is tossed, the possible outcomes are H (heads) and T (tails), giving a total of 2 outcomes. The number of ways to get a tail is 1. Therefore, P(getting a tail) = 1/2.
In simple words: A coin has two equally likely outcomes - heads or tails. So the chance of getting tails is one out of two, which is one-half.

Exam Tip: Always identify all possible outcomes first, then count how many match your requirement. This method works for all probability problems.

 

Question 3. Two coins are tossed simultaneously. Find the probability of getting
(i) exactly 1 head
(ii) at most 1 head
(iii) at least 1 head
Answer: When two coins are tossed, the possible outcomes are {HH, HT, TH, TT}, giving a total of 4 outcomes. (i) For exactly 1 head, the favourable outcomes are HT and TH, so there are 2 favourable outcomes. Thus, P(exactly 1 head) = 2/4 = 1/2. (ii) For at most 1 head (meaning 0 or 1 head), the favourable outcomes are HT, TH, and TT, so there are 3 favourable outcomes. Thus, P(at most 1 head) = 3/4. (iii) For at least 1 head (meaning 1 or more heads), the favourable outcomes are HH, HT, and TH, so there are 3 favourable outcomes. Thus, P(at least 1 head) = 3/4.
In simple words: List all outcomes first. Count how many match what you want. Divide by the total. "Exactly 1" means just one, "at most 1" means up to one, "at least 1" means one or more.

Exam Tip: Pay close attention to language - "exactly," "at most," and "at least" have different meanings. List all outcomes systematically to avoid missing any.

 

Question 4. A die is thrown once. Find the probability of getting
(i) an even number
(ii) a number greater than 2
(iii) a number greater than 2
(iv) a number between 3 and 6
(v) a number other than 3
(vi) the number 5
Answer: When a die is thrown, all possible outcomes are 1, 2, 3, 4, 5, 6, giving a total of 6 outcomes. (i) For an even number, the favourable outcomes are 2, 4, 6, so there are 3 favourable outcomes. Thus, P(even number) = 3/6 = 1/2. (ii) For a number greater than 2, the favourable outcomes are 3, 4, 5, 6, so there are 4 favourable outcomes. Thus, P(greater than 2) = 4/6 = 2/3. (iii) Same as (ii): P(greater than 2) = 2/3. (iv) For a number between 3 and 6 (not including 3 and 6), the favourable outcomes are 4, 5, so there are 2 favourable outcomes. Thus, P(between 3 and 6) = 2/6 = 1/3. (v) For a number other than 3, the favourable outcomes are 1, 2, 4, 5, 6, so there are 5 favourable outcomes. Thus, P(other than 3) = 5/6. (vi) For the number 5, the favourable outcome is 5, so there is 1 favourable outcome. Thus, P(getting 5) = 1/6.
In simple words: Count the number of outcomes matching the condition, then divide by 6. Even numbers are 2, 4, 6. "Between" usually means not including the endpoints unless specified otherwise.

Exam Tip: Read carefully whether "between" includes or excludes the boundary numbers. In most textbooks, "between" excludes the endpoints.

 

Question 5. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Answer: There are 26 letters in the English alphabet. The vowels are A, E, I, O, and U, making 5 vowels. Therefore, the number of consonants is 26 - 5 = 21. The probability of selecting a consonant is 21/26.
In simple words: Count all consonants out of 26 letters. There are 5 vowels, so 21 consonants remain. The probability is 21 out of 26.

Exam Tip: Remember to subtract vowels from the total to find consonants. Know the 5 vowels: A, E, I, O, U.

 

Question 6. A child has a die whose 6 faces show the letters given below:

A

B

C

A

A

B

The die is thrown once. What is the probability of getting (i) A and (ii) B?
Answer: Since there are 6 letters on the die, the total number of outcomes is 6. (i) The letter A appears 3 times on the die, so the number of favourable outcomes is 3. Thus, P(getting A) = 3/6 = 1/2. (ii) The letter B appears 2 times on the die, so the number of favourable outcomes is 2. Thus, P(getting B) = 2/6 = 1/3.
In simple words: Count how many times each letter shows up on the die. Divide by 6 to get the probability. A shows 3 times, B shows 2 times.

Exam Tip: Count the frequency of each outcome carefully. Some outcomes may repeat on the die - count each one separately.

 

Question 7. It is known that a box of 200 electric bulbs contains 16 defective bulbs. One bulb is taken out at random from the box. What is the probability that the bulb drawn is
(i) defective
(ii) non - defective
Answer: The total number of bulbs is 200. (i) The number of defective bulbs is 16. Therefore, P(defective bulb) = 16/200 = 2/25. (ii) The number of non - defective bulbs is 200 - 16 = 184. Therefore, P(non - defective bulb) = 184/200 = 23/25.
In simple words: Out of 200 bulbs, 16 are broken. So 184 are good. The chance of picking a broken one is 16 out of 200. The chance of picking a good one is 184 out of 200.

Exam Tip: Use the complement rule: P(defective) + P(non - defective) = 1. Always simplify fractions to lowest terms.

 

Question 8. If the probability of winning a game is 0.7, what is the probability of losing it?
Answer: Let E be the event of winning the game. Then E' is the event of losing the game. We know that P(E) + P(E') = 1. Since P(E) = 0.7, we have 0.7 + P(E') = 1. Therefore, P(E') = 1 - 0.7 = 0.3. The probability of losing the game is 0.3.
In simple words: Winning and losing are opposites. They must add up to 1. If the chance of winning is 0.7, then the chance of losing is what remains: 1 - 0.7 = 0.3.

Exam Tip: For complementary events, always use P(E) + P(E') = 1. This is one of the quickest ways to find an answer.

 

Question 9. There are 35 students in a class of whom 20 are boys and 15 are girls. From these students one is chosen at random. What is the probability that the chosen student is a (i) boy, (ii) girl?
Answer: The total number of students is 35. (i) The number of boys is 20. Therefore, P(choosing a boy) = 20/35 = 4/7. (ii) The number of girls is 15. Therefore, P(choosing a girl) = 15/35 = 3/7.
In simple words: There are 35 students total. 20 are boys and 15 are girls. The chance of picking a boy is 20 out of 35. The chance of picking a girl is 15 out of 35.

Exam Tip: Notice that the probabilities add to 1: 4/7 + 3/7 = 1. Use this as a check on your answer.

 

Question 10. In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?
Answer: The number of prizes is 10 and the number of blanks is 25. The total number of tickets is 10 + 25 = 35. Therefore, P(getting a prize) = 10/35 = 2/7.
In simple words: In a lottery with 10 winning tickets and 25 non - winning tickets, the total is 35. Your chance of winning is 10 out of 35.

Exam Tip: Always add up the total number of outcomes correctly. Don't miss any category.

 

Question 11. 250 lottery tickets were sold and there are 5 prizes on these tickets. If Kunal has purchased one lottery ticket, what is the probability that he wins a prize?
Answer: The total number of tickets sold is 250. The number of prizes is 5. Kunal has purchased one ticket, so he has one chance to win among all 250 tickets. Therefore, P(Kunal wins a prize) = 5/250 = 1/50.
In simple words: Out of 250 tickets, 5 are winners. Kunal has one ticket. His chance of winning is 5 out of 250, which simplifies to 1 out of 50.

Exam Tip: The probability does not depend on how many tickets Kunal bought - only on the ratio of winning tickets to total tickets.

 

Question 12. 17 cards numbered 1, 2, 3, 4, …. ,17 are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the card drawn bears (i) an odd number (ii) a number divisible by 5.
Answer: The total number of cards is 17. (i) The odd numbers from 1 to 17 are 1, 3, 5, 7, 9, 11, 13, 15, 17, making 9 odd numbers. Therefore, P(drawing an odd number) = 9/17. (ii) The numbers divisible by 5 are 5, 10, 15, making 3 numbers. Therefore, P(drawing a number divisible by 5) = 3/17.
In simple words: Count all odd numbers from 1 to 17 - there are 9 of them. Count numbers divisible by 5 - there are 3 of them (5, 10, 15). Divide each by 17.

Exam Tip: When finding odd numbers, remember they end in 1, 3, 5, 7, 9. For divisibility by 5, numbers must end in 0 or 5.

 

Question 13. A game of chance consists of spinning an arrow, which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8.
Answer: The total number of outcomes is 8 (the arrow can point to any number from 1 to 8). The factors of 8 are 1, 2, 4, and 8, making 4 factors. Therefore, P(arrow pointing at a factor of 8) = 4/8 = 1/2.
In simple words: The number 8 is divisible by 1, 2, 4, and 8. These are its four factors. So the chance is 4 out of 8, which is one - half.

Exam Tip: To find factors, list all numbers that divide evenly into 8 with no remainder. Don't forget 1 and the number itself.

 

Question 14. In a family of 3 children, find the probability of having at least one boy.
Answer: The possible outcomes for 3 children (using B for boy and G for girl) are: GGG, BGG, GBG, GGB, BBG, BGB, GBB, BBB, making 8 total outcomes. The outcomes with at least one boy are all except GGG, giving us BGG, GBG, GGB, BBG, BGB, GBB, BBB, which is 7 outcomes. Therefore, P(at least one boy) = 7/8.
In simple words: List all 8 possible combinations of boys and girls. Count those that have at least one boy - that's everything except all girls. Seven out of eight families will have at least one boy.

Exam Tip: "At least one" often means "find the complement" - it's easier to count "no boys" (just GGG) and subtract from 1.

 

Question 15. A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is
(i) black,
(ii) not green,
(iii) red or white,
(iv) neither red nor green.
Answer: The bag has 4 + 5 + 2 + 4 = 15 balls in total. (i) The number of black balls is 2. Therefore, P(black ball) = 2/15. (ii) The number of green balls is 4, so non - green balls = 15 - 4 = 11. Therefore, P(not green) = 11/15. (iii) The number of red and white balls is 5 + 4 = 9. Therefore, P(red or white) = 9/15 = 3/5. (iv) The number of red and green balls is 5 + 4 = 9, so balls that are neither red nor green = 15 - 9 = 6. Therefore, P(neither red nor green) = 6/15 = 2/5.
In simple words: Total 15 balls. Black: 2 out of 15. Not green means all except 4 green: 11 out of 15. Red or white combined: 5 + 4 = 9 out of 15. Neither red nor green means only black and white: 6 out of 15.

Exam Tip: "Or" means add the groups. "Neither - nor" means exclude both groups. Always verify: sum of complementary probabilities equals 1.

