OP Malhotra Class 10 Maths Solutions Chapter 20 Probability Exercise 20 (B)

Question 1. Find the probability of getting a number less than 5 in a single throw of a die.
Answer: When you throw a standard die, the possible numbers you can get are 1, 2, 3, 4, 5, and 6. So, there are 6 total possible outcomes. The numbers that are less than 5 are 1, 2, 3, and 4. This means there are 4 favourable outcomes. To find the probability, you divide the number of favourable outcomes by the total number of possible outcomes.
\( \text{Total possible outcomes} = 6 \)
\( \text{Favourable outcomes (numbers less than 5)} = \{1, 2, 3, 4\} = 4 \)
\( P(\text{number} < 5) = \frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}} = \frac{4}{6} = \frac{2}{3} \). The probability is a fraction that shows how likely an event is to happen.
In simple words: When you roll a die, there are 6 possible results. Only 4 of these are less than 5. So, the chance of getting a number less than 5 is 4 out of 6, which simplifies to 2 out of 3.

🎯 Exam Tip: Always list out the total possible outcomes and the favourable outcomes clearly before calculating the probability. Simplify the fraction if possible.

Question 2. A jar contains 3 white, 4 blue, 5 red and 2 green marbles. If a marble is drawn at random from the jar, what is the probability that the marble is (i) white ?, (ii) blue ?, (iii) red ?, (iv) green ?, (v) not white ?, (vi) not red?
Answer: First, we need to find the total number of marbles in the jar. There are 3 white marbles, 4 blue marbles, 5 red marbles, and 2 green marbles. The total number of marbles is 3 + 4 + 5 + 2 = 14 marbles. This is the total number of possible outcomes when drawing one marble.
\( \text{Total number of marbles} = 3 + 4 + 5 + 2 = 14 \)
The probability of an event (E) is calculated as: \( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} \)
(i) Probability of the marble being white:
There are 3 white marbles.
\( P(\text{white}) = \frac{3}{14} \)
(ii) Probability of the marble being blue:
There are 4 blue marbles.
\( P(\text{blue}) = \frac{4}{14} = \frac{2}{7} \)
(iii) Probability of the marble being red:
There are 5 red marbles.
\( P(\text{red}) = \frac{5}{14} \)
(iv) Probability of the marble being green:
There are 2 green marbles.
\( P(\text{green}) = \frac{2}{14} = \frac{1}{7} \)
(v) Probability of the marble not being white:
The number of marbles that are not white is \( 14 - 3 = 11 \) (blue, red, and green marbles).
\( P(\text{not white}) = \frac{11}{14} \)
(vi) Probability of the marble not being red:
The number of marbles that are not red is \( 14 - 5 = 9 \) (white, blue, and green marbles).
\( P(\text{not red}) = \frac{9}{14} \)
Probability helps us understand the chance of picking a specific color from a mix.
In simple words: Add up all the marbles to get the total. Then, for each question, count how many marbles fit the description. Divide that count by the total marbles to get the probability. For "not white," count all marbles that are not white.

🎯 Exam Tip: When calculating "not E" probabilities, it's often easiest to find P(E) first and then subtract it from 1, i.e., \( P(\text{not E}) = 1 - P(E) \). Alternatively, count the items that are not E.

Question 3. If a card is drawn at random from a standard deck of playing cards, what is the probability that it is (i) a nine? (ii) an ace? (iii) black? (iv) red? (v) a heart? (vi) a diamond? (vii) a face card? (viii) not a face card? (ix) the ace of diamonds? (x) a red face card? (xi) a black ace? (xii) a red king?
Answer: A standard deck of playing cards has 52 cards. There are 4 suits: hearts, diamonds, spades, and clubs. Hearts and diamonds are red, while spades and clubs are black. Each suit has 13 cards, which include an Ace, King, Queen, Jack, and numbers 2 through 10. For this problem, face cards are considered King, Queen, Jack, and Ace, making 4 face cards per suit, or 16 total face cards in the deck.
\( \text{Total possible outcomes} = 52 \)
The probability of drawing any card is \( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} \).
(i) Probability of drawing a nine:
There are four 9s in a deck (one for each suit).
\( P(\text{a nine}) = \frac{4}{52} = \frac{1}{13} \)
(ii) Probability of drawing an ace:
There are four aces in a deck (one for each suit).
\( P(\text{an ace}) = \frac{4}{52} = \frac{1}{13} \)
(iii) Probability of drawing a black card:
There are 26 black cards (13 spades + 13 clubs).
\( P(\text{black}) = \frac{26}{52} = \frac{1}{2} \)
(iv) Probability of drawing a red card:
There are 26 red cards (13 hearts + 13 diamonds).
\( P(\text{red}) = \frac{26}{52} = \frac{1}{2} \)
(v) Probability of drawing a heart:
There are 13 cards in the heart suit.
\( P(\text{a heart}) = \frac{13}{52} = \frac{1}{4} \)
(vi) Probability of drawing a diamond:
There are 13 cards in the diamond suit.
\( P(\text{a diamond}) = \frac{13}{52} = \frac{1}{4} \)
(vii) Probability of drawing a face card:
There are 4 face cards in each of the 4 suits (Ace, King, Queen, Jack). So, \( 4 \times 4 = 16 \) face cards in total.
\( P(\text{a face card}) = \frac{16}{52} = \frac{4}{13} \)
(viii) Probability of not drawing a face card:
The number of cards that are not face cards is \( 52 - 16 = 36 \).
\( P(\text{not a face card}) = \frac{36}{52} = \frac{9}{13} \)
(ix) Probability of drawing the ace of diamonds:
There is only one ace of diamonds in the deck.
\( P(\text{ace of diamonds}) = \frac{1}{52} \)
(x) Probability of drawing a red face card:
Red suits are hearts and diamonds. Each has 4 face cards. So, there are \( 2 \times 4 = 8 \) red face cards.
\( P(\text{a red face card}) = \frac{8}{52} = \frac{2}{13} \)
(xi) Probability of drawing a black ace:
There are two black aces (Ace of Spades and Ace of Clubs).
\( P(\text{a black ace}) = \frac{2}{52} = \frac{1}{26} \)
(xii) Probability of drawing a red king:
There are two red kings (King of Hearts and King of Diamonds).
\( P(\text{a red king}) = \frac{2}{52} = \frac{1}{26} \)
In simple words: For each part, count how many cards in a standard 52-card deck fit the description (like how many nines there are). Then, divide that count by 52. Remember that red suits are hearts and diamonds, and black suits are spades and clubs.

