CBSE Class 10 Probability Sure Shot Questions Set G

Points to Remember

• Probability of an event \( = \frac{\text{Number of favourable events}}{\text{Number of all possible outcomes of the experiment}} \)
• \( P(E) + P(\bar{E}) = 1 \)

Multiple Choice Questions

Question. Two different coins are tossed simultaneously. The probability of getting at least one head is :
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{8} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{7}{8} \)
Answer: (c) Possible outcomes = \( HH, HT, TH, TT \)
Hence, total number of possible outcomes = 4
The favourable outcomes are \( HH, HT, TH \)
Hence, the total number of favourable outcomes = 3
∴ Probability \( = \frac{3}{4} \)
So, the correct option is (c).

Question. The probability of getting an even number when a die is thrown once is :
(a) \( \frac{1}{2} \)
(b) \( \frac{1}{3} \)
(c) \( \frac{1}{6} \)
(d) \( \frac{5}{6} \)
Answer: (a) Possible outcomes = 1, 2, 3, 4, 5, 6
Hence, total number of possible outcomes = 6
The favourable outcomes are 2, 4, 6.
Hence, the total number of favourable outcomes = 3
∴ Probability = \( \frac{3}{6} = \frac{1}{2} \)
So, the correct option is (a).

Question. If P(A) denotes the probability of an event A, then :
(a) P(A) < 0
(b) P(A) > 1
(c) 0 ≤ P(A) ≤ 1
(d) – 1 ≤ P(A) ≤ 1
Answer: (c) The probability of a sure event is always 1.
The probability of an impossible event is always 0.
The sum of the probabilities of all the outcomes of elementary events is 1.
Thus, P(A) cannot be less than 0 and more than 1.
So, the correct option is (c).

Question. The probability of throwing a number greater than 2 with a single die is :
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{6} \)
(c) \( \frac{1}{3} \)
(d) \( \frac{2}{5} \)
Answer: (a) Total number of possible outcomes = 6
Favourable outcomes = 3, 4, 5, 6.
Hence, the total number of favourable outcomes = 4
∴ Probability = \( \frac{4}{6} = \frac{2}{3} \)
So, the correct option is (a).

Question. Which of the following cannot be the probability of an event ?
(a) 1.5
(b) \( \frac{3}{5} \)
(c) 25%
(d) 0.3
Answer: (a) The sum of the probabilities of all the outcomes of elementary events is 1.
Thus, probability cannot be less than 0 and more than 1.
So, the correct option is (a).

Fill in the Blanks

Question. The probability of an impossible event is .....................
Answer: Zero

Question. Number of face cards in a pack of 52 cards is ............
Answer: 12

Question. When a digit is choosen at random from the digits 1 to 9, then the probability of this chosen digit to be a prime number is ................
Answer: \( \frac{4}{9} \)

Question. An unbiased die is rolled once. The probability of getting an event prime number is .............
Answer: \( \frac{1}{6} \)

Question. The probability of getting a number which is neither prime nor composite in single throw of die is ..................
Answer: \( \frac{1}{6} \)

True/False

Question. For an event E, \( P(\bar{E}) = 1 – P(E) \).
Answer: True

Question. Probability of any event strictly between 0 and 1.
Answer: 0 ≤ P(E) ≤ 1. So, the given statement is false.

Question. A die is rolled once. The probability of getting a number which is a multiple of 3 is \( \frac{1}{2} \).
Answer: Total number of outcomes = 6. Multiple of 3 = {3, 6} ⇒ Favourable no. of outcomes = 2. ⇒ P(multiple of 3) = \( \frac{2}{6} = \frac{1}{3} \). So, the given statement is false.

Question. If P(E) = a and \( P(\bar{E}) = b \), then ab = 1.
Answer: We know, \( P(E) + P(\bar{E}) = 1 \). ⇒ \( a + b = 1 \). So, the given statement is false.

Question. Total number of outcomes in a single throw of three coins is 8.
Answer: True

Very Short Answer Type Questions

Question. If P(E) = 0.05, what is the probability of ‘not E’?
Answer: We know, P(E) + P(not E) = 1
⇒ 0.05 + P(not E) = 1
⇒ P(not E) = 1 – 0.05 = 0.95

Question. What is the probability that a randomly taken leap year has 52 Sundays?
Answer: Probability that a randomly taken leap year has 52 Sundays is 1 because there are 52 Sundays in every leap year. So, it is a sure event.