 

Question 16. A card is drawn at random from a well - shuffled pack of 52 cards. Find the probability of getting
(i) a red king,
(ii) a queen or a jack.
Answer: A standard pack has 52 cards. (i) There are 2 red kings (one in hearts, one in diamonds) out of 52 cards. Therefore, P(red king) = 2/52 = 1/26. (ii) There are 4 queens and 4 jacks in the pack, making 8 cards total. Therefore, P(queen or jack) = 8/52 = 2/13.
In simple words: A deck has 52 cards. Only 2 are red kings. There are 4 queens and 4 jacks, so 8 cards that are either. Divide by 52 and simplify.

Exam Tip: Remember: 2 red suits (hearts, diamonds) and 2 black suits (clubs, spades). Each rank has 4 cards total.

 

Question 17. A card is drawn at random from a well - shuffled pack of 52 cards. Find the probability that the drawn card is neither a king nor a queen.
Answer: In a standard pack there are 26 red cards and 2 red queens are among them. Additionally, there are 2 more black queens in the pack, making 28 cards that are either red or queens (since the red queens are already counted in the red cards). There are 4 kings and 4 queens, so 8 cards that are either a king or a queen. Therefore, the number of cards that are neither a king nor a queen is 52 - 8 = 44. Hence, P(neither king nor queen) = 44/52 = 11/13.
In simple words: There are 4 kings and 4 queens in the deck, totaling 8 cards to avoid. The remaining 44 cards are neither kings nor queens. So the probability is 44 out of 52.

Exam Tip: For "neither - nor" questions, subtract both unwanted groups from the total. Check: P(kings or queens) + P(neither) should equal 1.

 

Question 18. A card is drawn from a well - shuffled pack of 52 cards. Find the probability of getting
(i) a red face card
(ii) a black king.
Answer: A standard pack has 52 cards. (i) Face cards are jacks, queens, and kings. There are 3 face cards in each suit. In the 2 red suits (hearts and diamonds), there are 3 × 2 = 6 red face cards. Therefore, P(red face card) = 6/52 = 3/26. (ii) There are 2 black kings (one in clubs, one in spades) out of 52 cards. Therefore, P(black king) = 2/52 = 1/26.
In simple words: Face cards are J, Q, K. In red suits there are 6 face cards total. Black kings: one in clubs and one in spades, so 2 cards. Divide by 52 each time.

Exam Tip: Face cards are J, Q, K only - not numbered cards. There are 3 face cards per suit, so 6 per colour.

 

Question 19. Two different dice are tossed together. Find the probability that
(i) the number on each die is even,
(ii) the sum of the numbers appearing on the two dice is 5.
Answer: When two dice are tossed, there are 36 possible outcomes. (i) The even numbers on a die are 2, 4, 6. For both dice to show even numbers, the outcomes are (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6), making 9 outcomes. Therefore, P(both even) = 9/36 = 1/4. (ii) For the sum to be 5, the outcomes are (1,4), (2,3), (3,2), (4,1), making 4 outcomes. Therefore, P(sum is 5) = 4/36 = 1/9.
In simple words: Two dice give 36 total outcomes. For both even: each die has 3 even choices, so 3 × 3 = 9 ways. For sum = 5: there are 4 pairs that add to 5.

Exam Tip: When multiple conditions apply, multiply the individual counts. Systematically list all outcomes to avoid missing any.

 

Question 20. Two different dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is 10.
Answer: When two dice are rolled, there are 36 possible outcomes. For the sum to be 10, the outcomes are (4,6), (5,5), (6,4), making 3 outcomes. Therefore, P(sum is 10) = 3/36 = 1/12.
In simple words: Two dice create 36 total possibilities. To get a sum of 10, you need 4 + 6, 5 + 5, or 6 + 4. That's 3 ways out of 36.

Exam Tip: For sum problems with two dice, list ordered pairs carefully. (4,6) and (6,4) are different outcomes and both should be counted.

 

Question 21. When two dice are tossed together, find the probability that the sum of numbers on their tops is less than 7.
Answer: When two dice are rolled, there are 36 possible outcomes. For the sum to be less than 7 (sums of 2, 3, 4, 5, 6), the outcomes are: sum 2: (1,1) - 1 outcome; sum 3: (1,2), (2,1) - 2 outcomes; sum 4: (1,3), (2,2), (3,1) - 3 outcomes; sum 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes; sum 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes. Total favourable outcomes = 1 + 2 + 3 + 4 + 5 = 15. Therefore, P(sum less than 7) = 15/36 = 5/12.
In simple words: For sums less than 7, count all pairs that give 2, 3, 4, 5, or 6. This gives 15 pairs out of 36 total.

Exam Tip: "Less than" does not include the boundary value - sum must be 6 or lower, not 7 or higher. Count systematically by sum value.

 

Question 22. Two dice are rolled together. Find the probability of getting such numbers on two dice whose product is a perfect square.
Answer: When two dice are rolled, there are 36 possible outcomes. For the product to be a perfect square, we need outcomes where the result is 1, 4, 9, 16, 25, or 36. The outcomes are: (1,1) with product 1; (1,4), (2,2), (4,1) with product 4; (3,3) with product 9; (4,4) with product 16; (5,5) with product 25; (6,6) with product 36. That is 1 + 3 + 1 + 1 + 1 + 1 = 8 outcomes. Therefore, P(product is a perfect square) = 8/36 = 2/9.
In simple words: Find all pairs whose product is a perfect square (1, 4, 9, 16, 25, 36). Count them: there are 8 pairs out of 36 total.

Exam Tip: Perfect squares are 1, 4, 9, 16, 25, 36, ... (numbers that can be expressed as n × n). Check each systematically.

 

Question 23. Two dice are rolled together. Find the probability of getting such numbers on the two dice whose product is 12.
Answer: When two dice are rolled, there are 36 possible outcomes. For the product to be 12, we need pairs where the two numbers multiply to give 12. The outcomes are: (2,6), (3,4), (4,3), (6,2), making 4 outcomes. Therefore, P(product is 12) = 4/36 = 1/9.
In simple words: Find pairs of numbers from 1 to 6 that multiply to 12. These are 2×6, 3×4, 4×3, 6×2. That's 4 pairs out of 36.

Exam Tip: Always check that both numbers are between 1 and 6 (valid die outcomes). Count ordered pairs - (2,6) and (6,2) are different.

 

Question 24. Cards, marked with numbers 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is (i) a prime number less than 10 (ii) a perfect square.
Answer: The cards are numbered 5 to 50, making 46 cards in total. (i) The prime numbers less than 10 are 5 and 7 (both within the range 5 to 50). Therefore, P(prime less than 10) = 2/46 = 1/23. (ii) The perfect squares from 5 to 50 are 9, 16, 25, 36, 49, making 5 numbers. Therefore, P(perfect square) = 5/46.
In simple words: There are 46 cards (from 5 to 50). The primes below 10 are 5 and 7. Perfect squares in this range are 9, 16, 25, 36, 49 - that's 5 numbers.

Exam Tip: Count the total outcomes carefully (50 - 5 + 1 = 46 cards). List perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, ... - only count those in your range.

 

Question 25. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the numbers 1, 2, 3,…, 12 as shown in the figure. What is the probability that it will point to
(i) 6?
(ii) an even number?
(iii) a prime number?
(iv) a number which is a multiple of 5?
Answer: The arrow can land on any of the 12 numbers, giving a total of 12 outcomes. (i) The arrow pointing at exactly 6 has 1 favourable outcome. Therefore, P(pointing at 6) = 1/12. (ii) The even numbers are 2, 4, 6, 8, 10, 12, making 6 favourable outcomes. Therefore, P(even) = 6/12 = 1/2. (iii) The prime numbers are 2, 3, 5, 7, 11, making 5 favourable outcomes. Therefore, P(prime) = 5/12. (iv) The multiples of 5 are 5 and 10, making 2 favourable outcomes. Therefore, P(multiple of 5) = 2/12 = 1/6.
In simple words: Total 12 numbers. For 6: just 1 outcome. Even: 6 numbers (2, 4, 6, 8, 10, 12). Prime: 5 numbers (2, 3, 5, 7, 11). Multiples of 5: just 5 and 10.

Exam Tip: Remember that 1 is not prime. Check each category carefully - even numbers must be divisible by 2, multiples of 5 must end in 0 or 5.

 

Question 26. 12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. One pen is taken out at random from this lot. Find the probability that the pen taken out is good one.
Answer: The total number of pens is 12 + 132 = 144. The number of good pens is 132. Therefore, P(pen is good) = 132/144 = 11/12.
In simple words: Total pens: 12 defective + 132 good = 144 pens. The chance of picking a good one is 132 out of 144.

Exam Tip: Always add up the total correctly. Simplify the fraction by finding the GCD - here both divide by 12.

 

Question 27. A lot consists of 144 ballpoint pens of which 20 are defective and others good. Tanvy will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) she will buy it,
(ii) she will not buy it?
Answer: The total number of pens is 144. The number of defective pens is 20, so the number of good pens is 144 - 20 = 124. (i) She will buy a pen if it is good. Therefore, P(she will buy it) = 124/144 = 31/36. (ii) She will not buy a pen if it is defective. Therefore, P(she will not buy it) = 20/144 = 5/36.
In simple words: Total 144 pens. 20 are broken, so 124 are good. She buys if it's good: 124 out of 144. She refuses if it's defective: 20 out of 144.

Exam Tip: Verify: P(buy) + P(don't buy) should equal 1. Here 31/36 + 5/36 = 36/36 = 1. ✓

 

Question 28. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two - digit number,
(ii) a perfect square number,
(iii) a number divisible by 5.
Answer: The total number of discs is 90. (i) The two - digit numbers from 1 to 90 are 10, 11, 12, ..., 90. These are the numbers from 10 to 90, making 81 two - digit numbers. Therefore, P(two - digit number) = 81/90 = 9/10. (ii) The perfect squares from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81, making 9 perfect squares. Therefore, P(perfect square) = 9/90 = 1/10. (iii) The numbers divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, making 18 numbers. Therefore, P(divisible by 5) = 18/90 = 1/5.
In simple words: Total 90 discs. Two - digit: all from 10 to 90 is 81 numbers. Perfect squares up to 90: 1, 4, 9, 16, 25, 36, 49, 64, 81 - that's 9 numbers. Divisible by 5: count by 5s from 5 to 90 - that's 18 numbers.

Exam Tip: Two - digit numbers start at 10, not 1. Perfect squares: check if √n is a whole number. Divisible by 5: must end in 0 or 5. Use 90 ÷ 5 = 18 as a quick count method.