🎯 Exam Tip: Familiarize yourself with the composition of a standard 52-card deck (suits, colors, ranks, and special cards like face cards) to quickly identify favourable outcomes.

Question 4. A bag contains 3 red bails, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is : (i) White (ii) red (iii) black?
Answer: First, calculate the total number of balls in the bag. There are 3 red balls, 5 black balls, and 4 white balls. So, the total number of balls is \( 3 + 5 + 4 = 12 \). This means there are 12 total possible outcomes when one ball is drawn.
\( \text{Total number of balls} = 12 \)
(i) Probability of drawing a white ball:
There are 4 white balls.
\( P(\text{white ball}) = \frac{4}{12} = \frac{1}{3} \)
(ii) Probability of drawing a red ball:
There are 3 red balls.
\( P(\text{red ball}) = \frac{3}{12} = \frac{1}{4} \)
(iii) Probability of drawing a black ball:
There are 5 black balls.
\( P(\text{black ball}) = \frac{5}{12} \)
Understanding the probability of picking each color helps predict outcomes in real-world scenarios, like lottery draws.
In simple words: Add all the balls to get the total number. For each question, count the number of balls of that specific color. Then divide the count of that color by the total number of balls to find the probability.

🎯 Exam Tip: Always state the total number of outcomes clearly at the beginning of your solution. Make sure to simplify all probability fractions to their lowest terms.

Question 5. 1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Salman has purchased on lottery ticket, what is the probability of (i) winning a prize? (ii) not winning a prize?
Answer: In this lottery, a total of 1000 tickets were sold. Among these, there are 5 tickets that have prizes. Salman bought one ticket. This means the total number of possible outcomes for Salman's ticket is 1000.
\( \text{Total number of tickets} = 1000 \)
\( \text{Number of prizes} = 5 \)
(i) Probability of winning a prize:
There are 5 prize tickets, so these are the favourable outcomes for winning.
\( P(\text{winning a prize}) = \frac{\text{Number of prizes}}{\text{Total number of tickets}} = \frac{5}{1000} = \frac{1}{200} = 0.005 \)
(ii) Probability of not winning a prize:
The number of tickets that do not have a prize is the total tickets minus the prize tickets: \( 1000 - 5 = 995 \). These are the favourable outcomes for not winning.
\( P(\text{not winning a prize}) = \frac{\text{Number of non-prize tickets}}{\text{Total number of tickets}} = \frac{995}{1000} = \frac{199}{200} = 0.995 \)
The sum of the probability of winning and not winning should always be 1.
In simple words: There are 1000 tickets in total. 5 of them can win. So, the chance of winning is 5 out of 1000. The chance of not winning is the rest, which is 995 out of 1000.

🎯 Exam Tip: Remember that the probability of an event happening plus the probability of it not happening always adds up to 1. This can be used to check your calculations or find one probability if the other is known.

Question 6. The numbers from 1-15 are each printed on a counter and the 15 counters are placed in a bag. If one counter is picked at random, what is the probability that the number on it is : (i) an odd number (ii) an even number (iii) a prime number (iv) a multiple of 4
Answer: There are 15 counters in a bag, numbered from 1 to 15. When one counter is picked at random, the total number of possible outcomes is 15.
\( \text{Total possible outcomes} = 15 \)
The probability of an event (E) is \( P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} \).
(i) Probability of picking an odd number:
The odd numbers between 1 and 15 are 1, 3, 5, 7, 9, 11, 13, 15. There are 8 odd numbers.
\( P(\text{odd number}) = \frac{8}{15} \)
(ii) Probability of picking an even number:
The even numbers between 1 and 15 are 2, 4, 6, 8, 10, 12, 14. There are 7 even numbers.
\( P(\text{even number}) = \frac{7}{15} \)
(iii) Probability of picking a prime number:
The prime numbers between 1 and 15 are 2, 3, 5, 7, 11, 13. There are 6 prime numbers.
\( P(\text{prime number}) = \frac{6}{15} = \frac{2}{5} \)
(iv) Probability of picking a multiple of 4:
The multiples of 4 between 1 and 15 are 4, 8, 12. There are 3 multiples of 4.
\( P(\text{multiple of 4}) = \frac{3}{15} = \frac{1}{5} \)
Identifying prime numbers, odd numbers, and multiples correctly is key to solving these probability problems.
In simple words: Count the total numbers from 1 to 15. Then, for each part, count how many numbers fit the description (like being odd or prime). Divide this count by 15.

🎯 Exam Tip: Always list out the numbers that fit each category (odd, even, prime, multiples) to avoid counting errors. Remember, 1 is not a prime number.

Question 7. the letters in the word CHANCE were stuck on a die. Find the probability of rolling: (i) letter C (ii) a vowel (iii) a consonant (iv) any letter except C.
Answer: The word "CHANCE" has 6 letters. These letters are C, H, A, N, C, E. When this die is thrown once, there are 6 possible outcomes, one for each face. We need to find the probability of rolling specific types of letters.
\( \text{Total possible outcomes} = 6 \)
(i) Probability of rolling the letter C:
The letter C appears 2 times in the word CHANCE.
\( P(\text{letter C}) = \frac{2}{6} = \frac{1}{3} \)
(ii) Probability of rolling a vowel:
The vowels in the word CHANCE are A and E. There are 2 vowels.
\( P(\text{a vowel}) = \frac{2}{6} = \frac{1}{3} \)
(iii) Probability of rolling a consonant:
The consonants in the word CHANCE are C, H, N, C. There are 4 consonants.
\( P(\text{a consonant}) = \frac{4}{6} = \frac{2}{3} \)
(iv) Probability of rolling any letter except C:
The letters in CHANCE that are not C are H, A, N, E. There are 4 such letters.
\( P(\text{any letter except C}) = \frac{4}{6} = \frac{2}{3} \)
This type of problem demonstrates how probability applies to non-numeric outcomes, like letters in a word.
In simple words: Count all the letters in CHANCE (there are 6). Then count how many times each type of letter appears, like how many C's, how many vowels (A, E), or how many letters are not C. Divide these counts by 6.

