Probability: 1 mark Questions (CBSE Class 10)
133 different 1 mark questions on Probability from CBSE Class 10 Maths board exams 2022–2026, newest first.
Probability of getting an irrational number at random from the numbers 3 , 4 , 3 9 , 3 8 , 5 , 0 , 4 2 3 is :
(A) 0(B) 7 4 (C) 7 3 (D) 1
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Answer: (C) 7 3
4 = 2 , 3 9 = 9 , 4 2 3 = 8 and 0 are rational.3 , 3 8 = 6 2 and 5 are irrational: 3 of 7 numbers.P(irrational) = 7 3 .
A card is drawn from a well shuffled deck of 52 cards. The probability of getting an ace or a ten is :
(A) 13 1 (B) 13 2 (C) 26 1 (D) 13 4
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Answer: (B) 13 2
There are 4 aces and 4 tens: 8 favourable cards. P = 52 8 = 13 2 .
Assertion (A) : The probability of an event can not be 0.9 1 . Reason (R) : 0 ≤ P ( E ) ≤ 1 for an event E.
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
0.9 1 = 9 10 > 1 .Since every probability lies between 0 and 1, it cannot be a probability. So A is true, R is true and R explains A.
A card is drawn from a well shuffled deck of 52 cards. The probability that it is not a diamond card is :
(A) 4 1 (B) 0(C) 2 1 (D) 4 3
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Answer: (D) 4 3
There are 13 diamonds, so 52 – 13 = 39 cards are not diamonds. P = 52 39 = 4 3 .
The probability of getting sum greater than 10, when two dice are rolled together, is
(A) 9 1 (B) 18 1 (C) 12 1 (D) 1
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Answer: (C) 12 1
Total outcomes = 36. Sum greater than 10: (5, 6), (6, 5), (6, 6), i.e. 3 outcomes. Probability = 36 3 = 12 1
A bag contains some red and some white balls. A ball is drawn at random from the bag. If the probability of getting a red ball is 7 2 , then the probability of getting a white ball is
(A) 14 1 (B) 7 5 (C) 7 1 (D) 7 2
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Answer: (B) 7 5
The ball is either red or white. P(white) = 1 − 7 2 = 7 5
A card is drawn from a well-shuffled deck of 52 playing cards. The probability of getting a queen of spade is
(A) 26 1 (B) 52 1 (C) 0(D) 4 1
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Answer: (B) 52 1
There is only one queen of spades in the deck. Probability = 52 1
Three coins are tossed together. The probability of getting exactly two tails is
(A) 8 2 (B) 2 1 (C) 8 3 (D) 1
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Answer: (C) 8 3
Total outcomes = 8. Exactly two tails: HTT, THT, TTH, i.e. 3 outcomes. Probability = 8 3
A die is thrown once. Probability of getting a number other than 3 is :
(A) 6 1 (B) 6 3 (C) 6 5 (D) 1
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Answer: (C) 6 5
Total outcomes = 6. Numbers other than 3: 1, 2, 4, 5, 6, i.e. 5 outcomes. P = 6 5 .
Assertion (A) : The probability that a leap year has 53 Mondays is 7 2 . Reason (R) : The probability that a non-leap year has 53 Mondays is 7 5 .
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (C) Assertion (A) is true, but Reason (R) is false.
A leap year has 366 days = 52 weeks + 2 days. The 2 extra days can be (Sun, Mon), (Mon, Tue), ..., (Sat, Sun): 7 cases, 2 of which contain Monday. So P = 7 2 ; A is true. A non-leap year has 365 days = 52 weeks + 1 day; the extra day is Monday in 1 of 7 cases, so P = 7 1 , not 7 5 . R is false.
For an event E, P ( E ) + P ( E ) = x , then the value of x 2 − 3 is :
(A) − 2 (B) 2(C) 1(D) − 1
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Answer: (A) − 2
For any event, P ( E ) + P ( E ) = 1 , so x = 1 . x 2 − 3 = 1 − 3 = − 2 .
The probability for a randomly selected number out of 1, 2, 3, 4, ..., 25 to be a composite number is :
(A) 25 15 (B) 25 10 (C) 25 11 (D) 25 9
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Answer: (A) 25 15
Primes up to 25: 2, 3, 5, 7, 11, 13, 17, 19, 23 (9 numbers). The number 1 is neither prime nor composite. Composite numbers = 25 - 9 - 1 = 15. P(composite) = 25 15
Two dice are rolled together. The probability that the sum of the numbers obtained is divisible by 6, is :
(A) 6 1 (B) 36 11 (C) 12 1 (D) 4 1
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Answer: (A) 6 1
Total outcomes = 36 Sum 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), i.e. 5 outcomes Sum 12: (6, 6), i.e. 1 outcome P = 36 6 = 6 1
Three coins are tossed together. The probability of getting exactly one head, is :
(A) 8 1 (B) 4 3 (C) 2 1 (D) 8 3
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Answer: (D) 8 3
Total outcomes = 2 3 = 8 Exactly one head: HTT, THT, TTH, i.e. 3 outcomes P = 8 3
Two different dice are rolled together. The probability that both the obtained numbers are less than 4, is
(A) 9 2 (B) 36 7 (C) 4 1 (D) 3 2
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Answer: (C) 4 1
Total outcomes = 36. Each die must show 1, 2 or 3: favourable outcomes = 3 × 3 = 9 . P = 36 9 = 4 1
Assertion (A) : If probability of happening of an event is 0.2p, p > 0 , then p can't be more than 5. Reason (R) : P ( E ) = 1 − P ( E ) for an event E.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (B) Both A and R are true, but R is not the correct explanation of A.