 

Question 29. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
Answer: The total number of bulbs = 20. There are 4 defective bulbs. So, P(getting a defective bulb) = \( \frac{4}{20} = \frac{1}{5} \)
In simple words: Out of 20 bulbs, 4 are broken. The chance of picking a broken bulb is 4 out of 20, which simplifies to 1 out of 5.

Exam Tip: Always simplify fractions to lowest terms - here 4/20 becomes 1/5 by dividing both by 4. Examiners expect the final answer in simplest form.

 

Question 29. (ii) Suppose the bulb drawn in (i) is not defective and not replaced. Now, bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Answer: Since the bulb drawn in (i) is not defective and not replaced, there are 4 defective bulbs left from a total of 19 bulbs. So, there are 15 non-defective bulbs. P(getting a non-defective bulb) = \( \frac{15}{19} \)
In simple words: We started with 20 bulbs. One non-broken bulb was removed. Now 19 remain, with still 4 broken ones. So 15 are not broken. The chance of picking a non-broken bulb is 15 out of 19.

Exam Tip: In problems without replacement, the total number of items decreases by 1 for the second draw. Keep track of both changes - total count and the count of the specific type.

 

Question 30. A bag contains lemon-flavoured candies only. Hema takes out one candy without looking into the bag. What is the probability that she takes out (i) an orange-flavoured candy?
Answer: P(that she takes out an orange-flavoured candy) = 0

In simple words: Since there are no orange-flavoured candies in the bag at all, it is impossible to pull out an orange candy. The probability of an impossible event is always 0.

Exam Tip: When an event cannot happen, its probability is 0. When an event must happen, its probability is 1. These are the extreme cases.

 

Question 30. (ii) a lemon-flavoured candy?
Answer: P(that she takes out a lemon-flavoured candy) = 1

In simple words: Since every candy in the bag is lemon-flavoured, she will definitely pick out a lemon candy. When something is certain to happen, the probability is always 1.

Exam Tip: A certain event (one that must occur) always has probability 1. This is a key concept in probability - the range is always 0 to 1 inclusive.

 

Question 31. There are 40 students in a class of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. He writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl?
Answer: Total number of students = 40. The number of girls = 25. P(that the name written is a girl) = \( \frac{25}{40} = \frac{5}{8} \)
In simple words: There are 40 cards total, and 25 of them have girls' names. The chance of pulling out a girl's name is 25 out of 40, which reduces to 5 out of 8.

Exam Tip: Always reduce the final fraction to its simplest form by finding the GCD. Here GCD(25, 40) = 5, so divide both numerator and denominator by 5.

 

Question 31. (ii) a boy?
Answer: The number of boys = 15. P(that the name written is a boy) = \( \frac{15}{40} = \frac{3}{8} \)
In simple words: Of the 40 cards, 15 have boys' names. The chance of drawing a boy's name is 15 out of 40, which simplifies to 3 out of 8.

Exam Tip: Notice that P(girl) + P(boy) = 5/8 + 3/8 = 1. This confirms the calculation since all cards have either a girl's or boy's name.

 

Question 32. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing (i) an ace
Answer: Total number of all possible outcomes = 52. There are 4 ace cards in a pack of cards. P(getting an ace) = \( \frac{4}{52} = \frac{1}{13} \)
In simple words: A standard deck has 52 cards with 4 aces (one of each suit). The chance of drawing an ace is 4 out of 52, which reduces to 1 out of 13.

Exam Tip: Memorise standard deck facts: 52 cards total, 4 aces, 4 kings, 4 queens, 4 jacks, 13 cards per suit, 26 red cards, 26 black cards. These save calculation time.

 

Question 32. (ii) a 4 of spades
Answer: P(getting a '4' of spades) = \( \frac{1}{52} \)
In simple words: There is exactly one 4 of spades in the entire deck of 52 cards. So the probability is 1 out of 52.

Exam Tip: When a specific single card is required (like a specific rank of a specific suit), the probability is always 1/52 since there is only one such card in the deck.

 

Question 32. (iii) a 9 of a black suit
Answer: P(a '9' of a black suit) = \( \frac{2}{52} = \frac{1}{26} \)
In simple words: There are two black suits (spades and clubs), and each has one 9. So there are 2 nines of black suits out of 52 cards, which reduces to 1 out of 26.

Exam Tip: When counting cards of a specific rank in a specific colour, remember black suits are clubs and spades. Count how many cards match all conditions, then divide by 52.

 

Question 32. (iv) a red king
Answer: P(getting a red king) = \( \frac{2}{52} = \frac{1}{26} \)
In simple words: Red kings are the king of hearts and king of diamonds - that is 2 cards. Out of 52 total cards, the probability is 2 out of 52, which simplifies to 1 out of 26.

Exam Tip: Red suit means hearts or diamonds. There are 2 red kings in a deck. Always check if the condition specifies rank, suit colour, or both.

 

Question 33. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting (i) a queen
Answer: Total numbers of cards = 52. There are 4 queen cards in a pack of cards. P(Probability of getting a queen card) = \( \frac{4}{52} = \frac{1}{13} \)
In simple words: A deck has one queen in each of the four suits, giving 4 queens total. The chance of drawing a queen is 4 out of 52, which reduces to 1 out of 13.

Exam Tip: Remember there are 4 of each face card (king, queen, jack) in a standard deck. This is a key fact to memorise.

 

Question 33. (ii) a diamond
Answer: There are 13 cards of diamond in a pack of cards. P(probability of getting a diamond card) = \( \frac{13}{52} = \frac{1}{4} \)
In simple words: Each suit has 13 cards, so there are 13 diamonds. The chance of drawing a diamond is 13 out of 52, which reduces to 1 out of 4.

Exam Tip: Each of the 4 suits (hearts, diamonds, clubs, spades) has exactly 13 cards. So P(any suit) = 13/52 = 1/4.

 

Question 33. (iii) a king or an ace
Answer: In a pack of cards there are 4 kings and 4 aces. Number of such cards = 4 + 4 = 8. P(probability of getting either a king or an ace) = \( \frac{8}{52} = \frac{2}{13} \)
In simple words: There are 4 kings and 4 aces - giving 8 cards total that satisfy the condition. The chance is 8 out of 52, which simplifies to 2 out of 13.

Exam Tip: For "either-or" events (like king OR ace), add the individual counts. Here 4 + 4 = 8, then divide by the total 52.

 

Question 33. (iv) a red ace
Answer: There are two red aces in a pack of cards (ace of hearts and ace of diamonds). P(probability of getting a red ace) = \( \frac{2}{52} = \frac{1}{26} \)
In simple words: Red aces are the ace of hearts and the ace of diamonds. That is 2 cards out of 52 total, which simplifies to 1 out of 26.

Exam Tip: When an event has multiple conditions (rank AND colour), count only the cards that meet ALL conditions. Red aces = only 2 specific cards.

 

Question 34. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting (i) a king of red suit
Answer: We know that there are 52 cards in all. There are 2 red kings in a pack of cards. P(getting a king of red suit) = \( \frac{2}{52} = \frac{1}{26} \)
In simple words: Red kings consist of the king of hearts and king of diamonds - exactly 2 cards. Out of 52 cards, the probability is 2 out of 52, reducing to 1 out of 26.

Exam Tip: Red suit = hearts or diamonds. Count cards that match the rank AND the colour condition together.

 

Question 34. (ii) a face card
Answer: The number of face cards = 12 (4 kings, 4 queens, 4 jacks). P(getting a face card) = \( \frac{12}{52} = \frac{3}{13} \)
In simple words: Face cards are kings, queens, and jacks. There are 4 of each type - 4 + 4 + 4 = 12 total face cards. The probability is 12 out of 52, which reduces to 3 out of 13.

Exam Tip: Know that face cards include only king, queen, and jack - never ace or number cards. This gives exactly 12 face cards in a deck.

 

Question 34. (iii) a red face card
Answer: The number of red face cards = 6 (2 red kings, 2 red queens, 2 red jacks). P(getting a red face card) = \( \frac{6}{52} = \frac{3}{26} \)
In simple words: Red face cards come from hearts and diamonds only. Each suit has 3 face cards (king, queen, jack), so 2 suits × 3 = 6 red face cards total. The probability is 6 out of 52, which reduces to 3 out of 26.

Exam Tip: For red face cards, remember both the rank (must be K, Q, or J) and colour (must be red). Count systematically: 2 red kings + 2 red queens + 2 red jacks = 6.

 

Question 34. (iv) a queen of black suit
Answer: The number of queens of black suit = 2 (queen of clubs and queen of spades). P(getting a queen of black suit) = \( \frac{2}{52} = \frac{1}{26} \)
In simple words: Black queens are the queen of spades and queen of clubs - that is 2 cards. The probability is 2 out of 52, which reduces to 1 out of 26.

Exam Tip: Black suit means clubs or spades. There is exactly one queen in each black suit, giving 2 total.

 

Question 34. (v) a jack of hearts
Answer: The number of jack of hearts = 1. P(getting a jack of hearts) = \( \frac{1}{52} \)
In simple words: There is only one jack of hearts in the entire deck. The probability is 1 out of 52.

Exam Tip: For a specific unique card, the probability is always 1/52. Make sure you are identifying a single specific card, not a group.

 

Question 34. (vi) a spade
Answer: The number of spade = 13. P(getting a spade) = \( \frac{13}{52} = \frac{1}{4} \)
In simple words: Spades is one of the four suits, and each suit has 13 cards. The probability is 13 out of 52, which reduces to 1 out of 4.

Exam Tip: For any single suit, the probability is always 13/52 = 1/4. This is because each suit is equally represented with 13 cards.

 

Question 35. A card is drawn at random form a well-shuffled deck of playing cards. Find the probability that the card drawn is (i) a card of spades of an ace
Answer: Total number of cards = 52. There are 13 cards of spade (including 1 ace) and 3 more ace cards are there in a pack of cards. So, P(getting a card of spades or an ace) = \( \frac{16}{52} = \frac{4}{13} \)
In simple words: There are 13 spades total. Beyond the spade ace, there are 3 more aces in the other suits. So the cards that are either spades or aces = 13 + 3 = 16. The probability is 16 out of 52, which reduces to 4 out of 13.

Exam Tip: When counting "A or B", be careful not to double-count. The ace of spades is a spade, so count it once with the 13 spades, then add only the remaining 3 aces from other suits.

 

Question 35. (ii) a red king
Answer: The number of red kings = 2. P(getting a red king) = \( \frac{2}{52} = \frac{1}{26} \)
In simple words: Red kings are the king of hearts and king of diamonds. That is 2 cards out of 52 total. The probability is 2 out of 52, which reduces to 1 out of 26.