🎯 Exam Tip: Be careful to count repeated letters correctly when determining the number of favourable outcomes, as each face of the die represents a distinct outcome.

Question 8. A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is : (i) white (ii) red or black? (iii) not blue (iv) neither white nor blue.
Answer: First, calculate the total number of balls in the bag. There are 5 white balls, 7 red balls, 4 black balls, and 2 blue balls. The total number of balls is \( 5 + 7 + 4 + 2 = 18 \). This is the total number of possible outcomes.
\( \text{Total number of balls} = 18 \)
(i) Probability of being a white ball:
There are 5 white balls.
\( P(\text{white ball}) = \frac{5}{18} \)
(ii) Probability of being a red or black ball:
The number of red balls is 7 and the number of black balls is 4. So, the number of balls that are either red or black is \( 7 + 4 = 11 \).
\( P(\text{red or black ball}) = \frac{11}{18} \)
(iii) Probability of being not a blue ball:
The number of blue balls is 2. So, the number of balls that are not blue is \( 18 - 2 = 16 \).
\( P(\text{not a blue ball}) = \frac{16}{18} = \frac{8}{9} \)
(iv) Probability of a ball which is neither white nor blue:
This means the ball must be either red or black. The number of red balls is 7 and black balls is 4. So, \( 7 + 4 = 11 \) balls are neither white nor blue. Another way to calculate this is \( \text{Total balls} - (\text{white balls} + \text{blue balls}) = 18 - (5 + 2) = 18 - 7 = 11 \).
\( P(\text{neither white nor blue}) = \frac{11}{18} \)
This problem demonstrates how to find probabilities for combined and complementary events.
In simple words: Add up all the balls to find the total. Then, for each part, count how many balls fit that description. Divide the count for each part by the total number of balls. For "not blue" or "neither white nor blue," count the balls that are left over.

🎯 Exam Tip: For "neither A nor B" events, find the sum of outcomes for all categories *other* than A and B. For "A or B" events, sum the outcomes for A and B. Ensure there's no overlap if they are mutually exclusive.

Question 9. Find the probability of drawing an ace or a jack from a pack of 52 cards.
Answer: A standard deck of playing cards contains 52 cards in total. We want to find the probability of drawing either an ace or a jack. These are distinct types of cards. There are 4 aces in a deck (one from each suit) and 4 jacks in a deck (one from each suit).
\( \text{Total number of possible outcomes} = 52 \)
\( \text{Number of aces} = 4 \)
\( \text{Number of jacks} = 4 \)
Since drawing an ace and drawing a jack are mutually exclusive events (they cannot happen at the same time), the number of favourable outcomes for drawing an ace or a jack is the sum of the number of aces and the number of jacks.
\( \text{Number of favourable outcomes (ace or jack)} = 4 \text{ aces} + 4 \text{ jacks} = 8 \)
\( P(\text{ace or jack}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{8}{52} = \frac{2}{13} \)
This shows how the probability of either one event or another event occurring can be calculated by summing their individual favourable outcomes.
In simple words: In a deck of 52 cards, there are 4 aces and 4 jacks. So, there are 8 cards that are either an ace or a jack. The chance of drawing one of them is 8 out of 52, which simplifies to 2 out of 13.

🎯 Exam Tip: When calculating the probability of "A or B" for mutually exclusive events, simply add the number of outcomes for A and the number of outcomes for B, then divide by the total outcomes. No need to subtract any overlap.

Question 10. Tickets numbered 1 to 25 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of both 2 and 3?
Answer: There are 25 tickets in total, numbered from 1 to 25. When one ticket is drawn randomly, the total number of possible outcomes is 25. We need to find the probability that the number on the ticket is a multiple of both 2 and 3. A number that is a multiple of both 2 and 3 is also a multiple of their least common multiple, which is 6.
\( \text{Total number of tickets} = 25 \)
The numbers between 1 and 25 that are multiples of 6 are: 6, 12, 18, 24.
\( \text{Number of favourable outcomes (multiples of 6)} = 4 \)
\( P(\text{multiple of both 2 and 3}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{4}{25} \)
Finding the least common multiple is a key step in solving problems involving "multiples of both" several numbers.
In simple words: There are 25 tickets. We need to find numbers that can be divided by both 2 and 3, which means numbers that can be divided by 6. The numbers are 6, 12, 18, and 24. There are 4 such numbers. So, the chance is 4 out of 25.

🎯 Exam Tip: When a question asks for a number that is a multiple of two different numbers (e.g., 2 and 3), always find the least common multiple (LCM) of those numbers first (e.g., LCM of 2 and 3 is 6). Then, list the multiples of the LCM within the given range.