Since 0 ≤ P ( E ) ≤ 1 , 0.2 p ≤ 1 ⇒ p ≤ 5 . So A is true. P ( E ) = 1 − P ( E ) is true for any event E, so R is true.But A follows from P ( E ) ≤ 1 , not from R. So R does not explain A.
Two dice are rolled together. The probability that sum of the numbers obtained is atmost 12, is
(A) 1(B) 0(C) 2 1 (D) 36 35
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Answer: (A) 1
The largest possible sum with two dice is 6 + 6 = 12. So every one of the 36 outcomes has sum at most 12: a sure event. Probability = 1
Which of the following can not be the probability of an event ?
(A) 100 39 (B) 20 0.001 (C) 0.2 10 (D) 10%
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Answer: (C) 0.2 10
Probability of an event lies between 0 and 1. 0.2 10 = 50 > 1 , so it cannot be a probability.The others are 0.39, 0.00005 and 0.1, all between 0 and 1.
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is
(A) 26 1 (B) 13 2 (C) 13 1 (D) 26 8
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Answer: (B) 13 2
Number of tens = 4, number of kings = 4, so favourable outcomes = 8. P(ten or king) = 52 8 = 13 2
Meena calculates that the probability of her winning the first prize in a lottery is 0.08. If total 800 tickets were sold, the number of tickets bought by her, is
(A) 64(B) 640(C) 100(D) 10
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Answer: (A) 64
P(winning) = 800 tickets bought by her = 0.08 Tickets bought = 0.08 × 800 = 64
Two dice are rolled together. The probability of getting an outcome ( x , y ) where x > y , is
(A) 12 5 (B) 6 5 (C) 1(D) 0
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Answer: (A) 12 5
Total outcomes = 36. Outcomes with x = y : 6. By symmetry, outcomes with x > y = 2 36 − 6 = 15 . P = 36 15 = 12 5
In a random experiment of throwing a die, which of the following is a sure event ?
(A) Getting a number between 1 and 6(B) Getting an odd number < 7(C) Getting an even number < 7(D) Getting a natural number < 7
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Answer: (D) Getting a natural number < 7
The possible outcomes are 1, 2, 3, 4, 5, 6. Every outcome is a natural number less than 7, so this event has probability 1 and is sure.
A bag contains 3 red, 4 white and 7 green balls. A ball is drawn at random. The probability that the ball drawn is not of red colour is :
(A) 11 1 (B) 14 3 (C) 14 11 (D) 11 3
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Answer: (C) 14 11
Total balls = 3 + 4 + 7 = 14 ; balls that are not red = 4 + 7 = 11 . P(not red) = 14 11 .
In an experiment of throwing a pair of dice, the probability of not getting a doublet is :
(A) 6 1 (B) 6 5 (C) 5 1 (D) 30 1
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Answer: (B) 6 5
Total outcomes = 36 ; doublets: (1, 1), (2, 2), ..., (6, 6), i.e. 6 outcomes. P(doublet) = 36 6 = 6 1 . P(not a doublet) = 1 − 6 1 = 6 5 .
The total number of outcomes in the experiment of simultaneous throw of three dice is :
(A) 6 (B) 18 (C) 36 (D) 216
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Answer: (D) 216
Each die has 6 outcomes. Total outcomes = 6 × 6 × 6 = 216 .
A pair of dice is thrown simultaneously. Let E denote the event that “The sum of numbers obtained on both dice is at least 9.” The number of outcomes in favour of event E is :
(A) 4(B) 6(C) 10(D) 26
Show answer & solution
Answer: (C) 10
Sum 9: (3, 6), (4, 5), (5, 4), (6, 3), which is 4 outcomes. Sum 10: (4, 6), (5, 5), (6, 4), which is 3 outcomes. Sum 11: (5, 6), (6, 5), which is 2 outcomes. Sum 12: (6, 6), which is 1 outcome. Total = 4 + 3 + 2 + 1 = 10 .
Three coins are tossed together. The probability that only one coin shows tail, is :
(A) 2 1 (B) 8 3 (C) 8 7 (D) 1
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Answer: (B) 8 3
Total outcomes = 2 3 = 8 . Exactly one tail: THH, HTH, HHT, i.e. 3 outcomes. Probability = 8 3 .
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that drawn card shows number ‘9’ is :
(A) 26 1 (B) 13 4 (C) 52 1 (D) 13 1
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Answer: (D) 13 1
There are four 9s in a deck (one in each suit). Probability = 52 4 = 13 1 .
Assertion (A) : If E is an event such that P ( E ) = 999 1 , then P ( E ) = 0.001 . Reason (R) : P ( E ) + P ( E ) = 1
(A) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (D) Assertion (A) is false, but Reason (R) is true.
R is true: an event and its complement have probabilities adding to 1. P ( E ) = 1 − 999 1 = 999 998 = 0.001 , so A is false.
Two dice are rolled together. The probability that at least one of them shows a six, is :
(A) 36 12 (B) 36 5 (C) 36 11 (D) 36 6
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Answer: (C) 36 11
Total outcomes = 36 . Outcomes with no six = 5 × 5 = 25 . P(at least one six) = 1 − 36 25 = 36 11 .
A black card is lost from a deck of 52 playing cards. Rest of the cards are shuffled and one card is drawn at random from the available cards. The probability that drawn card is 'king of hearts', is
(A) 52 1 (B) 4 1 (C) 51 1 (D) 26 1
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Answer: (C) 51 1
51 cards remain. The king of hearts is a red card, so it is still in the deck. P(king of hearts) = 51 1 .