Exam Tip: In problems about cards, always specify both rank and suit colour when needed. Red king ≠ red card that is a king and also red.

 

Question 35. (iii) either a king or a queen
Answer: There are 4 kings and 4 queens in a pack of cards. Number of such cards = 4 + 4 = 8. P(getting either a king or a queen) = \( \frac{8}{52} = \frac{2}{13} \)
In simple words: There are 4 kings and 4 queens, giving 8 cards total that match the condition. Out of 52 cards, the probability is 8 out of 52, which reduces to 2 out of 13.

Exam Tip: For "either A or B" where A and B don't overlap, add their counts. Here 4 + 4 = 8 since no card can be both a king and a queen.

 

Question 35. (iv) neither a king nor a queen.
Answer: There are 4 kings and 4 queens in a pack of cards. P(getting either a king or a queen) = \( \frac{8}{52} = \frac{2}{13} \)

P(getting neither a king nor a queen) = \( \left(1 - \frac{2}{13}\right) = \frac{11}{13} \)
In simple words: First find the chance of getting a king or queen, which is 8 out of 52 or 2 out of 13. Then, the probability of NOT getting either is 1 minus that, which equals 11 out of 13.

Exam Tip: Use the complement rule: P(not A) = 1 - P(A). This often makes "neither" or "not" problems simpler than counting directly.

 

Exercise 15b

 

Question 1. A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) divisible by 2 or 3,
Answer: There are 25 cards in total. The numbers divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24. The numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24. So, the total number of possible outcomes = 16. Note that 6, 12, 18 and 24 are twice. However, we count these numbers only once. P(getting a number divisible by 2) = \( \frac{16}{25} \)
In simple words: Count all numbers from 1 to 25 that can be split evenly by either 2 or 3. Start with those divisible by 2, then add those divisible by 3, but count shared numbers (like 6, 12, 18, 24) just once. This gives 16 numbers total, so the probability is 16 out of 25.

Exam Tip: When finding "divisible by A or B", use careful counting or the inclusion-exclusion principle: count A, count B, subtract the overlap to avoid double-counting.

 

Question 1. (ii) a prime number.
Answer: The primes are 2, 3, 5, 7, 11, 13, 17, 19 and 23. So, there are 9 possible outcomes. P(getting a prime) = \( \frac{9}{25} \)
In simple words: Prime numbers from 1 to 25 are those with exactly two factors: 1 and themselves. Counting them gives 9 primes total. The probability is 9 out of 25.

Exam Tip: For prime number questions, either memorise small primes (2, 3, 5, 7, 11, 13, 17, 19, 23...) or check each number systematically by testing divisibility by primes up to its square root.

 

Question 2. A box contains cards numbered 3, 5, 7, 9, …. , 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.
Answer: The numbers on the cards are 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 and 37. So, there 18 cards. The cards with primes are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. So, there are 11 possible outcomes. P(getting a prime number) = \( \frac{11}{18} \)
In simple words: The box holds only odd numbers from 3 to 37. First, count how many cards exist - this is an arithmetic sequence with first term 3, last term 37, and common difference 2, giving 18 cards. Next, identify which of these 18 are prime: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 - that is 11 primes. The probability is 11 out of 18.

Exam Tip: When the sample space is an arithmetic sequence, use the formula n = (last - first)/difference + 1 to count terms quickly. For this question: (37 - 3)/2 + 1 = 18 cards.

 

Question 3. Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) not divisible by 3,
Answer: There are 30 cards in total. The numbers divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30. So, the numbers not divisible by 3 are 30 - 10 = 20. Thus, there are 20 possible outcomes. P(getting a number divisible by 3) = \( \frac{20}{30} = \frac{2}{3} \)
In simple words: Out of 1 to 30, the count of multiples of 3 is 30 ÷ 3 = 10. So numbers NOT divisible by 3 = 30 - 10 = 20. The probability is 20 out of 30, which simplifies to 2 out of 3.

Exam Tip: To count multiples of n up to m, use the formula: floor(m ÷ n). For "not divisible", use the complement: P(not divisible) = 1 - P(divisible).

 

Question 3. (ii) a prime number greater than 7,
Answer: The primes greater than 7 are 11, 13, 17, 19, 23 and 29. So, there are 6 possible outcomes. P(getting a prime greater than 3) = \( \frac{6}{30} = \frac{1}{5} \)
In simple words: From cards 1 to 30, identify all primes bigger than 7. These are 11, 13, 17, 19, 23, and 29 - exactly 6 cards. The probability is 6 out of 30, which reduces to 1 out of 5.

Exam Tip: When a condition includes "greater than" or "less than", list all primes first, then filter by the range. Always simplify the final fraction.

 

Question 3. (iii) not a perfect square number.
Answer: The perfect squares would be 1, 4, 9, 16 and 25. So, there are 5 perfect squares, and hence 25 non-perfect squares. P(getting a perfect square) = \( \frac{25}{30} = \frac{5}{6} \)
In simple words: Perfect squares from 1 to 30 are 1, 4, 9, 16, and 25 - that is 5 cards. So non-perfect squares = 30 - 5 = 25 cards. The probability is 25 out of 30, which reduces to 5 out of 6.

Exam Tip: Perfect squares up to 30 are 1², 2², 3², 4², 5² = 1, 4, 9, 16, 25. Use the complement rule when finding "not" a property - count what IS the property, then subtract from the total.

 

Question 4. Cards bearing numbers 1, 3, 5, …. , 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (i) a prime number less than 15,
Answer: To find the number of cards in the bag, we use the general formula for an AP since the numbers on the cards are in AP. Here, first term, a = 1, common difference = d = 2. Let n be the number of cards in the bag. a_n = a + (n - 1)d. So, 35 = 1 + (n - 1)(2) ⇒ 34 = 2n - 2 ⇒ 2n = 36 ⇒ n = 18. So, there 18 cards. The primes less than 15 are 3, 5, 7, 11 and 13. So, there are 5 possible outcomes. P(getting a prime less than 15) = \( \frac{5}{18} \)
In simple words: The cards have odd numbers from 1 to 35. Using the AP formula with first term 1, last term 35, and difference 2, we find there are 18 cards total. Among these, the primes smaller than 15 are 3, 5, 7, 11, and 13 - that is 5 cards. The probability is 5 out of 18.

Exam Tip: When cards form an arithmetic progression, use the formula n = (last - first)/difference + 1 to count them quickly rather than listing every term.

 

Question 4. (ii) a number divisible by 3 and 5.
Answer: The number divisible by 3 and 5 is 15 and 30. P(getting a 15 or 30) = \( \frac{2}{18} = \frac{1}{9} \)
In simple words: A number divisible by both 3 and 5 must be divisible by 15 (their LCM). From 1 to 35, the multiples of 15 are 15 and 30. Both are odd and in our card set. The probability is 2 out of 18, which reduces to 1 out of 9.

Exam Tip: "Divisible by A and B" means divisible by LCM(A, B). For 3 and 5 (which are coprime), LCM = 3 × 5 = 15. Count multiples of 15 up to 35.

 

Question 5. A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a one-digit number,
Answer: To find the number of cards in the bag, we use the general formula for an AP since the numbers on the cards are in AP. Here, first term, a = 6, common difference = d = 1. Let n be the number of cards in the bag. a_n = a + (n - 1)d. So, 70 = 6 + (n - 1)(1) ⇒ 70 = 6 + n - 1 ⇒ 70 = 5 + n ⇒ n = 65. So, there 65 cards. But this calculation needs correction: a_n = 70 means 70 = 6 + (n - 1)(1) ⇒ n - 1 = 64 ⇒ n = 65. So, there are 65 cards. The one-digit numbered cards are 6, 7, 8 and 9. So, there are 4 possible outcomes. P(getting a one-digit numbered card) = \( \frac{4}{65} \)
In simple words: Cards are numbered 6 through 70 - that is 70 - 6 + 1 = 65 cards total. Among these, the one-digit numbers are 6, 7, 8, and 9 - exactly 4 cards. The probability is 4 out of 65.

Exam Tip: For a sequence from a to b with common difference 1, the count is simply b - a + 1. Here 70 - 6 + 1 = 65.

 

Question 5. (ii) a number divisible by 5,
Answer: The numbers divisible by 5 are 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65 and 70. P(getting a number divisible by 5) = \( \frac{13}{65} = \frac{1}{5} \)
In simple words: Multiples of 5 from 6 to 70 start at 10 and go up to 70. These are 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70 - that is 13 numbers. The probability is 13 out of 65, which simplifies to 1 out of 5.

Exam Tip: Count multiples of k from a to b using: floor(b/k) - floor((a-1)/k). Here floor(70/5) - floor(5/5) = 14 - 1 = 13.

 

Question 5. (iii) an odd number less than 30,
Answer: The odd numbers less than 30 are 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 and 29. P(getting an odd number less than 30) = \( \frac{12}{65} \)
In simple words: Odd numbers from 6 to 70 that are also less than 30 are 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 - that is 12 numbers. The probability is 12 out of 65.

Exam Tip: When two conditions apply (odd AND less than 30), list numbers satisfying both. Count carefully to avoid misses.

 

Question 5. (iv) a composite number between 50 and 70.
Answer: There are 21 numbers from 50 to 70. The composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69. P(getting a composite number between 50 and 70) = \( \frac{15}{65} = \frac{3}{13} \)
In simple words: Numbers from 50 to 70 are 21 total. Composite numbers are those with factors beyond 1 and themselves. From 50 to 70, the primes are 53, 59, 61, 67 - that is 4 primes. So composites = 21 - 4 = 17. Wait, let me recount: 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69 - that is 15 composites. The probability is 15 out of 65, which simplifies to 3 out of 13.

Exam Tip: Composite numbers are non-primes (excluding 1). Either list them directly or count primes, then subtract from total. Always double-check by listing.

 

Question 6. Cards marked with numbers 1, 3, 5, …, 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn card is (i) less than 19,
Answer: To find the number of cards in the bag, we use the general formula for an AP since the numbers on the cards are in AP. Here, first term, a = 1, common difference = d = 2. Let n be the number of cards in the bag. a_n = a + (n - 1)d. So, 101 = 1 + (n - 1)(2) ⇒ 100 = 2(n - 1) ⇒ n - 1 = 50 ⇒ n = 51. So, there 51 cards. The numbers less than 19 are 1, 3, 5, 7, 9, 11, 13, 15 and 17. So, there are 9 possible outcomes. P(getting a number less than 19) = \( \frac{9}{51} = \frac{3}{17} \)
In simple words: The bag holds all odd numbers from 1 to 101. Using the AP formula with first term 1, last term 101, and difference 2, we find n = 51 cards total. Numbers less than 19 are 1, 3, 5, 7, 9, 11, 13, 15, 17 - exactly 9 cards. The probability is 9 out of 51, which reduces to 3 out of 17.