Question 11. A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is (i) black (ii) red and a queen (iii) neither heart nor a king (iv) an ace, jack or a king (v)'10' of diamonds (vi) club or an ace
Answer: A standard deck of playing cards has 52 cards. When one card is drawn at random, the total number of possible outcomes is 52. We will calculate the probability for each specific event.
\( \text{Total number of possible outcomes} = 52 \)
(i) Probability of drawing a black card:
There are 26 black cards (13 spades + 13 clubs) in the deck.
\( P(\text{black card}) = \frac{26}{52} = \frac{1}{2} \)
(ii) Probability of drawing a red and a queen:
There are two red suits (hearts and diamonds), and each suit has one queen. So, there are 2 red queens (Queen of Hearts and Queen of Diamonds).
\( P(\text{red and a queen}) = \frac{2}{52} = \frac{1}{26} \)
(iii) Probability of drawing neither a heart nor a king:
First, find the number of cards that *are* hearts or kings. There are 13 heart cards. There are 4 kings in total, but the King of Hearts is already counted. So, there are 3 more kings (King of Spades, King of Clubs, King of Diamonds) not in hearts. The total number of hearts or kings is \( 13 + 3 = 16 \).
The number of cards that are neither heart nor king is \( 52 - 16 = 36 \).
\( P(\text{neither heart nor king}) = \frac{36}{52} = \frac{9}{13} \)
(iv) Probability of drawing an ace, jack, or a king:
There are 4 aces, 4 jacks, and 4 kings in a deck. These are distinct sets of cards.
\( \text{Number of favourable outcomes} = 4 + 4 + 4 = 12 \)
\( P(\text{ace, jack or king}) = \frac{12}{52} = \frac{3}{13} \)
(v) Probability of drawing a '10' of diamonds:
There is only one card that is the 10 of Diamonds.
\( P(\text{10 of diamonds}) = \frac{1}{52} \)
(vi) Probability of drawing a club or an ace:
There are 13 club cards. There are 4 aces in the deck, but one of them (the Ace of Clubs) is already counted within the 13 club cards. So, we add the 3 aces from the other suits (Ace of Hearts, Ace of Diamonds, Ace of Spades).
\( \text{Number of favourable outcomes} = 13 \text{ (clubs)} + 3 \text{ (other aces)} = 16 \)
\( P(\text{club or an ace}) = \frac{16}{52} = \frac{4}{13} \)
This problem shows how to handle probabilities for single events, combined events, and "not" events in a deck of cards.
In simple words: For each part, count the specific cards from a 52-card deck. For example, there are 26 black cards. Divide this count by 52. When it's "neither X nor Y," find how many cards are X or Y, then subtract that from 52. For "X or Y," add the counts, being careful not to count any card twice.

🎯 Exam Tip: When dealing with "A or B" probabilities, always check for overlap. If the events are not mutually exclusive (e.g., "club or ace"), use the formula \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \), or carefully count unique outcomes.

Question 12. A box contains 20 balls bearing numbers 1, 2, 3, 4,...............20. A ball is drawn at random from the box. What is the probability that the number on the ball is (i) an odd number (ii) divisible by 2 or 3 (iii) prime number (iv) not divisible by 10.
Answer: A box contains 20 balls, each numbered from 1 to 20. When a ball is drawn at random, the total number of possible outcomes is 20.
\( \text{Total number of possible outcomes} = 20 \)
(i) Probability of drawing an odd number:
The odd numbers between 1 and 20 are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19. There are 10 odd numbers.
\( P(\text{odd number}) = \frac{10}{20} = \frac{1}{2} \)
(ii) Probability of drawing a number divisible by 2 or 3:
Numbers divisible by 2 (even numbers): 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 (10 numbers).
Numbers divisible by 3: 3, 6, 9, 12, 15, 18 (6 numbers).
Numbers divisible by both 2 and 3 (multiples of 6): 6, 12, 18 (3 numbers).
Number of outcomes divisible by 2 or 3 = (Numbers div by 2) + (Numbers div by 3) - (Numbers div by 6)
\( = 10 + 6 - 3 = 13 \)
\( P(\text{divisible by 2 or 3}) = \frac{13}{20} \)
(iii) Probability of drawing a prime number:
The prime numbers between 1 and 20 are: 2, 3, 5, 7, 11, 13, 17, 19. There are 8 prime numbers.
\( P(\text{prime number}) = \frac{8}{20} = \frac{2}{5} \)
(iv) Probability of drawing a number not divisible by 10:
First, find numbers divisible by 10: 10, 20. There are 2 such numbers.
The numbers not divisible by 10 are \( \text{Total numbers} - (\text{Numbers divisible by 10}) = 20 - 2 = 18 \).
\( P(\text{not divisible by 10}) = \frac{18}{20} = \frac{9}{10} \)
Understanding divisibility rules and prime numbers is essential for accurately calculating these probabilities.
In simple words: Out of 20 balls, count how many are odd, how many are divisible by 2 or 3 (remember to not double-count numbers like 6, 12, 18), how many are prime, and how many are not divisible by 10. Then divide each count by 20.

🎯 Exam Tip: When calculating "divisible by A or B", use the Principle of Inclusion-Exclusion: Count(A) + Count(B) - Count(A and B). For "not divisible by X", calculate P(divisible by X) and subtract from 1.

Question 13. A die is rolled. If the outcome is an even number, what is the probability that it is a prime number?
Answer: When a standard die is rolled, the possible outcomes are {1, 2, 3, 4, 5, 6}. The question states that the outcome *is* an even number. This means our new, reduced sample space (the set of all possible outcomes under this condition) consists only of the even numbers from the die. So, the conditional sample space is {2, 4, 6}.
\( \text{Conditional total possible outcomes (even numbers)} = 3 \)
Now, we need to find the numbers in this conditional sample space that are also prime numbers. From {2, 4, 6}, the only prime number is 2.
\( \text{Favourable outcomes (prime numbers within the even numbers)} = 1 \)
The probability is then calculated from this reduced sample space.
\( P(\text{prime | even}) = \frac{\text{Number of prime even numbers}}{\text{Total number of even numbers}} = \frac{1}{3} \)
This demonstrates conditional probability, where the sample space changes based on new information.
In simple words: If you know the number rolled on a die is even, the only possible numbers are 2, 4, or 6. Out of these three, only one (2) is a prime number. So, the chance is 1 out of 3.

🎯 Exam Tip: For conditional probability problems, the most crucial step is to correctly identify the *reduced sample space* based on the given condition before counting favourable outcomes.