Question. A die is thrown once. What is the probability of getting a prime number?
Answer: When a die is thrown once, Total number of outcomes = 6. Also, prime numbers from 1 to 6 = {2, 3, 5} ⇒ Number of favourable outcomes = 3. ∴ P(prime number) = \( \frac{3}{6} = \frac{1}{2} \)

Question. A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant.
Answer: Total number of all possible outcomes = 26. Number of consonants = 21. ∴ P (getting a consonant) = \( \frac{21}{26} \)

Question. Two friends were born in the year 2000. What is the probability that they have the same birthday ?
Answer: Since year 2000 is a leap year. ∴ Total possible cases = 366. Let E be the event of getting the birthdays of two friends on same day. So, favourable case = 1. ∴ \( P(E) = \frac{1}{366} \)

Question. What is the probability that two friends have different birthdays ?
Answer: Total number of possible outcomes = 365 \(\times\) 365. Let E be the event of getting two friends have different birthdays. The total number of favourable outcomes = 365 \(\times\) (365 – 1) = 365 \(\times\) 364. ∴ \( P(E) = \frac{365 \times 364}{365 \times 365} = \frac{364}{365} \)

Question. From a well-shuffled pack of 52 cards, one is drawn out at random. Find the probability of getting a black queen.
Answer: The number of maximum possible outcomes = 52. The number of favourable outcomes = 2. ∴ P(black queen) = \( \frac{2}{52} = \frac{1}{26} \)

Question. Cards marked with numbers 3, 4, 5, ....., 50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number.
Answer: Total outcomes = 3, 4, 5, ......, 50. ∴ Total no. of outcomes = 48. Let E be the event of getting a perfect square number. So, favourable outcomes = 4, 9, 16, 25, 36, 49. ⇒ No. of favourable outcomes = 6. ∴ \( P(E) = \frac{6}{48} = \frac{1}{8} \)

Question. If a number x is chosen at random from the numbers – 3, – 2, – 1, 0, 1, 2, 3, then find the probability of \( x^2 < 4 \).
Answer: Total outcomes = {– 3, – 2, – 1, 0, 1, 2, 3}. ⇒ Number of total outcomes, n(S) = 7. Let A be the event of chosing a number whose square is less than 4. ∴ A = {– 1, 0, 1} ⇒ n(A) = 3. ∴ \( P(x^2 < 4) = \frac{n(A)}{n(S)} = \frac{3}{7} \)

Question. What is the probability that a number selected at random from the numbers 3, 4, 5, 6, 7, 8, 9 is a multiple of 4 ?
Answer: The maximum no. of possible outcomes = 7. Favourable outcomes are 4, 8. Thus, no. of favourable outcomes = 2. ∴ Probability = \( \frac{2}{7} \)

Question. If three different coins are tossed together, then find the probability of getting two heads.
Answer: Possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. ⇒ Total number of possible outcomes = 8. Favourable outcomes = {HHT, HTH, THH}. ⇒ Number of favourable outcomes = 3. ∴ Required probabiltiy = \( \frac{3}{8} \)

Question. Rahim tosses two different coins simultaneously. Find the probability of getting at least one tail.
Answer: Total number of possible outcomes = HH, HT, TH, TT. Favourable outcomes are HT, TH, TT. Hence, the total number of favourable outcomes = 3. ∴ Probability = \( \frac{3}{4} \)

Question. A box contains 3 blue, 2 white and 4 red marbles. If a marble is drawn out at random from the box, what is the probability that it will not be a white marble?
Answer: Total no. of possible outcomes = 3 + 2 + 4 = 9. If the marble drawn out is not to be white then the favourable outcomes are 3 blue marbles and 4 red marbles. Hence, no. of favourable outcomes = 3 + 4 = 7. ∴ Probability = \( \frac{7}{9} \)

Question. A bag contains 3 red and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is not red ?
Answer: Total number of outcomes = 3 + 5 = 8. Number of favourable outcomes = 5. ∴ Required probability = \( \frac{5}{8} \)

Short Answer Type Questions-I

Question. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out.
(i) an orange flavoured candy ?
(ii) a lemon flavoured candy ?
Answer: (i) As the bag contains only lemon flavoured candies, so, the event related to the experiment of taking out an orange flavoured candy is an impossible event. So, its probability is 0.
(ii) As the bag contains only lemon flavoured candies, so, the event related to the experiment of taking out lemon flavoured candies is a certain event. So, its probability is 1.