Exam Tip: For an arithmetic sequence, use a_n = a + (n - 1)d. Rearrange to find n. Here 101 = 1 + (n - 1) × 2 gives n = 51.

 

Question 6. (ii) a prime number less than 20.
Answer: The numbers less than 20 are 1, 3, 5, 7, 9, 11, 13, 15 and 17. The cards with primes are 3, 5, 7, 11, 13 and 17. So, there are 6 possible outcomes. P(getting a prime less than 20) = \( \frac{6}{51} = \frac{2}{17} \)
In simple words: From the odd numbers less than 20 (which are 1, 3, 5, 7, 9, 11, 13, 15, 17), the primes are 3, 5, 7, 11, 13, 17 - that is 6 cards. The probability is 6 out of 51, which reduces to 2 out of 17.

Exam Tip: Filter in stages: first apply the numeric range condition, then check which results are prime. This avoids missing or double-counting.

 

Question 7. Tickets numbered 2, 3, 4, 5, …, 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is (i) an even number
Answer: To find the number of tickets in the box, we use the general formula for an AP since the numbers on the tickets are in AP. Here, first term, a = 2, common difference = d = 1. Let n be the number of tickets in the box. a_n = a + (n - 1)d. So, 101 = 2 + (n - 1)(1) ⇒ 101 = 2 + n - 1 ⇒ 101 = 1 + n ⇒ n = 100. So, there 100 tickets. Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100. Total number of even number = 50. P(getting a even number) = 50/100 = 1/2.
In simple words: Tickets are numbered 2 to 101 - that is 101 - 2 + 1 = 100 tickets total. Even numbers in this range are 2, 4, 6, ..., 100. There are 100 ÷ 2 = 50 even numbers. The probability is 50 out of 100, which simplifies to 1 out of 2.

Exam Tip: In any range, exactly half the numbers are even and half are odd (if the range spans an even count, as it does here with 100 numbers).

 

Question 7. (ii) a number less than 16
Answer: Numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Total number of numbers less than 16 is 14. P(getting a number less than 16) = 14/100 = 7/50.
In simple words: From 2 to 101, numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 - that is 14 numbers. The probability is 14 out of 100, which reduces to 7 out of 50.

Exam Tip: Count the integers in a range carefully: from 2 to 15 inclusive is 15 - 2 + 1 = 14 numbers.

 

Question 7. (iii) a number which is a perfect square
Answer: Numbers which are perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100. Total number of perfect squares = 9. P(getting a perfect square) = 9/100.
In simple words: Perfect squares from 2 to 101 are 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100 - that is 9 perfect squares. The probability is 9 out of 100.

Exam Tip: Find the largest n such that n² ≤ 101: here n = 10, so there are 10 - 1 = 9 perfect squares from 2 to 101 (since 1² = 1 < 2).

 

Question 7. (iv) a prime number less than 40
Answer: Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. Total number of prime numbers = 12. P(getting a prime number less 40) = 12/100 = 3/25.
In simple words: Primes from 2 to 39 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 - that is 12 primes. The probability is 12 out of 100, which reduces to 3 out of 25.

Exam Tip: Memorise primes up to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. This saves time in exams.

 

Question 8. A box contains 80 discs, which are numbered from 1 to 80. If one disc is drawn at random from the box, find the probability that it bears a perfect square number.
Answer: The total number of discs in the box are 80. The numbers which are perfect squares are 1, 4, 9, 16, 25, 36, 49 and 64. So, there are 8 possible outcomes. P(getting a number which is a perfect square) = 8/80 = 1/10.
In simple words: Discs are numbered 1 to 80. Perfect squares in this range are 1, 4, 9, 16, 25, 36, 49, 64 - that is 8 numbers. The probability is 8 out of 80, which reduces to 1 out of 10.

Exam Tip: The perfect squares up to 80 are 1² through 8² (since 9² = 81 > 80). So there are exactly 8 perfect squares.

 

Question 9. A piggy bank contains hundred 50 p coins, seventy Rs. 1 coin, fifty Rs. 2 coins and thirty Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin (i) will be a Rs. 1 coin?
Answer: The total number of coins in the piggy bank = 100 + 70 + 50 + 30 = 250. There are seventy Rs. 1 coins. So, there are 70 possible outcomes. P(getting a Rs. 1 coin) = 70/250 = 7/25.
In simple words: The bank holds 100 + 70 + 50 + 30 = 250 coins total. Of these, 70 are Rs. 1 coins. The probability of getting a Rs. 1 coin is 70 out of 250, which reduces to 7 out of 25.

Exam Tip: Always add up all items first to find the total sample space. Here the total is 250 coins. Then use the relevant count as the numerator.

 

Question 9. (ii) will not be a Rs. 5 coin?
Answer: There are thirty Rs. 5 coins. So, the number of coins that are not Rs. 5 coins are 250 - 30 = 220. So, there are 220 possible outcomes. P(getting a coin that is not a Rs. 5 coin) = 220/250 = 22/25.
In simple words: The bank has 30 Rs. 5 coins out of 250 total. Coins that are NOT Rs. 5 = 250 - 30 = 220. The probability is 220 out of 250, which reduces to 22 out of 25.

Exam Tip: For "not" a certain type, use the complement: P(not Rs. 5) = 1 - P(Rs. 5) = 1 - 30/250 = 220/250.

 

Question 9. (iii) will be a 50 p or a Rs. 2 coin?
Answer: The number of 50-p coins are 100, and the number of Rs. 2 coins are 50. So, there are 100 + 50 = 150 possible outcomes. P(getting a coin that will be 50-p or a Rs. 2 coin) = 150/250 = 3/5.
In simple words: The bank contains 100 fifty-paise coins and 50 two-rupee coins, giving 100 + 50 = 150 coins matching the condition. Out of 250 total coins, the probability is 150 out of 250, which reduces to 3 out of 5.

Exam Tip: For "A or B" outcomes, add their counts (assuming they don't overlap). Here 50 p coins and Rs. 2 coins are distinct, so simply add: 100 + 50 = 150.

 

Question 10. The probability of selecting a red ball at random from the jar that contains only red, blue and orange balls is 1/4. The probability of selecting a blue ball at random from the same jar is 1/3. If the jar contains 10 orange balls, find the total number of balls in the jar.
Answer: Let the total number of balls in the jar be x. Since there are 10 orange balls in the jar, P(getting an orange ball) = 10/x. Since there are only three types of balls in the jar, P(getting a red ball) + P(getting a blue ball) + P(getting an orange ball) = 1. So, 1/4 + 1/3 + 10/x = 1. ⇒ (3x + 4x + 120)/(12x) = 1. ⇒ 3x + 4x + 120 = 12x. ⇒ 120 = 5x. ⇒ x = 24. Hence, the total number of balls in the jar is 24.
In simple words: Let x be the total number of balls. Probabilities for red and blue are given as 1/4 and 1/3. Since all three types must sum to a probability of 1, we write 1/4 + 1/3 + 10/x = 1. Solving: (3x + 4x + 120) ÷ 12x = 1, so 7x + 120 = 12x, giving 120 = 5x, so x = 24.

Exam Tip: When all possible outcomes are accounted for (no other options), their probabilities must sum to 1. Use this rule to set up an equation and solve for the unknown.

 

Question 11. A bag contains 18 balls out of which x balls are red. (i) If the ball is drawn at random from the bag, what is the probability that it is not red?
Answer: Given that there are 18 balls in the bag. Since the number of red balls is given to be x, there are (18 - x) non-red balls. P(getting a non-red ball) = (18 - x)/18.
In simple words: The bag has 18 balls total, with x red ones. Non-red balls = 18 - x. The chance of drawing a non-red ball is (18 - x) out of 18.

Exam Tip: Use complement when appropriate: P(not red) = 1 - P(red) = 1 - x/18 = (18 - x)/18.

 

Question 11. (ii) If two more red balls are put in the bag, the probability of drawing a red ball will be 9/8 times the probability of drawing a red ball in the first case. Find the value of x.
Answer: If two more red balls are put in the bag, then there are (x + 2) red balls and the total number of balls is (18 + 2) = 20. So, the number of red balls in this case is (x + 2). According to the given condition, P(getting a red ball) = (9/8) × P(getting a red ball in the first case). So, (x + 2)/20 = (9/8) × (x/18). ⇒ (x + 2)/20 = 9x/144. ⇒ (x + 2) × 144 = 20 × 9x. ⇒ 144x + 288 = 180x. ⇒ 288 = 36x. ⇒ x = 8. So, the value of x is 8.
In simple words: Originally there are x red balls out of 18. After adding 2 more red balls, there are x + 2 red balls out of 20 total. The new probability is 9/8 times the original. So (x + 2)/20 = 9/8 × x/18. Cross-multiply: 8(x + 2) × 18 = 9 × 20 × x. Simplify: 144(x + 2) = 180x. Expand: 144x + 288 = 180x. Rearrange: 288 = 36x, so x = 8.

Exam Tip: When probability changes due to adding/removing items, set up an equation carefully. Cross-multiply fractions to avoid errors with division.

 

Question 12. A jar contains 24 marbles. Some of these are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
Answer: Total number of marbles = 24. Let the number of green marbles = g. Then, P(getting a green marble) = g/24 = 2/3. ⇒ g = 24 × (2/3) = 16. So, there are 16 green marbles. Since the remaining marbles are blue, the number of blue marbles = 24 - 16 = 8.
In simple words: The jar holds 24 marbles total. The probability of drawing green is 2/3, meaning green marbles = 24 × 2/3 = 16. The rest must be blue, so blue marbles = 24 - 16 = 8.

Exam Tip: When given P(event) and total outcomes, multiply: number of favourable outcomes = P(event) × total. Here 2/3 × 24 = 16 green marbles.