Question 14. A single letter is selected at random from the word “PROBABILITY”. What is.the probability of it being a vowel?
Answer: The word is "PROBABILITY". To find the probability of selecting a vowel, we first need to count the total number of letters in the word and then identify the number of vowels.
The letters in "PROBABILITY" are P, R, O, B, A, B, I, L, I, T, Y.
\( \text{Total number of letters} = 11 \)
The vowels in the English alphabet are A, E, I, O, U. From the word "PROBABILITY", the vowels are O, A, I, I.
\( \text{Number of vowels} = 4 \)
The probability of selecting a vowel is the number of vowels divided by the total number of letters.
\( P(\text{vowel}) = \frac{\text{Number of vowels}}{\text{Total number of letters}} = \frac{4}{11} \)
Identifying all occurrences of vowels in a word, even repeated ones, is important for accuracy.
In simple words: Count all the letters in the word "PROBABILITY" (there are 11). Then count all the vowel letters (A, E, I, O, U) in it (there are 4: O, A, I, I). The chance of picking a vowel is 4 out of 11.

🎯 Exam Tip: Carefully list out all letters in the word and then systematically identify the vowels to ensure none are missed or double-counted incorrectly. Treat each instance of a letter as a separate outcome.

Question 15. A survey of 500 families shows the following results: No. of children in the family 1 2 3 0 No. of families 250 125 75 50 Out of these, one family is chosen at random. Find the probability that the chosen family has 2 children.
Answer: A survey was conducted among 500 families. This means the total number of possible outcomes when choosing one family at random is 500. We are interested in the probability that the chosen family has 2 children. The survey results tell us how many families fall into each category.
\( \text{Total number of families} = 500 \)
From the given table, the number of families that have exactly 2 children is 125.
\( \text{Number of families with 2 children} = 125 \)
The probability is calculated by dividing the number of favourable outcomes by the total number of outcomes.
\( P(\text{family with 2 children}) = \frac{\text{Number of families with 2 children}}{\text{Total number of families}} = \frac{125}{500} = \frac{1}{4} \)
This is an example of using data from a survey to calculate empirical probability.
In simple words: There are 500 families in total. We know that 125 of these families have 2 children. So, the chance of picking a family with 2 children is 125 out of 500, which simplifies to 1 out of 4.

🎯 Exam Tip: When data is presented in a table, make sure to correctly identify the total number of items (in this case, families) and the specific count for the desired event from the table rows/columns.

Question 16. If a spinner illustrated is spun, what is the probability of getting (i) an even number? (ii) a 3 or a 5? (iii) a number greater than 4? (iv) a multiple of 2?
Answer: The spinner has numbers from 1 to 8. When the spinner is spun, the total number of possible outcomes is 8. We will find the probability for each requested event.
\( \text{Total possible outcomes} = 8 \)
(i) Probability of getting an even number:
The even numbers on the spinner are 2, 4, 6, 8. There are 4 even numbers.
\( P(\text{even number}) = \frac{4}{8} = \frac{1}{2} \)
(ii) Probability of getting a 3 or a 5:
The numbers that are either 3 or 5 are 3 and 5. There are 2 such numbers.
\( P(\text{3 or 5}) = \frac{2}{8} = \frac{1}{4} \)
(iii) Probability of getting a number greater than 4:
The numbers greater than 4 are 5, 6, 7, 8. There are 4 such numbers.
\( P(\text{number} > 4) = \frac{4}{8} = \frac{1}{2} \)
(iv) Probability of getting a multiple of 2:
Multiples of 2 are numbers that can be divided by 2. These are the even numbers: 2, 4, 6, 8. There are 4 multiples of 2.
\( P(\text{multiple of 2}) = \frac{4}{8} = \frac{1}{2} \)
Spinner problems are great for understanding basic probability concepts with a visual aid.
In simple words: The spinner has 8 numbers. For each question, count how many of those numbers fit the description (like being even, or being 3 or 5). Then, divide that count by 8. Simplify the fraction if you can.

🎯 Exam Tip: For spinner problems, always clearly identify the numbers on the spinner first. Listing them helps in correctly counting the favourable outcomes for each part of the question.

Question 17. A fete is being held in a large hall which has 9 doors. By mistake, the caretaker has left 3 of them locked. If someone tries a door at random, what is the probability that it will be : (i) locked? (ii) not locked
Answer: In the hall, there are 9 doors in total. When someone tries a door at random, the total number of possible outcomes is 9. We are told that 3 of these doors are locked.
\( \text{Total number of doors} = 9 \)
\( \text{Number of locked doors} = 3 \)
The number of doors that are not locked is \( 9 - 3 = 6 \).
(i) Probability of the door being locked:
There are 3 locked doors, which are the favourable outcomes.
\( P(\text{locked door}) = \frac{3}{9} = \frac{1}{3} \)
(ii) Probability of the door not being locked:
There are 6 doors that are not locked.
\( P(\text{not locked door}) = \frac{6}{9} = \frac{2}{3} \)
This problem demonstrates simple probability with real-world scenarios, making it easy to understand the concepts of favorable and total outcomes.
In simple words: There are 9 doors. 3 are locked. So, the chance of picking a locked door is 3 out of 9. The other 6 doors are not locked, so the chance of picking an unlocked door is 6 out of 9. Simplify both fractions.

🎯 Exam Tip: Clearly distinguish between the number of favourable outcomes (e.g., locked doors) and the total number of outcomes (total doors). Always reduce fractions to their simplest form.

Question 18. A child has a block in the shape of a cube with one letter written on each face as shown. A B C D E A The cube is thrown once. What is the probability of getting (i) A ? (ii) D?
Answer: The child has a cube-shaped block with letters on its faces: A, B, C, D, E, A. Since a cube has 6 faces, and each face has one letter, the total number of possible outcomes when the cube is thrown once is 6.
\( \text{Letters on the cube faces} = \{\text{A, B, C, D, E, A}\} \)
\( \text{Total number of possible outcomes} = 6 \)
(i) Probability of getting the letter A:
The letter A appears 2 times on the faces of the cube.
\( P(\text{getting A}) = \frac{2}{6} = \frac{1}{3} \)
(ii) Probability of getting the letter D:
The letter D appears 1 time on the faces of the cube.
\( P(\text{getting D}) = \frac{1}{6} \)
This problem helps in understanding how probability works when there are repeated outcomes, like the letter 'A' here.
In simple words: The cube has 6 sides. Two sides show the letter A, so the chance of getting A is 2 out of 6. Only one side shows the letter D, so the chance of getting D is 1 out of 6.