If E is an event such that P(E) = 0.1, then P(E ) is equal to
(A) 0.9(B) 2 1 (C) 0.99(D) –1
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Answer: (A) 0.9
P(E ) = 1 − P(E) = 1 − 0.1 = 0.9 .
Two dice are rolled together. The probability that only one die shows number 4, is
(A) 36 11 (B) 3 1 (C) 18 5 (D) 4 1
Show answer & solution
Answer: (C) 18 5
Total outcomes = 36 . First die 4, second not 4: 5 outcomes; second die 4, first not 4: 5 outcomes. Favourable = 10 , so probability = 36 10 = 18 5 .
If E is an event such that P(E) = 1%, then P(E ) is equal to
(A) 0.09(B) 0.99(C) 99 1 (D) 0.90
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Answer: (B) 0.99
P(E) = 1% = 0.01, so P(E ) = 1 − 0.01 = 0.99 .
Two dice are rolled together. The probability of getting a sum more than 9 is
(A) 6 5 (B) 18 5 (C) 6 1 (D) 2 1
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Answer: (C) 6 1
Total outcomes = 36. Sum 10: (4, 6), (5, 5), (6, 4); sum 11: (5, 6), (6, 5); sum 12: (6, 6). That is 6 outcomes. P(sum more than 9) = 36 6 = 6 1 .
If probability of happening of an event is 57%, then probability of non-happening of the event is
(A) 0.43(B) 0.57(C) 53%(D) 57 1
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Answer: (A) 0.43
P(E) = 57% = 0.57. P(not E) = 1 − 0.57 = 0.43.
Three coins are tossed together. The probability that exactly one coin shows head, is
(A) 8 1 (B) 4 1 (C) 1(D) 8 3
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Answer: (D) 8 3
Total outcomes = 8. Exactly one head: HTT, THT, TTH, i.e. 3 outcomes. Probability = 8 3 .
Two dice are rolled together. The probability of getting an outcome (a, b) such that b = 2a, is
(A) 6 1 (B) 12 1 (C) 36 1 (D) 9 1
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Answer: (B) 12 1
Total outcomes = 36. b = 2a: (1, 2), (2, 4), (3, 6), i.e. 3 outcomes. Probability = 36 3 = 12 1 .
Three coins are tossed together. The probability that at least one head comes up, is
(A) 8 3 (B) 8 7 (C) 8 1 (D) 4 3
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Answer: (B) 8 7
Total outcomes = 8; only TTT has no head. P(at least one head) = 1 − 8 1 = 8 7 .
A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is :
(A) 13 3 (B) 13 2 (C) 2 1 (D) 26 3
Show answer & solution
Answer: (D) 26 3
Red face cards: J, Q, K of hearts and of diamonds = 6 . P = 52 6 = 26 3 .
Assertion (A) : The probability of selecting a number at random from the numbers 1 to 20 is 1. Reason (R) : For any event E, if P(E) = 1, then E is called a sure event.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
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Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
When a number is selected from 1 to 20, it is certain that the number selected is one of the numbers 1 to 20, so this is a sure event and its probability is 1. Assertion is true. Reason states the definition of a sure event: if P ( E ) = 1 , E is a sure event. Reason is true. The assertion holds because the event is a sure event, so R correctly explains A.
A card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a spade or a king ?
(A) 13 1 (B) 13 2 (C) 13 4 (D) 13 9
Show answer & solution
Answer: (C) 13 4
Spades: 13 cards. Kings that are not spades: 3 cards. Favourable outcomes = 13 + 3 = 16 . P = 52 16 = 13 4 .
The probability of drawing an even prime number out of numbers from 1 to 30 is :
(A) 30 1 (B) 15 4 (C) 30 7 (D) 0
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Answer: (A) 30 1
Total outcomes = 30 The only even prime is 2, so favourable outcomes = 1 P(even prime) = 30 1
In a cricket match, a batsman hits the boundary 7 times out of the 42 balls he plays. The probability of his not hitting a boundary is :
(A) 7 1 (B) 7 2 (C) 6 5 (D) 6 1
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Answer: (C) 6 5
Balls without a boundary = 42 − 7 = 35 P(not a boundary) = 42 35 = 6 5
If all the red face cards are removed from the deck of 52 playing cards, then the probability of getting a black jack from the remaining cards is :
(A) 46 2 (B) 52 2 (C) 48 4 (D) 23 2
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Answer: (A) 46 2
Red face cards = 6 (J, Q, K of hearts and diamonds), so remaining cards = 52 − 6 = 46 Black jacks = 2 (spades and clubs) P(black jack) = 46 2 ( = 23 1 )
If for any event E, P ( E ) + P ( E ) = q , then the value of q 2 − 3 is :
(A) 0(B) − 2 (C) 2(D) 1
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Answer: (B) − 2
For any event, P ( E ) + P ( E ) = 1 , so q = 1 q 2 − 3 = 1 − 3 = − 2
A die is thrown once. The probability of getting a number which is not a factor of 36, is :
(A) 2 1 (B) 3 2 (C) 6 1 (D) 6 5
Show answer & solution
Answer: (C) 6 1
Outcomes: 1, 2, 3, 4, 5, 6. Factors of 36 among them: 1, 2, 3, 4, 6. Only 5 is not a factor of 36. Probability = 6 1 .