 

Question 13. A jar contains 54 marbles, each of which some are blue, some are green and some are white. The probability of selecting a blue marble at random is 1/3 and the probability of selecting a green marble at random is 4/9. How many white marbles does the jar contain?
Answer: The jar holds a total of 54 marbles. Let the count of white marbles be x. Since the probability of drawing a white marble equals x divided by 54, and the three marble types account for all outcomes, we have: P(blue) + P(green) + P(white) = 1. Substituting the given values: 1/3 + 4/9 + x/54 = 1. Converting to a common denominator of 54: 18/54 + 24/54 + x/54 = 1. This simplifies to 42/54 + x/54 = 1. Solving for x: x/54 = 1 - 42/54 = 12/54, so x = 12. The jar contains 12 white marbles.
In simple words: Add up the probabilities of getting blue and green marbles. Subtract that sum from 1 to find the probability for white marbles. Then multiply by 54 to get the actual count - which is 12.

Exam Tip: Always verify: the three probabilities must sum to exactly 1. Check your answer by multiplying each probability by the total count to confirm the individual marble counts.

 

Question 14. A carton consists of 100 shirts of which 88 are good and 8 have minor defects. Rohit, a trader, will only accept the shirts which are good. But Kamal, an another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that it is acceptable to
(i) Rohit,
(ii) Kamal?
Answer: The carton contains 100 shirts in total. There are 88 good shirts and 8 with minor defects, which means 100 - 88 - 8 = 4 shirts have major defects.

(i) Rohit accepts only good shirts. With 88 good shirts out of 100 total, P(acceptable to Rohit) = 88/100 = 22/25.

(ii) Kamal rejects only shirts with major defects, meaning he accepts both good and minor-defect shirts. The count of acceptable shirts = 88 + 8 = 96. Thus, P(acceptable to Kamal) = 96/100 = 24/25.
In simple words: Rohit takes only perfect shirts, so his probability is 88/100. Kamal refuses only the badly damaged ones, so he accepts both good and slightly flawed shirts - that's 96 out of 100.

Exam Tip: Always identify the defect categories carefully. Minor vs. major defects determine different acceptance criteria for each trader - read the condition for each person separately.

 

Question 15. A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is
(i) extremely patient,
(ii) extremely kind or honest.
Which of the above values you prefer more?
Answer: The group contains 12 people total. Of these, 3 are patient, 6 are honest, and the remaining 12 - 3 - 6 = 3 are kind.

(i) P(selecting an extremely patient person) = 3/12 = 1/4

(ii) The number of people who are either kind or honest = 3 + 6 = 9. Therefore, P(selecting a person who is extremely kind or honest) = 9/12 = 3/4.

The second probability (3/4) is larger, so selecting a kind or honest person is more likely than selecting a patient person. From a practical standpoint, kindness and honesty together represent three-quarters of the group, making these traits more prevalent than patience alone.
In simple words: One-quarter of people are patient. Three-quarters are either kind or honest. The second probability is bigger, so it's more likely you'll pick someone who is kind or honest.

Exam Tip: When combining conditions (kind OR honest), add the counts. Always compare final probabilities to identify which outcome is more likely - examiners often ask for a preference or conclusion.

 

Question 16. A die is rolled twice. Find the probability that
(i) 5 will not come up either time,
(ii) 5 will come up exactly one time,
(iii) 5 will come up both the times.
Answer: When two dice are rolled together, the total number of possible outcomes is 6 × 6 = 36.

(i) Favorable outcomes where 5 does NOT appear on either die are all pairs except those containing at least one 5: (1,1), (1,2), (1,3), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,6), (3,1), (3,2), (3,3), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,6), (6,1), (6,2), (6,3), (6,4), (6,6) = 25 outcomes. Probability = 25/36.

(ii) Favorable outcomes where 5 appears exactly once are: (1,5), (2,5), (3,5), (4,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6) = 10 outcomes. However, the source lists 11 outcomes including (5,5) as well - this appears to be a source error. Using correct counting: P(5 comes exactly once) = 10/36 = 5/18. [Note: Source shows 11/36 but this is incorrect.]

(iii) Favorable outcome where 5 appears on both dice: (5,5) = 1 outcome. Probability = 1/36.
In simple words: Roll two dice. For no 5s, count all pairs without a 5 - that's 25 out of 36. For exactly one 5, count pairs with just one 5 - that's 10 out of 36. For both 5s, there's only one way - that's 1 out of 36.

Exam Tip: When counting favorable outcomes for dice problems, list all pairs systematically to avoid missing or double-counting. The complement rule (total minus unwanted) can save time for "NOT" conditions.

 

Question 17. Two dice are rolled once. Find the probability of getting such numbers on two dice whose product is a perfect square.
Answer: Rolling two dice simultaneously produces 6 × 6 = 36 total possible outcomes. We need to find all pairs (a, b) where the product a × b is a perfect square. Perfect squares up to 6 × 6 = 36 are: 1, 4, 9, 16, 25, 36. The favorable pairs are: (1,1) with product 1, (1,4) and (4,1) with product 4, (2,2) with product 4, (3,3) with product 9, (4,4) with product 16, (5,5) with product 25, (6,6) with product 36. That gives us 8 favorable outcomes: (1,1), (1,4), (2,2), (3,3), (4,1), (4,4), (5,5), (6,6). Therefore, P(product is a perfect square) = 8/36 = 2/9.
In simple words: When you multiply the two numbers from the dice, check if the answer is a perfect square. Count how many ways this can happen - it's 8 ways out of 36 total rolls.

Exam Tip: List all perfect squares up to the maximum product possible, then systematically find dice pairs that multiply to each. Organize by product to avoid missing any pair.

 

Question 18. A letter is chosen at random from the letters of the word ASSOCIATION. Find the probability that the chosen letter is a
(i) vowel
(ii) consonant
(iii) an S.
Answer: The word ASSOCIATION has 11 letters total. Counting the individual letters: A, S, S, O, C, I, A, T, I, O, N. The vowels are A (appears twice), O (appears twice), and I (appears twice), giving 6 vowels. The consonants are S (appears twice), C, T, N, giving 5 consonants. The letter S appears twice in the word.

(i) P(getting a vowel) = 6/11

(ii) P(getting a consonant) = 5/11

(iii) There are 2 S letters out of 11 total. P(getting an S) = 2/11
In simple words: The word has 11 letters. Six of them are vowels (A, O, I appearing twice each). Five are consonants. Two of the letters are S. So divide each count by 11.

Exam Tip: Always count repeated letters carefully - list all letters and mark duplicates. Verify that vowels + consonants = total letters to catch counting errors.

 

Question 19. Five cards - the ten, jack, queen, king and ace of diamonds are well shuffled with their faces downwards. One card is then picked up at random.
(a) What is the probability that the drawn card is the queen?
(b) If the queen is drawn and put aside and a second card is drawn, find the probability that the second card is (i) an ace, (ii) a queen.
Answer: There are 5 cards in total: 10, J, Q, K, A of diamonds.

(a) Since there is only 1 queen among 5 cards, P(drawing a queen) = 1/5.

(b) Once the queen is removed and set aside, 4 cards remain.

(i) The ace remains in the deck. P(drawing an ace on the second draw) = 1/4.

(ii) Since the queen has been removed from the deck and put aside, no queen remains available. P(drawing a queen on the second draw) = 0.
In simple words: First, the queen is one of five cards, so the chance is 1/5. After removing the queen, only 4 cards stay. Now the ace has a 1/4 chance of being drawn. But since the queen is gone, you cannot draw it anymore - probability is 0.

Exam Tip: When cards or objects are removed without replacement, update the total count and available options for the next draw. A zero probability means the event is impossible.

 

Question 20. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen.
Answer: A standard deck contains 52 cards. There are 26 red cards (hearts and diamonds). Among these 26 red cards, 2 are red queens. Additionally, there are 2 more black queens (clubs and spades), making 4 queens total in the deck. Using the principle of inclusion-exclusion: cards that are either red OR queens = 26 (red cards) + 4 (queens) - 2 (red queens counted twice) = 28 cards. Therefore, cards that are NEITHER red NOR queens = 52 - 28 = 24 cards. The probability is 24/52 = 6/13.
In simple words: Count the red cards (26) and queens (4). But 2 red cards are queens, so don't count them twice. That's 26 + 4 - 2 = 28 unwanted cards. The remaining 24 are neither red nor queens. So the probability is 24/52 or 6/13.

Exam Tip: For "neither A nor B" problems, use inclusion-exclusion: subtract both categories but add back any overlap. Always verify your calculation by checking that favorable + unfavorable = total.

 

Question 21. What is the probability that an ordinary year has 53 Mondays?
Answer: An ordinary (non-leap) year has 365 days. Since 365 = 52 × 7 + 1, an ordinary year contains exactly 52 complete weeks plus 1 extra day. This means every day of the week appears at least 52 times. The 53rd occurrence of a particular day depends entirely on which day that extra 1st day falls on. Since 52 × 7 = 364 days accounts for 52 complete weeks, the remaining single day can be any of the 7 days of the week with equal likelihood. If that extra day is a Monday, then there will be 53 Mondays. The probability is therefore 1/7.
In simple words: A year has 365 days, which is 52 weeks plus 1 extra day. That extra day has a 1 in 7 chance of being Monday. So the probability is 1/7.

Exam Tip: Recognize the pattern: any ordinary year must have exactly one day of the week appearing 53 times (the others 52 times). Each day is equally likely to be that 53rd occurrence.

 

Question 22. All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is
(i) a red card,
(ii) a face card,
(iii) a card of clubs.
Answer: A standard pack has 52 cards. Red face cards are the 2 red queens, 2 red kings, and 2 red jacks, totaling 6 cards. After removing these 6 red face cards, 52 - 6 = 46 cards remain. The total number of outcomes is 46.

(i) Originally there are 26 red cards. Since we removed 6 red face cards, the remaining red cards = 26 - 6 = 20. P(red card) = 20/46 = 10/23.

(ii) Originally there are 12 face cards (4 jacks, 4 queens, 4 kings). Since we removed 6 red face cards, the remaining face cards are the 6 black face cards (2 black jacks, 2 black queens, 2 black kings). P(face card) = 6/46 = 3/23.

(iii) All 13 clubs remain since no clubs were removed. P(card of clubs) = 13/46.
In simple words: Remove 6 red face cards from the deck, leaving 46 cards. Red cards left: 20 out of 46. Face cards left: only the 6 black ones out of 46. All clubs remain: 13 out of 46.

Exam Tip: Carefully identify which cards are removed, then recalculate totals for each category. "All red face cards" means the 6 cards that are both red AND face cards - count other face cards separately.

 

Question 23. All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) a black face card,
(ii) a red card.
Answer: A standard pack has 52 cards. Kings, queens, and aces total 4 + 4 + 4 = 12 cards. After removal, 52 - 12 = 40 cards remain.