🎯 Exam Tip: Count each face of the die as a distinct outcome, even if the letter printed on it is the same as another face. If a letter appears multiple times, include each instance in your count of favourable outcomes.

Question 19. A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing (i) a face card (ii) a red face card
Answer: A well-shuffled deck of playing cards has a total of 52 cards. When one card is drawn at random, the total number of possible outcomes is 52. We need to find the probability of drawing specific types of cards.
\( \text{Total number of cards} = 52 \)
(i) Probability of drawing a face card:
In a standard deck, the face cards are King, Queen, and Jack. There are 4 suits, so \( 3 \text{ face cards/suit} \times 4 \text{ suits} = 12 \) face cards in total. Note: Some problem definitions might include Ace as a face card; here we follow the common interpretation of K, Q, J.
\( P(\text{face card}) = \frac{12}{52} = \frac{3}{13} \)
(ii) Probability of drawing a red face card:
The red suits are hearts and diamonds. Each red suit has 3 face cards (King, Queen, Jack). So, there are \( 2 \text{ red suits} \times 3 \text{ face cards/suit} = 6 \) red face cards in total.
\( P(\text{red face card}) = \frac{6}{52} = \frac{3}{26} \)
This problem helps solidify the understanding of specific card categories within a deck and their respective probabilities.
In simple words: In a 52-card deck, face cards are Kings, Queens, and Jacks. There are 12 of them. So, the chance of drawing a face card is 12 out of 52. Out of these, 6 are red face cards (3 from hearts, 3 from diamonds), so the chance is 6 out of 52.

🎯 Exam Tip: Always confirm the definition of "face card" being used in the context of the question (e.g., does it include Ace?). If not specified, King, Queen, Jack is the standard assumption.

Question 20. A bag contains 5 red balls and some bine balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.
Answer: Let's assume the question meant "blue" balls instead of "bine" balls. We are given that there are 5 red balls in the bag. Let the number of blue balls be \( x \).
\( \text{Number of red balls} = 5 \)
\( \text{Number of blue balls} = x \)
The total number of balls in the bag is the sum of red and blue balls: \( 5 + x \).
The probability of drawing a blue ball is \( P(\text{blue}) = \frac{\text{Number of blue balls}}{\text{Total balls}} = \frac{x}{5+x} \).
The probability of drawing a red ball is \( P(\text{red}) = \frac{\text{Number of red balls}}{\text{Total balls}} = \frac{5}{5+x} \).
According to the problem, the probability of drawing a blue ball is double that of a red ball.
\( P(\text{blue}) = 2 \times P(\text{red}) \)
\( \frac{x}{5+x} = 2 \times \frac{5}{5+x} \)
\( \frac{x}{5+x} = \frac{10}{5+x} \)
Since the denominators are equal and positive (as \( x \) must be a count of balls), the numerators must also be equal.
\( x = 10 \)
Therefore, the number of blue balls in the bag is 10. The probabilities must add up to a logical total, and the number of balls cannot be negative.
In simple words: There are 5 red balls and 'x' blue balls. The chance of picking a blue ball is twice the chance of picking a red ball. This means there must be twice as many blue balls as red balls. So, \( x = 2 \times 5 = 10 \) blue balls.

🎯 Exam Tip: When setting up equations for probability, ensure the total number of outcomes (denominator) accurately reflects all items in the bag. Carefully translate the given relationship (e.g., "double that of") into a correct mathematical equation.

Question 21. The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. One card is selected from the remaining cards. Find the probability of getting (i) a heart, (ii) a king, (iii) a club, (iv) the '10′ of hearts.
Answer: A standard deck has 52 cards. From these, the king, queen, and jack of clubs (3 cards) are taken out. So, the total number of cards left is \( 52 - 3 = 49 \).
(i) Number of heart cards: There are 13 hearts in a deck, and none of them are clubs, so all 13 hearts remain.
\( P(heart) = \frac{\text{Number of heart cards}}{\text{Total remaining cards}} = \frac{13}{49} \)
(ii) Number of kings: There are 4 kings in a full deck. Since the king of clubs was removed, 3 kings are left (king of spades, king of diamonds, king of hearts).
\( P(king) = \frac{\text{Number of kings}}{\text{Total remaining cards}} = \frac{3}{49} \)
(iii) Number of club cards: There are 13 clubs originally. Since 3 club cards (king, queen, jack) were removed, \( 13 - 3 = 10 \) club cards are left.
\( P(club) = \frac{\text{Number of club cards}}{\text{Total remaining cards}} = \frac{10}{49} \)
(iv) Number of '10' of hearts: There is only one 10 of hearts in a deck, and it was not removed.
\( P(10 \text{ of hearts}) = \frac{\text{Number of 10 of hearts}}{\text{Total remaining cards}} = \frac{1}{49} \)
In simple words: First, figure out how many cards are left after some are removed. Then, for each part, count how many of the desired cards are still in the deck and divide by the total number of remaining cards.

🎯 Exam Tip: Always clearly state the reduced sample space (total possible outcomes) when cards are removed from a deck before calculating individual probabilities.

Question 22. If a coin is tossed twice, what is the probability of getting 'head' at least once?
Answer: When a coin is tossed twice, all the possible outcomes are: Head-Head (HH), Head-Tail (HT), Tail-Head (TH), and Tail-Tail (TT). There are 4 total possible outcomes.
The outcomes where 'head' appears at least once are HH, HT, and TH. These are 3 favorable outcomes.
The probability is calculated as the number of favorable outcomes divided by the total number of possible outcomes.
\( P(\text{head at least once}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{3}{4} \)
In simple words: When you flip a coin twice, there are four possible results. We want to know the chance of getting at least one head, which means HH, HT, or TH. So, 3 out of 4 times, you will get at least one head.

🎯 Exam Tip: List all possible outcomes clearly to avoid missing any, especially for "at least once" type questions, as it helps identify all favorable events.