If in a lottery, there are 10 prizes and 30 blanks, then the probability of winning a prize is :
(A) 4 1 (B) 3 1 (C) 4 3 (D) 3 2
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Answer: (A) 4 1
Total tickets = 10 + 30 = 40 . P ( prize ) = 40 10 = 4 1 .
The number of red balls in a bag is 10 more than the number of black balls. If the probability of drawing a red ball at random from this bag is 5 3 , then the total number of balls in the bag is :
(A) 50(B) 60(C) 80(D) 40
Show answer & solution
Answer: (A) 50
Let black balls = b ; red = b + 10 ; total = 2 b + 10 2 b + 10 b + 10 = 5 3 ⇒ 5 b + 50 = 6 b + 30 ⇒ b = 20 Total = 2 ( 20 ) + 10 = 50
A pair of dice is thrown. The probability that sum of numbers appearing on top faces is at most 10 is :
(A) 11 1 (B) 11 10 (C) 6 5 (D) 12 11
Show answer & solution
Answer: (D) 12 11
Total outcomes = 36 Sum more than 10: sum 11 {( 5 , 6 ) , ( 6 , 5 )} , sum 12 {( 6 , 6 )} , i.e. 3 outcomes P(sum at most 10) = 1 − 36 3 = 36 33 = 12 11
A pair of dice is thrown once. The probability that sum of numbers appearing on top faces is at least 4 is :
(A) 11 1 (B) 11 10 (C) 6 5 (D) 12 11
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Answer: (D) 12 11
Total outcomes = 36 Sum less than 4: ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 1 ) , i.e. 3 outcomes P(sum at least 4) = 1 − 36 3 = 36 33 = 12 11
Letters A to F are mentioned on six faces of a die such that each face has a different letter. Two such dice are thrown simultaneously. The probability that vowels turn up on both the dice is :
(A) 4 1 (B) 3 1 (C) 9 1 (D) 36 1
Show answer & solution
Answer: (C) 9 1
Vowels among A to F: A and E, so 2 of 6 faces. Favourable outcomes = 2 × 2 = 4 out of 6 × 6 = 36 Probability = 36 4 = 9 1
A bag contains red balls and black balls in the ratio 3 : 7. A ball is drawn at random. The probability that ball so drawn is black in colour, is
(A) 7 3 (B) 0.3(C) 0.7(D) 7 1
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Answer: (C) 0.7
Let red = 3 k , black = 7 k ; total = 10 k . P ( black ) = 10 k 7 k = 0.7
The probability of getting a composite number greater than 3 on throwing a die is
(A) 6 1 (B) 3 1 (C) 2 1 (D) 3 2
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Answer: (B) 3 1
Composite numbers greater than 3 on a die: 4, 6 (2 outcomes). P = 6 2 = 3 1
A piggy bank contains ₹ 1 coins and ₹ 2 coins in the ratio 9 : 11 respectively. The piggy bank is accidently dropped and a coin pops out of it. The probability that it is a ₹ 2 coin is
(A) 11 9 (B) 0.45(C) 0.55(D) 11 1
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Answer: (C) 0.55
Let ₹ 1 coins = 9 k and ₹ 2 coins = 11 k ; total = 20 k . P ( ₹ 2 coin ) = 20 k 11 k = 0.55
A bag contains red coloured, blue coloured and green coloured balls in the ratio 2 : 3 : 4. A ball is drawn at random from the given bag. The probability that the ball so drawn being not of blue colour is
(A) 9 1 (B) 3 1 (C) 3 2 (D) 9 8
Show answer & solution
Answer: (C) 3 2
Let red = 2 k , blue = 3 k , green = 4 k ; total = 9 k . P ( not blue ) = 9 k 9 k − 3 k = 9 6 = 3 2
Two coins are tossed simultaneously. The probability of getting atleast one head is
(A) 4 1 (B) 2 1 (C) 4 3 (D) 1
Show answer & solution
Answer: (C) 4 3
Outcomes: HH, HT, TH, TT (4 outcomes). At least one head: HH, HT, TH (3 outcomes). Probability = 4 3
In an experiment of throwing a die, Assertion (A) : Event E 1 : getting a number less than 3 and Event E 2 : getting a number greater than 3 are complementary events. Reason (R) : If two events E and F are complementary events, then P ( E ) + P ( F ) = 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (D) Assertion (A) is false, but Reason (R) is true.
E 1 = { 1 , 2 } , E 2 = { 4 , 5 , 6 } . Together they miss the outcome 3.P ( E 1 ) + P ( E 2 ) = 6 2 + 6 3 = 6 5 = 1 , so they are not complementary. A is false.R is the definition-property of complementary events, so R is true.
A card is drawn from a well shuffled deck of 52 playing cards. The probability that drawn card is a red queen, is :
(A) 13 1 (B) 13 2 (C) 52 1 (D) 26 1
Show answer & solution
Answer: (D) 26 1
There are 2 red queens (hearts and diamonds). P(red queen) = 52 2 = 26 1 .
Three coins are tossed together. The probability of getting exactly one tail, is :
(A) 8 1 (B) 4 1 (C) 8 7 (D) 8 3
Show answer & solution
Answer: (D) 8 3
There are 2 3 = 8 equally likely outcomes. Exactly one tail: HHT, HTH, THH, i.e. 3 outcomes. P = 8 3 .
On a throw of a die, if getting 6 is considered success then probability of losing the game is :
(A) 0(B) 1(C) 6 1 (D) 6 5
Show answer & solution
Answer: (D) 6 5
P(getting 6) = 6 1 . P(losing) = 1 − 6 1 = 6 5 .