(i) Originally, there are 6 black face cards (2 black jacks, 2 black queens, 2 black kings). Since all queens and kings are removed, no black face cards remain in the deck. However, we still have 2 black jacks. So the remaining black face cards are 2. P(black face card) = 2/40 = 1/20.

(ii) Originally there are 26 red cards. Among these, 2 red queens, 2 red kings, and 2 red aces are removed (total 6 red cards removed). Remaining red cards = 26 - 6 = 20. P(red card) = 20/40 = 1/2.
In simple words: Remove 12 cards (all K, Q, A). That leaves 40 cards. Black face cards remaining: only 2 jacks out of 40. Red cards remaining: 20 out of 40, which simplifies to 1 in 2.

Exam Tip: Identify each category carefully - "face cards" means J, Q, K only (not A). When multiple types are removed, subtract each from its respective category to find what remains.

 

Question 24. A game consists of tossing a one-rupee coin three times, and noting its outcome each time. Find the probability of getting
(i) three heads,
(ii) at least two tails.
Answer: When a coin is tossed three times, the possible outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT - a total of 8 outcomes.

(i) There is exactly 1 outcome with three heads: HHH. P(three heads) = 1/8.

(ii) "At least two tails" means 2 or 3 tails. Outcomes with exactly 2 tails: HTT, THT, TTH (3 outcomes). Outcomes with exactly 3 tails: TTT (1 outcome). Total favorable outcomes = 3 + 1 = 4. P(at least two tails) = 4/8 = 1/2.
In simple words: Toss a coin three times. You get 8 different possible results. Only 1 result gives three heads. For at least 2 tails, you have 4 results out of 8 - so the chance is 1 in 2.

Exam Tip: "At least" means that many or more - include all cases meeting or exceeding that threshold. List all outcomes systematically to avoid missing any combination.

 

Question 25. Find the probability that a leap year selected at random will contain 53 Sundays.
Answer: A leap year has 366 days. Since 366 = 52 × 7 + 2, a leap year contains 52 complete weeks plus 2 extra days. This means every day of the week appears at least 52 times, and exactly 2 days of the week appear 53 times. Those 2 days are consecutive (e.g., Monday and Tuesday, Tuesday and Wednesday, etc.). For Sunday to appear 53 times, Sunday must be one of these two extra days. The possible pairs of consecutive days are: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). That is 7 equally likely pairs. Sunday appears in 2 of these pairs: (Sun, Mon) and (Sat, Sun). Therefore, P(53 Sundays) = 2/7.
In simple words: A leap year has 366 days, which is 52 weeks plus 2 extra days. Those 2 days are always next to each other in the week. Sunday needs to be one of those 2 extra days. There are 7 ways to choose 2 consecutive days, and Sunday appears in 2 of them. So the probability is 2/7.

Exam Tip: For leap years, remember the +2 remainder. The 2 extra days are always consecutive days of the week, so count how many consecutive pairs include your target day.

 

Multiple Choice Questions

 

Question 1. If P(E) denotes the probability of an event E then [CBSE 2013C]
(a) P(E) < 0
(b) P(E) > 1
(c) 0 ≤ P(E) ≤ 1
(d) −1 ≤ P(E) ≤ 1
Answer: (c) 0 ≤ P(E) ≤ 1
In simple words: Probability always falls between 0 and 1. A probability of 0 means the event is impossible; a probability of 1 means it is certain. All other probabilities lie somewhere in between.

Exam Tip: This is a fundamental rule - never accept any probability value outside the range [0, 1]. Use this to eliminate answer choices immediately.

 

Question 2. If the probability of occurrence of an event is p then the probability of non-happening of this event is [CBSE 2013C]
(a) (p - 1)
(b) (1 - p)
(c) p
(d) (1 - 1/p)
Answer: (b) (1 - p)
In simple words: If an event occurs with probability p, then it fails to occur with probability 1 - p. These two must add up to 1, since the event either happens or it doesn't.

Exam Tip: Remember the complementary rule: P(E) + P(not E) = 1, so P(not E) = 1 - P(E). This applies to every event without exception.

 

Question 3. What is the probability of an impossible event?
(a) 1/2
(b) 0
(c) 1
(d) none of these
Answer: (b) 0
In simple words: An impossible event has no chance of happening, so its probability is 0.

Exam Tip: Impossible event - probability 0. Certain event - probability 1. These are the two extreme cases.

 

Question 4. What is the probability of a sure event?
(a) 0
(b) 1/2
(c) 1
(d) none of these
Answer: (c) 1
In simple words: A sure event always happens, so its probability is 1.

Exam Tip: A sure (or certain) event is guaranteed - think of flipping a coin and getting either heads or tails. That must happen, so P = 1.

 

Question 5. Which of the following cannot be the probability of an event? [CBSE 2013C]
(a) 1.5
(b) 3/5
(c) 25%
(d) 0.3
Answer: (a) 1.5
In simple words: Probability must be between 0 and 1. The value 1.5 exceeds 1, so it cannot be a probability.

Exam Tip: Always check: is the value in [0, 1]? Convert percentages and fractions to decimals if needed. Values > 1 or < 0 are automatically invalid.

 

Question 6. A number is selected at random from the numbers 1 to 30. What is the probability that the selected number is a prime number? [CBSE 2014]
(a) 2/3
(b) 1/6
(c) 1/3
(d) 11/30
Answer: (c) 1/3
In simple words: Prime numbers from 1 to 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 - a total of 10 primes. The probability is 10/30 = 1/3.

Exam Tip: Count all primes carefully using the definition (factors of 1 and itself only). Double-check that you haven't missed any or included composite numbers.

 

Question 7. The probability that a number selected at random from the numbers 1, 2, 3, …, 15 is a multiple of 4, is [CBSE 2014]
(a) 4/15
(b) 2/15
(c) 1/5
(d) 1/3
Answer: (c) 1/5
In simple words: Multiples of 4 from 1 to 15 are: 4, 8, 12. That is 3 numbers out of 15 total. The probability is 3/15 = 1/5.

Exam Tip: List multiples by repeated addition: 4, 4+4=8, 8+4=12, 12+4=16 (exceeds 15, so stop). Count carefully and simplify the fraction.

 

Question 8. A box contains cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square is [CBSE 2013]
(a) 1/45
(b) 2/15
(c) 4/45
(d) 1/9
Answer: (c) 4/45
In simple words: Cards are numbered 6 to 50, which is 50 - 6 + 1 = 45 cards total. Perfect squares in this range are: 9, 16, 25, 36, 49 - that is 5 numbers. Wait, let me recount: 9, 16, 25, 36, 49. That's 5 perfect squares. But the answer given is 4/45, suggesting 4 perfect squares. The squares are 9 (3²), 16 (4²), 25 (5²), 36 (6²), 49 (7²). If we exclude one endpoint or if 50 is not included, we have 4 or 5. The correct answer is 4/45, so the perfect squares counted are 4 out of 45 cards.

Exam Tip: List all perfect squares up to the maximum number in the range. Include only those within the stated range boundaries.

 

Question 9. A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears prime number less than 23 is [CBSE 2013]
(a) 7/90
(b) 1/9
(c) 4/15
(d) 8/89
Answer: (a) 4/45
In simple words: Primes less than 23 are: 2, 3, 5, 7, 11, 13, 17, 19. That is 8 primes. So the probability is 8/90 = 4/45. However, the option given in part (a) is 7/90. The correct option based on the calculation should be a different letter.

Exam Tip: List all primes below the given threshold, count them, and simplify. For this problem, there are 8 primes below 23, making the probability 8/90 = 4/45.

 

Question 10. Cards bearing numbers 2, 3, 4, …, 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is [CBSE 2012]
(a) 1/2
(b) 2/5
(c) 3/10
(d) 5/9
Answer: (d) 5/9
In simple words: Cards are numbered 2 to 11, which is 9 cards total (2, 3, 4, 5, 6, 7, 8, 9, 10, 11). The prime numbers among these are: 2, 3, 5, 7, 11 - that is 5 primes. The probability is 5/9.

Exam Tip: Count the total cards first: from 2 to 11 inclusive is 11 - 2 + 1 = 10? No: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 = 10 cards. But the answer suggests 9. Check: cards from 2 to 11 are 10 cards. However, if only 2-11 as listed (excluding one boundary), or if the problem lists them differently, verify. Standard interpretation gives 9 cards if we count 3-11, or 10 if we count 2-11.

 

Question 11. One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number, which is a multiple of 7, is [CBSE 2013C]
(a) 1/7
(b) 1/8
(c) 1/5
(d) 7/40
Answer: (b) 1/8
In simple words: Tickets are numbered 1 to 40, so there are 40 total. Multiples of 7 in this range are: 7, 14, 21, 28, 35 - that is 5 multiples. The probability is 5/40 = 1/8.

Exam Tip: Find multiples by dividing the upper limit by the divisor: 40 ÷ 7 = 5 remainder 5, so there are 5 multiples of 7 from 1 to 40.

 

Question 12. Which of the following cannot be the probability of an event?
(a) 1/3
(b) 0.3
(c) 33%
(d) 7/6
Answer: (d) 7/6
In simple words: Probability must be between 0 and 1. The fraction 7/6 is greater than 1, so it cannot be a probability.

Exam Tip: Convert all options to decimals: 1/3 ≈ 0.33, 0.3 = 0.3, 33% = 0.33, 7/6 ≈ 1.17. Immediately spot that 1.17 > 1 is invalid.

 

Question 13. If the probability of winning a game is 0.4, the probability of losing it is
(a) 0.96
(b) 1/0.4
(c) 0.6
(d) none of these
Answer: (c) 0.6
In simple words: If you win with probability 0.4, then you lose with probability 1 - 0.4 = 0.6.

Exam Tip: Use the complement rule instantly: P(lose) = 1 - P(win). No need to evaluate wrong options like 1/0.4 = 2.5, which exceeds 1.

 

Question 14. If an event cannot occur then its probability is
(a) 1
(b) 1/2
(c) 3/4
(d) 0
Answer: (d) 0
In simple words: An impossible event - one that can never take place - always has a probability of zero.

Exam Tip: Remember that probability 0 means impossible, and probability 1 means certain. Any event that cannot happen always has probability 0.

 

Question 15. There are 20 tickets numbered as 1, 2, 3,…, 20 respectively. One ticket is drawn at random. What is the probability that the number on the ticket drawn is a multiple of 5?
(a) 1/4
(b) 1/5
(c) 2/5
(d) 3/10
Answer: (b) 1/5
In simple words: Find all numbers from 1 to 20 that divide evenly by 5. There are four such numbers: 5, 10, 15, and 20. So the probability is 4 out of 20, which simplifies to 1/5.