Question 23. Fill in the blanks with the appropriate correct answer.
(i) Chance of throwing 6 with a single die is __________.
(ii) A pair of fair dice is thrown and one die shows a four. The probability that the other die shows 5 is __________.
(iii) Probability of a sure event is __________.
(iv) Probability of an impossible event is __________.
(v) The probability of an event (other than sure and an impossible event) lies between.
(vi) A die is rolled once. The probability of getting a prime number is __________.
Answer:
(i) \( \frac{1}{6} \)
(ii) \( \frac{1}{36} \)
(iii) \( 1 \)
(iv) \( 0 \)
(v) \( 0 \) and \( 1 \)
(vi) \( \frac{3}{6} = \frac{1}{2} \)
In simple words: For a single die, there is one '6' out of six sides. For two dice, the chance of one specific outcome like (4,5) is 1 out of 36 total combinations. An event that must happen has a probability of 1, and an event that cannot happen has a probability of 0. All other probabilities are between 0 and 1. Prime numbers on a die are 2, 3, and 5.

🎯 Exam Tip: Remember the basic definitions: probability is always between 0 and 1. A sure event is 1, an impossible event is 0. Listing sample spaces helps with die questions.

Self-Evaluation And Revision (Latest Icse Questions)

Question 1. A die is thrown once. What is the probability that the
(i) number is even?
(ii) number is greater than 2?
Answer: When a die is thrown once, the possible outcomes (sample space) are {1, 2, 3, 4, 5, 6}. The total number of possible outcomes is 6.
(i) The even numbers in the sample space are {2, 4, 6}. There are 3 favorable outcomes.
\( P(\text{even number}) = \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} \)
(ii) The numbers greater than 2 in the sample space are {3, 4, 5, 6}. There are 4 favorable outcomes.
\( P(\text{number} > 2) = \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3} \)
In simple words: When you roll a die, there are 6 possible results. Half of these results (2, 4, 6) are even numbers. Four of these results (3, 4, 5, 6) are bigger than 2.

🎯 Exam Tip: Always clearly list the sample space first. This helps in correctly identifying favorable outcomes for different conditions like "even" or "greater than".

Question 2. Cards marked with numbers 1, 2, 3, 4, ....... 20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is :
(i) a prime number?
(ii) divisible by 3?
(iii) a perfect square?
Answer: Cards are numbered from 1 to 20. So, the total number of possible outcomes is 20.
(i) Prime numbers between 1 and 20 are numbers that are only divisible by 1 and themselves. These are {2, 3, 5, 7, 11, 13, 17, 19}. There are 8 prime numbers.
\( P(\text{prime number}) = \frac{\text{Number of prime numbers}}{\text{Total outcomes}} = \frac{8}{20} = \frac{2}{5} \)
(ii) Numbers divisible by 3 between 1 and 20 are {3, 6, 9, 12, 15, 18}. There are 6 such numbers.
\( P(\text{divisible by 3}) = \frac{\text{Number of multiples of 3}}{\text{Total outcomes}} = \frac{6}{20} = \frac{3}{10} \)
(iii) Perfect squares between 1 and 20 are numbers that can be obtained by squaring an integer. These are {1, 4, 9, 16} (since \( 1^2=1, 2^2=4, 3^2=9, 4^2=16 \)). There are 4 perfect squares.
\( P(\text{perfect square}) = \frac{\text{Number of perfect squares}}{\text{Total outcomes}} = \frac{4}{20} = \frac{1}{5} \)
In simple words: For cards numbered 1 to 20, find how many cards fit each rule. Then divide that count by 20. Prime numbers are only divided by 1 and themselves. Multiples of 3 are numbers you get when you multiply by 3. Perfect squares are numbers like 1, 4, 9, 16.

🎯 Exam Tip: When dealing with numbers, listing out the favorable outcomes (like primes or multiples) for the given range is a reliable method to ensure accuracy.

Question 3. From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is
(i) a face card (King, Jack or Queen)?
(ii) an even numbered red card?
Answer: A standard deck has 52 cards. Cards that are multiples of 3 are removed. These are 3s, 6s, and 9s from each of the 4 suits. So, \( 3 \text{ cards/suit} \times 4 \text{ suits} = 12 \) cards are removed.
The total number of remaining cards is \( 52 - 12 = 40 \).
(i) Face cards (King, Jack, or Queen): There are 4 Kings, 4 Queens, and 4 Jacks, making a total of 12 face cards. None of these cards are multiples of 3, so no face cards were removed.
\( P(\text{face card}) = \frac{\text{Number of face cards}}{\text{Total remaining cards}} = \frac{12}{40} = \frac{3}{10} \)
(ii) Even numbered red card: Red cards are hearts and diamonds. The even numbered cards are 2, 4, 6, 8, 10. So there are 5 even-numbered hearts and 5 even-numbered diamonds, making 10 even-numbered red cards in total. The cards removed were multiples of 3, which includes the 6 of hearts and 6 of diamonds (2 cards). Therefore, \( 10 - 2 = 8 \) even-numbered red cards remain.
\( P(\text{even numbered red card}) = \frac{10}{40} = \frac{1}{4} \)
In simple words: First, subtract the cards that are multiples of 3 (like 3, 6, 9 of each suit) from the total deck to find the new total. Then, count the face cards left (King, Queen, Jack) and divide by the new total. For red even cards, count how many were there originally, and then subtract any that were also multiples of 3, then divide by the new total.

🎯 Exam Tip: When cards are removed from a deck, make sure to adjust both the total number of cards (sample space) and the number of favorable outcomes for each specific event.

Question 4. Two coins are tossed once. Find the probability of getting :
(i) 2 heads
(ii) at least 1 tail.
Answer: When two coins are tossed once, the possible outcomes are Head-Head (HH), Head-Tail (HT), Tail-Head (TH), and Tail-Tail (TT). There are 4 total possible outcomes.
(i) To get 2 heads, only one outcome is favorable: HH.
\( P(2 \text{ heads}) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \frac{1}{4} \)
(ii) To get at least 1 tail, the favorable outcomes are HT, TH, and TT. There are 3 such outcomes.
\( P(\text{at least 1 tail}) = \frac{\text{Number of favourable outcomes}}{\text{Total possible outcomes}} = \frac{3}{4} \)
In simple words: When you flip two coins, there are four ways they can land. The chance of both being heads is one out of four. The chance of seeing at least one tail (meaning one or two tails) is three out of four.