If probability of winning a game is p , then probability of losing the game is :
(A) 1 + p (B) − p (C) p − 1 (D) 1 − p
Show answer & solution
Answer: (D) 1 − p
Winning and losing are complementary events. P(losing) = 1 − p .
Two dice are rolled together. The probability of getting a doublet is :
(A) 36 2 (B) 36 1 (C) 6 1 (D) 6 5
Show answer & solution
Answer: (C) 6 1
There are 36 equally likely outcomes. Doublets: (1, 1), (2, 2), ..., (6, 6), i.e. 6 outcomes. P = 36 6 = 6 1 .
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
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Answer: (C) 0 ≤ P ( A ) ≤ 1
Probability of any event lies between 0 and 1, both inclusive. So 0 ≤ P ( A ) ≤ 1 .
A die is thrown once. The probability of getting a number less than 6, is :
(A) 0(B) 6 5 (C) 6 1 (D) 1
Show answer & solution
Answer: (B) 6 5
Numbers less than 6: 1, 2, 3, 4, 5, i.e. 5 favourable outcomes out of 6. P(number less than 6) = 6 5 .
Two fair coins are tossed together. The probability of getting 2 heads, is :
(A) 2 1 (B) 4 3 (C) 4 1 (D) 8 3
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Answer: (C) 4 1
Outcomes: HH, HT, TH, TT (4 equally likely). Only HH gives 2 heads, so the probability is 4 1 .
Which of the following relationship is correct ?
(A) P ( E ) = 1 + P ( E ) (B) P ( E ) − P ( E ) = 1 (C) P ( E ) + P ( E ) = 1 (D) P ( E ) = 2 P ( E )
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Answer: (C) P ( E ) + P ( E ) = 1
An event and its complement together cover all outcomes and cannot happen together. So P ( E ) + P ( E ) = 1 .
Assertion (A): The probability of getting number 8 on rolling a die is zero (0). Reason (R): The probability of an impossible event is zero (0).
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
A die shows only 1 to 6, so getting 8 is an impossible event. The probability of an impossible event is 0, so both A and R are true and R explains A. Answer (A).
If the probability of an event is ‘p ’, what is the probability of its complementary event ?
(A) 1 − p (B) p − 1 (C) p (D) p 1
Show answer & solution
Answer: (A) 1 − p
P ( E ) + P ( E ) = 1 , so P ( E ) = 1 − p .
A fair die is thrown once. The probability of getting a composite number less than 5 is :
(A) 3 1 (B) 3 2 (C) 0(D) 6 1
Show answer & solution
Answer: (D) 6 1
Outcomes: 1, 2, 3, 4, 5, 6 (6 outcomes). Composite numbers less than 5: only 4. P = 6 1 .
A die is rolled once. What is the probability of getting an odd prime number ?
(A) 6 1 (B) 3 1 (C) 3 2 (D) 4 3
Show answer & solution
Answer: (B) 3 1
Outcomes: 1, 2, 3, 4, 5, 6 (6 outcomes). Odd primes: 3 and 5 (2 outcomes). P(odd prime) = 6 2 = 3 1 .
A bag has 4 red balls and 2 yellow balls. A ball is drawn at random from the bag without looking into the bag. What is the probability of getting a yellow ball ?
(A) 6 1 (B) 3 2 (C) 3 1 (D) 1
Show answer & solution
Answer: (C) 3 1
Total balls = 4 + 2 = 6 ; yellow balls = 2 . P(yellow) = 6 2 = 3 1 .
A letter is chosen at random from the word 'MATHEMATICS'. What is the probability that it will be a 'vowel' ?
(A) 11 3 (B) 11 4 (C) 11 5 (D) 11 7
Show answer & solution
Answer: (B) 11 4
MATHEMATICS has 11 letters. Vowels: A, E, A, I, i.e. 4 letters. P(vowel) = 11 4 .
If the probability of a player winning a game is 0.79, then the probability of his losing the same game is :
(A) 1.79(B) 0.31(C) 0.21%(D) 0.21
Show answer & solution
Answer: (D) 0.21
Winning and losing are complementary events. P(losing) = 1 − 0.79 = 0.21 .
From the data 1, 4, 7, 9, 16, 21, 25, if all the even numbers are removed, then the probability of getting at random a prime number from the remaining is :
(A) 5 2 (B) 5 1 (C) 7 1 (D) 7 2
Show answer & solution
Answer: (B) 5 1
Removing the even numbers 4 and 16 leaves 1, 7, 9, 21, 25 (5 numbers). Only 7 is prime. Required probability = 5 1 .
Two dice are rolled together. The probability of getting sum of numbers on the two dice as 2, 3 or 5, is :
(A) 36 7 (B) 36 11 (C) 36 5 (D) 9 4
Show answer & solution
Answer: (A) 36 7
Total outcomes = 36 . Sum 2: (1, 1) — 1 outcome. Sum 3: (1, 2), (2, 1) — 2 outcomes. Sum 5: (1, 4), (2, 3), (3, 2), (4, 1) — 4 outcomes. Favourable = 7 , so probability = 36 7 .
Which of the following is not probability of an event ?
(A) 0.89(B) 52%(C) 13 1 % (D) 0.89 1
Show answer & solution
Answer: (D) 0.89 1
A probability must lie between 0 and 1. 0.89, 52% = 0.52 and 13 1 % all lie between 0 and 1. 0.89 1 > 1 , so it cannot be a probability.