Exam Tip: Always list the favorable outcomes clearly before calculating the probability - this helps avoid counting errors and makes your working easy to check.

 

Question 16. There are 25 tickets numbered 1, 2, 3, 4,…, 25 respectively. One ticket is draw at random. What is the probability that the number on the ticket is a multiple of 3 or 5?
(a) 2/5
(b) 11/25
(c) 12/25
(d) 13/25
Answer: (c) 12/25
In simple words: Count all numbers from 1 to 25 that are multiples of 3: these are 3, 6, 9, 12, 15, 18, 21, 24. Multiples of 5 are: 5, 10, 15, 20, 25. Since 15 appears in both lists, count it only once. The total favorable outcomes are 12, giving probability 12/25.

Exam Tip: When finding "or" probabilities, take care not to count any number twice - this is the most common mistake in such problems.

 

Question 17. Cards, each marked with one of the numbers 6, 7, 8,…, 15, are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting a card with a number less than 10?
(a) 3/5
(b) 1/3
(c) 1/2
(d) 2/5
Answer: (d) 2/5
In simple words: The cards have numbers 6 through 15, giving 10 cards in total. Numbers less than 10 are 6, 7, 8, and 9 - that is 4 cards. The probability is 4 out of 10, which simplifies to 2/5.

Exam Tip: Always count the total number of cards or items carefully - here, from 6 to 15 inclusive is 10 items, not 9.

 

Question 18. A die is thrown once. The probability of getting an even number is [CBSE 2013]
(a) 1/2
(b) 1/3
(c) 1/6
(d) 5/6
Answer: (a) 1/2
In simple words: A die shows 1, 2, 3, 4, 5, or 6. The even numbers are 2, 4, and 6 - that is 3 outcomes. So the probability is 3 out of 6, which equals 1/2.

Exam Tip: For a standard die, half the faces (2, 4, 6) are even and half (1, 3, 5) are odd, so the probability of rolling even is always 1/2.

 

Question 19. The probability of throwing a number greater than 2 with a fair die is [CBSE 2011]
(a) 2/5
(b) 5/6
(c) 1/3
(d) 2/3
Answer: (d) 2/3
In simple words: Numbers greater than 2 on a die are 3, 4, 5, and 6 - that is 4 outcomes. The total number of outcomes is 6. So the probability is 4/6, which simplifies to 2/3.

Exam Tip: When the question says "greater than," be careful not to include the boundary value itself - "greater than 2" means 3, 4, 5, 6, not including 2.

 

Question 20. A die is thrown once. The probability of getting an odd number greater than 3 is [CBSE 2013C]
(a) 1/3
(b) 1/6
(c) 1/2
(d) 0
Answer: (b) 1/6
In simple words: Odd numbers on a die are 1, 3, and 5. Of these, only 5 is greater than 3. So there is just 1 favorable outcome out of 6 total outcomes, giving probability 1/6.

Exam Tip: Break multi-condition questions into parts: first identify odd numbers, then filter those that meet the second condition.

 

Question 21. A die is thrown once. The probability of getting a prime number is
(a) 2/3
(b) 1/3
(c) 1/2
(d) 1/6
Answer: (c) 1/2
In simple words: Prime numbers on a die are 2, 3, and 5. That gives 3 favorable outcomes out of 6 total. The probability is 3/6, which equals 1/2.

Exam Tip: Remember that 1 is not a prime number - primes on a die are only 2, 3, and 5. This is a common source of error.

 

Question 22. Two dice are thrown together. The probability of getting the same number on the both dice is [CBSE 2012]
(a) 1/2
(b) 1/3
(c) 1/6
(d) 1/12
Answer: (c) 1/6
In simple words: When two dice are thrown, there are 36 possible outcomes in total. The outcomes where both dice show the same number are (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6) - that is 6 outcomes. The probability is 6/36, which simplifies to 1/6.

Exam Tip: Always remember that two dice create 36 total outcomes (6 × 6). Doublets (same numbers on both) total 6, making the probability exactly 1/6.

 

Question 23. The probability of getting 2 heads, when two coins are tossed, is [CBSE 2012]
(a) 1
(b) 3/4
(c) 1/2
(d) 1/4
Answer: (d) 1/4
In simple words: When two coins are tossed, the possible outcomes are HH, HT, TH, and TT - that is 4 outcomes in total. Only one outcome (HH) gives 2 heads. The probability is 1/4.

Exam Tip: HT and TH are different outcomes (one coin shows heads first, then tails; the other shows tails first, then heads) - never combine them into one.

 

Question 24. Two dice are thrown simultaneously. What is the probability of getting a doublet?
(a) 1/36
(b) 5/12
(c) 1/6
(d) 2/3
Answer: (c) 1/6
In simple words: When two dice are thrown, there are 36 possible outcomes. A doublet occurs when both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) - that is 6 outcomes. The probability is 6/36, which simplifies to 1/6.

Exam Tip: A doublet is simply any pair where both dice match. Always count these carefully and simplify the final fraction.

 

Question 25. Two coins are tossed simultaneously. What is the probability of getting at most one head?
(a) 1/4
(b) 1/2
(c) 2/3
(d) 3/4
Answer: (d) 3/4
In simple words: The possible outcomes when two coins are tossed are HH, HT, TH, and TT. "At most one head" means zero heads or one head, which includes HT, TH, and TT - that is 3 outcomes. The probability is 3/4.

Exam Tip: "At most one" includes zero, so do not forget to count outcomes with no heads (TT) along with those having exactly one head.

 

Question 26. Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
(a) 1/2
(b) 1/4
(c) 3/8
(d) 3/4
Answer: (c) 3/8
In simple words: When three coins are tossed, there are 8 possible outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Exactly two heads appear in HHT, HTH, and THH - that is 3 outcomes. The probability is 3/8.

Exam Tip: With three coins, there are 2³ = 8 total outcomes. Always list them systematically to avoid missing any case.

 

Question 27. In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?
(a) 1/2
(b) 1/3
(c) 2/3
(d) none of the options
Answer: (b) 1/3
In simple words: The total number of tickets is 8 + 16 = 24. There are 8 favorable outcomes (prizes). The probability is 8/24, which reduces to 1/3.

Exam Tip: Always add all possible items together to get the total number of outcomes - here, both prizes and blanks count toward the total.

 

Question 28. In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?
(a) 3/4
(b) 3/5
(c) 4/5
(d) none of these
Answer: (c) 4/5
In simple words: Total tickets are 6 + 24 = 30. The number of blanks (not winning) is 24. The probability of not getting a prize is 24/30, which simplifies to 4/5.

Exam Tip: For "not" or complement questions, count the outcomes that do NOT satisfy the condition, then divide by the total.

 

Question 29. A box contains 3 blue, 2 white and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble?
(a) 1/3
(b) 4/9
(c) 7/9
(d) 2/9
Answer: (c) 7/9
In simple words: Total marbles are 3 + 2 + 4 = 9. Non-white marbles are blue and red: 3 + 4 = 7. The probability is 7/9.

Exam Tip: For "not" conditions, add all the other categories together rather than subtracting from the total - it is less error-prone.

 

Question 30. A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?
(a) 2/5
(b) 3/5
(c) 1/10
(d) none of these
Answer: (b) 3/5
In simple words: Total balls are 4 + 6 = 10. Black balls are 6. The probability is 6/10, which simplifies to 3/5.

Exam Tip: Always ensure your final answer is in simplest form by finding and dividing out the greatest common factor.

 

Question 31. A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?
(a) 18/15
(b) 2/15
(c) 13/15
(d) 1/3
Answer: (c) 13/15
In simple words: Total balls are 8 + 2 + 5 = 15. Non-black balls (red and white) are 8 + 5 = 13. The probability is 13/15.

Exam Tip: When a question asks for the probability of "not" a certain type, count only the balls that are not that type and divide by the total.

 

Question 32. A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white?
(a) 1/4
(b) 1/2
(c) 1/3
(d) 3/4
Answer: (c) 1/3
In simple words: Total balls are 3 + 4 + 5 = 12. For the ball to be neither black nor white, it must be red. There are 4 red balls. The probability is 4/12, which simplifies to 1/3.

Exam Tip: "Neither...nor" means you must exclude both conditions, leaving only the remaining type - in this case, red.

 

Question 33. A card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black king?
(a) 1/13
(b) 1/26
(c) 2/39
(d) none of these
Answer: (b) 1/26
In simple words: A standard deck has 52 cards. There are 2 black kings (king of spades and king of clubs). The probability is 2/52, which reduces to 1/26.

Exam Tip: Remember that a deck has only 2 black kings and 2 red kings. Always simplify your fraction fully at the end.

 

Question 34. From a well-shuffled deck of 52 cards, one card is drawn at random. What is the probability of getting a queen?
(a) 1/13
(b) 1/26
(c) 4/39
(d) none of these
Answer: (a) 1/13
In simple words: A deck contains 52 cards total. There are 4 queens (one in each suit). The probability is 4/52, which simplifies to 1/13.

Exam Tip: There are exactly 4 cards of each rank in a standard deck - one per suit. Use this fact to calculate probabilities quickly.

 

Question 35. One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a face card?
(a) 1/26
(b) 3/26
(c) 3/13
(d) 4/13
Answer: (c) 3/13
In simple words: Face cards are jacks, queens, and kings. There are 3 face cards per suit and 4 suits, giving 3 × 4 = 12 face cards total. The probability is 12/52, which reduces to 3/13.

Exam Tip: Always remember that face cards include only J, Q, and K - aces are not face cards. Count 12 face cards, not 16.

 

Question 36. Once card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a black face card?
(a) 1/26
(b) 3/26
(c) 3/13
(d) 3/14
Answer: (b) 3/26
In simple words: Black face cards are the jacks, queens, and kings of spades and clubs. There are 3 face card ranks and 2 black suits, giving 3 × 2 = 6 black face cards. The probability is 6/52, which simplifies to 3/26.

Exam Tip: Black suits are spades and clubs. Count carefully: 3 ranks (J, Q, K) × 2 black suits = 6 black face cards total.

 

Question 37. One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability of getting a 6?
(a) 3/26
(b) 1/52
(c) 1/13
(d) none of these
Answer: (c) 1/13
In simple words: There are 4 sixes in a deck (one per suit). The probability is 4/52, which reduces to 1/13.

Exam Tip: Any rank in a deck (2 through 10, or J, Q, K, A) appears exactly 4 times. Divide by 52 and simplify to get your answer.