Question. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Answer: Here, total number of pens = 132 + 12 = 144. ∴ Total number of elementary outcomes = 144. Now, favourable number of elementary events = 132. ∴ Probability that a pen taken out is good one = \( \frac{132}{144} = \frac{11}{12} \)

Question. A child has a die whose six faces show the letters as given below :
A B C D E A
The die is thrown once. What is the probability of getting (i) A ? and (ii) D ?
Answer: The total number of elementary events associated with random experiment of throwing a die is 6.
(i) Let E be the event of getting a letter A. ∴ Favourable number of elementary events = 2. ∴ \( P(E) = \frac{2}{6} = \frac{1}{3} \).
(ii) Let E be the event of getting a letter D. ∴ Favourable number of elementary events = 1. ∴ \( P(E) = \frac{1}{6} \).

Question. If two dice are rolled together, find the probability of getting an even number on both the dice.
Answer: Possible outcomes are (1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6); (2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6); (3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6); (4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6); (5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6); (6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6). Hence, total number of possible outcomes = 36. Let E be the event of getting an even number on both the dice. Then, the favourable outcomes are (2, 2); (2, 4); (2, 6); (4, 2); (4, 4); (4, 6); (6, 2); (6, 4); (6, 6). Hence, the total number of favourable outcomes = 9. ∴ \( P(E) = \frac{9}{36} = \frac{1}{4} \)

Question. Two different dice are rolled simultaneously. Find the probability that the sum of the numbers appearing on the two dice is 10.
Answer: Possible outcomes are (1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6); (2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6); (3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6); (4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6); (5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6); (6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6). Hence, total number of possible outcomes = 36. Favourable outcomes are (4, 6); (5, 5); (6, 4). Hence, the total number of favourable outcomes = 3. ∴ Probability = \( \frac{3}{36} = \frac{1}{12} \)

Question. Two different dice are tossed together. Find the probability that (i) the number on each dice is even, and (ii) the sum of the numbers, appearing on the two dice is 5.
Answer: Possible outcomes are (1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6); (2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6); (3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6); (4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6); (5, 1); (5, 2); (5, 3); (5, 4); (5, 5); (5, 6); (6, 1); (6, 2); (6, 3); (6, 4); (6, 5); (6, 6). Hence, total number of possible outcomes = 36.
(i) Favourable outcomes are (2, 2); (2, 4); (2, 6); (4, 2); (4, 4); (4, 6); (6, 2); (6, 4); (6, 6). Hence, total number of favourable outcomes = 9. ∴ Probability = \( \frac{9}{36} = \frac{1}{4} \).
(ii) Favourable outcomes are (1, 4); (2, 3); (3, 2); (4, 1). Hence, total number of the favourable outcomes = 4. ∴ Probability = \( \frac{4}{36} = \frac{1}{9} \).

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Question. Two different dice are tossed together. Find the probability : (i) of getting a doublet. (ii) of getting a sum 10, of the numbers on the two dice.
Answer: Two dice tossed together.
⇒ Total outcome = 36.
i) doublet: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) → 6 possibilities.
Probability = \( \frac{\text{Favorable outcome}}{\text{Total outcome}} = \frac{6}{36} = \frac{1}{6} \).
ii) sum of 10: (4,6), (6,4), (5,5) → 3 possibilities.
Probability = \( \frac{\text{Favorable outcome}}{\text{Total outcome}} = \frac{3}{36} = \frac{1}{12} \).