🎯 Exam Tip: Always list all possible outcomes clearly for small experiments like coin tosses to correctly count favorable outcomes for "exactly" or "at least" conditions.

Question 5. A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box.
Answer: Let the number of black balls in the box be \( x \). The number of white balls is 30.
The total number of balls in the box is \( 30 + x \).
The probability of drawing a black ball is \( P(\text{black ball}) = \frac{x}{30+x} \).
The probability of drawing a white ball is \( P(\text{white ball}) = \frac{30}{30+x} \).
According to the problem, the probability of drawing a black ball is two-fifths of a white ball:
\( P(\text{black ball}) = \frac{2}{5} \times P(\text{white ball}) \)
\( \frac{x}{30+x} = \frac{2}{5} \times \frac{30}{30+x} \)
We can multiply both sides by \( (30+x) \) because it is a non-zero quantity (since \( x \) must be non-negative).
\( x = \frac{2}{5} \times 30 \)
\( x = 2 \times 6 \)
\( x = 12 \)
So, there are 12 black balls in the box.
In simple words: If you have some black balls and 30 white balls, and the chance of picking a black ball is \( \frac{2}{5} \) the chance of picking a white ball, then there must be 12 black balls. This keeps the chances balanced correctly.

🎯 Exam Tip: When probabilities are given as ratios or fractions of each other, set up an equation using \( x \) for the unknown quantity and solve algebraically.

Question 6. The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than -3.
(iii) the smallest integer.
Answer: The faces of the die are marked with the numbers {1, 2, 3, -1, -2, -3}. The total number of possible outcomes when the die is thrown once is 6.
(i) Positive integers: The numbers on the die that are positive are {1, 2, 3}. There are 3 favorable outcomes.
\( P(\text{positive integer}) = \frac{\text{Number of positive integers}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} \)
(ii) Integers greater than -3: The numbers on the die that are greater than -3 are {-2, -1, 1, 2, 3}. There are 5 favorable outcomes.
\( P(\text{integer} > -3) = \frac{\text{Number of integers greater than -3}}{\text{Total outcomes}} = \frac{5}{6} \)
(iii) The smallest integer: The smallest integer on the die is {-3}. There is 1 favorable outcome.
\( P(\text{smallest integer}) = \frac{\text{Number of smallest integer}}{\text{Total outcomes}} = \frac{1}{6} \)
In simple words: Look at the numbers on the die. Count how many are positive (bigger than zero), how many are bigger than minus three, and how many are the absolute smallest number. Then divide each count by the total of 6 sides to get the probability.

🎯 Exam Tip: Remember to correctly identify negative and positive integers and their order when comparing numbers, especially with a mixed set of face values on a die.

Question 7. A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball
(iii) is neither a green ball nor a white ball.
Answer: There are 5 white balls, 6 red balls, and 9 green balls in the bag.
The total number of balls in the bag is \( 5 + 6 + 9 = 20 \). This is the total number of possible outcomes.
(i) Probability of drawing a green ball: There are 9 green balls.
\( P(\text{green ball}) = \frac{\text{Number of green balls}}{\text{Total number of balls}} = \frac{9}{20} \)
(ii) Probability of drawing a white or a red ball: There are 5 white balls and 6 red balls, so \( 5 + 6 = 11 \) balls are either white or red.
\( P(\text{white or red ball}) = \frac{\text{Number of white balls} + \text{Number of red balls}}{\text{Total number of balls}} = \frac{5+6}{20} = \frac{11}{20} \)
(iii) Probability of drawing a ball that is neither a green ball nor a white ball: If the ball is not green and not white, it must be a red ball. There are 6 red balls.
\( P(\text{neither green nor white ball}) = P(\text{red ball}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{6}{20} = \frac{3}{10} \)
In simple words: First, add up all the balls to find the total. Then, for each question, count the balls that fit the description and divide that count by the total number of balls. If a ball is not green and not white, it has to be red.

🎯 Exam Tip: For "neither-nor" conditions, it's often easiest to find the complement – which is simply the remaining category. For "or" conditions, add the favorable outcomes of each type.

Question 8. A game of numbers has cards marked with 11, 12, 13, ......., 40. A card is drawn at random. Find the probability that the number on the card drawn is :
(i) A perfect square
(ii) Divisible by 7
Answer: The cards are marked from 11 to 40. The total number of cards is \( 40 - 11 + 1 = 30 \). So, there are 30 possible outcomes.
(i) A perfect square: We need to find the perfect squares between 11 and 40.
These are \( 4^2 = 16 \), \( 5^2 = 25 \), and \( 6^2 = 36 \). (Since \( 3^2=9 \) is too small and \( 7^2=49 \) is too large). There are 3 perfect squares.
\( P(\text{perfect square}) = \frac{\text{Number of perfect squares}}{\text{Total number of cards}} = \frac{3}{30} = \frac{1}{10} \)
(ii) Divisible by 7: We need to find the numbers between 11 and 40 that are divisible by 7.
These are \( 7 \times 2 = 14 \), \( 7 \times 3 = 21 \), \( 7 \times 4 = 28 \), and \( 7 \times 5 = 35 \). There are 4 such numbers.
\( P(\text{divisible by 7}) = \frac{\text{Number of cards divisible by 7}}{\text{Total number of cards}} = \frac{4}{30} = \frac{2}{15} \)
In simple words: First, count how many cards there are in total (from 11 to 40). Then, for perfect squares, list numbers like 16, 25, 36. For numbers divisible by 7, list numbers like 14, 21, 28, 35. Divide each count by the total number of cards to get the probability.

🎯 Exam Tip: Always calculate the total number of outcomes correctly for a given range (upper-lower+1). For specific conditions, carefully list all numbers that meet the criteria within that range.