One card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is a red ace card, is :
(A) 13 1 (B) 26 1 (C) 52 1 (D) 2 1
Show answer & solution
Answer: (B) 26 1
There are 2 red aces (hearts and diamonds) in 52 cards. Probability = 52 2 = 26 1 .
Two dice are rolled together. The probability of getting the sum of the two numbers to be more than 10, is
(A) 9 1 (B) 6 1 (C) 12 7 (D) 12 1
Show answer & solution
Answer: (D) 12 1
Total outcomes = 36. Sum more than 10 (11 or 12): (5, 6), (6, 5), (6, 6), i.e. 3 outcomes. P = 36 3 = 12 1
A box contains cards numbered 6 to 55. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square, is
(A) 50 7 (B) 55 7 (C) 10 1 (D) 49 5
Show answer & solution
Answer: (C) 10 1
Number of cards = 55 − 6 + 1 = 50. Perfect squares from 6 to 55: 9, 16, 25, 36, 49, i.e. 5 numbers. P = 50 5 = 10 1
Two dice are thrown together. The probability that they show different numbers is :
(A) 1/6 (B) 5/6 (C) 1/3 (D) 2/3
Show answer & solution
Answer: (B) 5/6
Total outcomes = 36. Outcomes with the same number on both dice (doublets) = 6. Outcomes with different numbers = 36 - 6 = 30. P(different numbers) = 36 30 = 6 5 .
The probability of guessing the correct answer to a certain test question is 6 x . If the probability of not guessing the correct answer to this question is 3 2 , then the value of x is :
(A) 2(B) 3(C) 4(D) 6
Show answer & solution
Answer: (A) 2
P(correct) + P(not correct) = 1. 6 x + 3 2 = 1 ⇒ 6 x = 3 1 .x = 2 .
If a digit is chosen at random from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9; then the probability that this digit is an odd prime number is :
(A) 3 1 (B) 3 2 (C) 9 4 (D) 9 5
Show answer & solution
Answer: (A) 3 1
Total outcomes = 9. Odd primes among them: 3, 5, 7, i.e. 3 outcomes. P(odd prime) = 9 3 = 3 1 .
Assertion (A) : In a cricket match, a batsman hits a boundary 9 times out of 45 balls he plays. The probability that in a given ball, he does not hit the boundary is 5 4 . Reason (R) : P(E) + P(not E) = 1
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
P(boundary) = 45 9 = 5 1 . P(no boundary) = 1 − 5 1 = 5 4 , so A is true. R is true, and it is exactly the rule used to get A, so R explains A. Answer: (A).
Two coins are tossed simultaneously. The probability of getting at most one tail is :
(A) 2 1 (B) 4 1 (C) 4 3 (D) 1
Show answer & solution
Answer: (C) 4 3
Outcomes: HH, HT, TH, TT (4 outcomes). At most one tail: HH, HT, TH (3 outcomes). P = 4 3 .
A bag contains 3 red balls, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is :
(A) 3 1 (B) 5 1 (C) 15 7 (D) 15 8
Show answer & solution
Answer: (A) 3 1
Total balls = 3 + 5 + 7 = 15. Neither red nor black means white: 5 balls. P = 15 5 = 3 1 .
The probability of getting a bad egg in a lot of 400 eggs is 0.045. The number of good eggs in the lot is :
(A) 18(B) 180(C) 382(D) 220
Show answer & solution
Answer: (C) 382
Number of bad eggs = 0.045 × 400 = 18 . Number of good eggs = 400 - 18 = 382.
Two dice are tossed simultaneously. The probability of getting odd numbers on both the dice is :
(A) 36 6 (B) 36 3 (C) 36 12 (D) 36 9
Show answer & solution
Answer: (D) 36 9
Total outcomes = 6 × 6 = 36 . Each die shows 1, 3 or 5: favourable outcomes = 3 × 3 = 9 . P(odd on both) = 36 9 .
For an event E, if P ( E ) + P ( E ) = q , then the value of q 2 − 4 is :
(A) − 3 (B) 3(C) 5(D) − 5
Show answer & solution
Answer: (A) − 3
P ( E ) + P ( E ) = 1 , so q = 1 .q 2 − 4 = 1 − 4 = − 3 .
From the letters of the word “MOBILE”, a letter is selected at random. The probability that the selected letter is a vowel, is :
(A) 7 3 (B) 6 1 (C) 2 1 (D) 3 1
Show answer & solution
Answer: (C) 2 1
MOBILE has 6 letters. Vowels: O, I, E (3 letters). P(vowel) = 6 3 = 2 1 .
The probability of throwing a number greater than 2 with a fair die is :
(A) 3 2 (B) 3 1 (C) 2 1 (D) 6 5
Show answer & solution
Answer: (A) 3 2
Numbers greater than 2: 3, 4, 5, 6 (4 outcomes out of 6). P = 6 4 = 3 2 .
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 :
(A) 7 1 (B) 8 1 (C) 5 1 (D) 40 7
Show answer & solution
Answer: (B) 8 1
Multiples of 7 from 1 to 40: 7, 14, 21, 28, 35, i.e. 5 numbers. P(multiple of 7) = 40 5 = 8 1 .
What is the probability that a number selected randomly from the numbers 1, 2, 3, ..., 15 is a multiple of 4 ?
(A) 15 4 (B) 15 6 (C) 15 3 (D) 15 5
Show answer & solution
Answer: (C) 15 3
Multiples of 4 from 1 to 15: 4, 8, 12, i.e. 3 numbers. P(multiple of 4) = 15 3 .