Question. An integer is chosed at random betwen 1 and 100. Find the probability that it is :
(i) divisible by 8.
(ii) not divisible by 8.
Answer: Section-B
Integers, 1 to 100. (between)
⇒ total = 98 possible outcomes.
i) divisible by 8 → 12 numbers. (8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96).
⇒ Probability = \( \frac{\text{Favorable outcome}}{\text{Total outcome}} = \frac{12}{98} = \frac{6}{49} \).
ii) not divisible by 8 ⇒ 98 – 12 = 86 numbers.
⇒ Probability = \( \frac{\text{Favorable outcome}}{\text{Total outcome}} = \frac{86}{98} = \frac{43}{49} \).

Question. A bag contains cards numbered 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the card is divisible by both 2 and 3.
Answer: Total numbers = 25.
Let the event of getting a number divisible by 2 and 3 be E.
Then, the favourable outcomes are 6, 12, 18, 24.
Hence, number of favourable outcomes = 4
∴ \( P(E) = \frac{4}{25} \).

Question. There are 100 cards in a bag on which numbers from 1 to 100 are written. A card is taken out from the bag at random. Find the probability that the number on the selected card (i) is divisible by 9 and is a perfect square, (ii) is a prime number greater than 80.
Answer: Number of possible outcomes = 100
(i) Let \( E_1 \) be the event of getting a number divisible by 9 and is a perfect square.
∴ Favourable outcomes = {9, 36, 81}
Number of favourable outcomes = 3
∴ \( P(E_1) = \frac{3}{100} \)
(ii) Let \( E_2 \) be the event of getting a prime number greater than 80.
∴ Favourable outcomes = {83, 89, 97}
Number of favourable outcomes = 3
∴ \( P(E_2) = \frac{3}{100} \).

Question. A box contains 90 discs numbered from 1 to 90. If one disc is drawn at random from the box, then find the probability that it bears a prime number less than 23.
Answer: Total number of possible outcomes = 90
Let E be the event of getting a prime number less than 23.
Then, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17, 19
Hence, the total number of favourable outcomes = 8.
∴ \( P(E) = \frac{8}{90} = \frac{4}{45} \).

Question. A card is drawn from a well shuffled pack of 52 playing cards. Find the probability of getting (i) a red-faced card and (ii) a black-king.
Answer: Total number of cards = 52
(i) Number of red-faced cards = 3 + 3 = 6
∴ \( P(E) = \frac{6}{52} = \frac{3}{26} \)
(ii) Number of black kings = 2
∴ \( P(E) = \frac{2}{52} = \frac{1}{26} \).

Question. A die is thrown once. Find the probability of getting a number which (i) is a prime number (ii) lies between 2 and 6.
Answer: In throwing a die
Total possible outcomes = 6
i.e., \( S = \{1, 2, 3, 4, 5, 6\} \)
Prime numbers 2, 3, 5
∴ \( P(\text{Prime No.}) = \frac{\text{favourable outcomes}}{\text{Total possible outcomes}} = \frac{3}{6} = \frac{1}{2} \)
Numbers between 2 and 6 are 3, 4, 5
\( P(\text{Numbers between 2 and 6}) = \frac{3}{6} = \frac{1}{2} \).

Question. A die is thrown twice. Find the probability that (i) 5 will come up at least once. (ii) 5 will not come up either time.
Answer: When two dice are thrown simultaneously, all possible outcomes are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
Total number of outcomes = 36
Total outcomes where 5 comes up at least once = 11
(i) Probability that 5 will come up at least once \( = \frac{\text{Total outcomes where 5 will come up}}{\text{Total number of outcomes}} = \frac{11}{36} \)
Total outcomes where 5 will not come up either time = 36 – 11 = 25
(ii) Probability that 5 will not come up either time \( = \frac{\text{Total outcomes where 5 will not come up}}{\text{Total number of outcomes}} = \frac{25}{36} \).

Question. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of losing the game.
Answer: When a coin is tossed three times, the set of all possible outcomes is given by,
\( S = \{HHH, HHT, HTH, HTT, TTT, TTH, THT, THH\} \)
Same result on all tosses = {HHH, TTT}
\( P(\text{losing game}) = \frac{\text{No. of favourable outcomes}}{\text{Total possible outcomes}} = \frac{6}{8} = \frac{3}{4} \).