A card is drawn at random from a well-shuffled deck of 52 cards. The probability of getting a red card is :
(A) 26 1 (B) 13 1 (C) 4 1 (D) 2 1
Show answer & solution
Answer: (D) 2 1
A deck has 26 red cards (13 hearts and 13 diamonds). P(red card) = 52 26 = 2 1
A die is thrown once. Find the probability of getting a number less than 7.
(A) 6 5 (B) 1(C) 6 1 (D) 0
Show answer & solution
Answer: (B) 1
All six outcomes 1, 2, 3, 4, 5, 6 are less than 7. P(number less than 7) = 6 6 = 1 (a sure event).
A card is drawn at random from a well-shuffled deck of 52 playing cards. The probability of getting an ace of spade is :
(A) 13 1 (B) 52 3 (C) 26 1 (D) 52 1
Show answer & solution
Answer: (D) 52 1
There is only one ace of spades in a deck of 52 cards. P(ace of spades) = 52 1
A die is thrown once. The probability of getting an odd prime number is
(A) 2 1 (B) 6 1 (C) 3 1 (D) 3 2
Show answer & solution
Answer: (C) 3 1
Possible outcomes: 1, 2, 3, 4, 5, 6. Odd prime numbers among them: 3 and 5 (2 outcomes). P(odd prime) = 6 2 = 3 1
A die is rolled once. The probability that a composite number comes up, is :
(A) 2 1 (B) 3 2 (C) 3 1 (D) 0
Show answer & solution
Answer: (c) 3 1
Composite numbers on a die: 4, 6, so 2 favourable outcomes out of 6. P(composite) = 6 2 = 3 1
Which of the following numbers cannot be the probability of an event ?
(A) 0.5(B) 5%(C) 0.5 1 (D) 14 0.5
Show answer & solution
Answer: (c) 0.5 1
A probability must lie between 0 and 1. 0.5, 5% = 0.05 and 14 0.5 all lie in this range. 0.5 1 = 2 > 1 , so it cannot be a probability.
Assertion (A) : The probability of getting a prime number, when a die is thrown once, is 3 2 . Reason (R): On the faces of a die, prime numbers are 2, 3, 5.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (d) Assertion (A) is false, but Reason (R) is true.
Prime numbers on a die are 2, 3, 5, so R is true. P(prime) = 6 3 = 2 1 = 3 2 , so A is false. Hence (d).
A card is drawn at random from a well-shuffled deck of 52 playing cards. The probability that it is a red king, is :
(A) 13 1 (B) 52 1 (C) 26 1 (D) 13 2
Show answer & solution
Answer: (c) 26 1
Red kings: king of hearts and king of diamonds, so 2 favourable outcomes. P(red king) = 52 2 = 26 1
Two coins are tossed together. The probability of getting atmost two heads, is :
(A) 2 1 (B) 4 1 (C) 4 3 (D) 1
Show answer & solution
Answer: (d) 1
Outcomes: HH, HT, TH, TT (4 outcomes). Every outcome has at most two heads, so all 4 are favourable. P(at most two heads) = 4 4 = 1
Which of the following cannot be the probability of an event ?
(A) 0.1(B) 3 5 (C) 3%(D) 3 1
Show answer & solution
Answer: (B) 3 5
A probability must lie between 0 and 1. 3 5 > 1 , so it cannot be a probability.
From a well-shuffled deck of 52 playing cards, a card is drawn at random. What is the probability of getting a red queen ?
(A) 52 1 (B) 26 1 (C) 13 1 (D) 13 12
Show answer & solution
Answer: (B) 26 1
There are 2 red queens (hearts and diamonds). P(red queen) = 52 2 = 26 1 .
Let E be an event such that P(not E) = 5 1 , then P(E) is equal to :
(A) 5 1 (B) 5 2 (C) 0(D) 5 4
Show answer & solution
Answer: (D) 5 4
P(E) + P(not E) = 1. P(E) = 1 − 5 1 = 5 4 .
From a well-shuffled deck of 52 cards, a card is drawn at random. What is the probability of getting king of hearts ?
(A) 52 1 (B) 26 1 (C) 13 1 (D) 13 12
Show answer & solution
Answer: (A) 52 1
There is exactly one king of hearts in 52 cards. P = 52 1 .
One card is drawn at random from a well shuffled deck of 52 playing cards. What is the probability of getting ‘4 of hearts’ ?
(A) 52 1 (B) 13 1 (C) 26 1 (D) 6 1
Show answer & solution
Answer: (A) 52 1
There is only one 4 of hearts in a deck of 52 cards. P(4 of hearts) = 52 1 .
Assertion (A) : When two coins are tossed together, the probability of getting no tail is 4 1 . Reason (R) : The probability P(E) of an event E satisfies 0 ≤ P(E) ≤ 1 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Outcomes: HH, HT, TH, TT. No tail means HH only. P(no tail) = 4 1 , so A is true. R is a true general fact, but it does not explain why the probability is 4 1 .
Two dice are rolled together. The probability that the sum of the numbers that appeared is 9, is :
(A) 36 5 (B) 9 1 (C) 12 1 (D) 6 1
Show answer & solution
Answer: (B) 9 1
Total outcomes = 36. Sum 9: (3, 6), (4, 5), (5, 4), (6, 3), i.e. 4 outcomes. P(sum 9) = 36 4 = 9 1 .
One card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is a red king is :
(A) 52 1 (B) 26 1 (C) 26 2 (D) 13 2
Show answer & solution
Answer: (B) 26 1
There are 2 red kings (king of hearts and king of diamonds). P(red king) = 52 2 = 26 1 .
One card is drawn at random from a well-shuffled deck of 52 playing cards. What is the probability of getting a black king ?
(A) 26 1 (B) 13 1 (C) 52 1 (D) 2 1
Show answer & solution
Answer: (A) 26 1
There are 2 black kings (spades and clubs) in 52 cards. P(black king) = 52 2 = 26 1 .
An unbiased die is thrown. The probability of getting an odd prime number is
(A) 6 1 (B) 2 1 (C) 3 2 (D) 3 1
Show answer & solution
Answer: (D) 3 1
Outcomes: 1, 2, 3, 4, 5, 6. Odd primes: 3 and 5. P(odd prime) = 6 2 = 3 1 .
Assertion (A) : The probability that a leap year has 53 Sundays is 7 2 . Reason (R) : The probability that a non-leap year has 53 Sundays is 7 1 .
(A) Both Assertion (A) and Reason (R) are true; and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true, but Reason (R) is false.(D) Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (B) Both A and R are true, but R is not the correct explanation of A.
A leap year has 366 days = 52 weeks + 2 days. The 2 extra days can be (Sun, Mon), (Mon, Tue), ..., (Sat, Sun): 7 cases, 2 contain a Sunday. P = 7 2 , so A is true. A non-leap year has 365 days = 52 weeks + 1 day; the extra day is Sunday in 1 of 7 cases. P = 7 1 , so R is true. R is about a non-leap year, so it does not explain A.
Probability of happening of an event is denoted by p and probability of non-happening of the event is denoted by q. Relation between p and q is
(A) p + q = 1(B) p = 1, q = 1(C) p = q – 1(D) p + q + 1 = 0
Show answer & solution
Answer: (A) p + q = 1
P(E) + P(not E) = 1. So p + q = 1.
A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought ?
(A) 40(B) 240(C) 480(D) 750
Show answer & solution
Answer: (C) 480
P(winning) = number of her tickets ÷ total tickets. Number of her tickets = 0.08 × 6000 = 480
In a group of 20 people, 5 can't swim. If one person is selected at random, then the probability that he/she can swim, is
(A) 4 3 (B) 3 1 (C) 1(D) 4 1
Show answer & solution
Answer: (A) 4 3
People who can swim = 20 – 5 = 15. P(can swim) = 20 15 = 4 3
In a survey, it is found that every fifth person has a vehicle. The probability of a person NOT having a vehicle, is
(A) 5 1 (B) 5%(C) 5 4 (D) 95%
Show answer & solution
Answer: (C) 5 4
P(having a vehicle) = 5 1 . P(not having a vehicle) = 1 − 5 1 = 5 4
A bag contains 100 cards numbered 1 to 100. A card is drawn at random from the bag. What is the probability that the number on the card is a perfect cube ?
(A) 20 1 (B) 50 3 (C) 25 1 (D) 100 7
Show answer & solution
Answer: (C) 25 1
Perfect cubes from 1 to 100: 1, 8, 27, 64, i.e. 4 numbers. P(perfect cube) = 100 4 = 25 1
If three coins are tossed simultaneously, what is the probability of getting at most one tail ?
(A) 8 3 (B) 8 4 (C) 8 5 (D) 8 7
Show answer & solution
Answer: (B) 8 4
There are 8 equally likely outcomes. At most one tail: HHH, HHT, HTH, THH, i.e. 4 outcomes. Probability = 8 4
In a single throw of two dice, the probability of getting 12 as a product of two numbers obtained is :
(A) 9 1 (B) 9 2 (C) 9 4 (D) 9 5
Show answer & solution
Answer: (A) 9 1
Total outcomes = 36. Product 12: (2, 6), (6, 2), (3, 4), (4, 3), i.e. 4 outcomes. Probability = 36 4 = 9 1
In a lottery, there are 5 prizes and 20 blanks. The probability of getting a prize is :
(A) 4 1 (B) 20 1 (C) 25 1 (D) 5 1
Show answer & solution
Answer: (D) 5 1
Total tickets = 5 + 20 = 25 . P(prize) = 25 5 = 5 1
Two dice are thrown together. The probability of getting the difference of numbers on their upper faces equals to 3 is :
(A) 9 1 (B) 9 2 (C) 6 1 (D) 12 1
Show answer & solution
Answer: (C) 6 1
Total outcomes = 36. Difference 3: (1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3), i.e. 6 outcomes. P = 36 6 = 6 1
A card is drawn at random from a well-shuffled pack of 52 cards. The probability that the card drawn is not an ace is :
(A) 13 1 (B) 13 9 (C) 13 4 (D) 13 12
Show answer & solution
Answer: (D) 13 12
There are 4 aces, so 48 cards are not aces. P(not an ace) = 52 48 = 13 12
Assertion (A) : The probability that a leap year has 53 Sundays is 7 2 . Reason (R) : The probability that a non-leap year has 53 Sundays is 7 5 .
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).(C) Assertion (A) is true but Reason (R) is false.(D) Assertion (A) is false but Reason (R) is true.
Show answer & solution
Answer: (C) Assertion (A) is true but Reason (R) is false.
A leap year has 366 days = 52 weeks + 2 days. The 2 extra days can be (Sun, Mon), (Mon, Tue), ..., (Sat, Sun): 7 cases, 2 of which contain a Sunday. P = 7 2 , so A is true. A non-leap year has 365 days = 52 weeks + 1 day; the extra day is a Sunday in 1 of 7 cases. P = 7 1 , so R is false.
A bag contains 5 red balls and n green balls. If the probability of drawing a green ball is three times that of a red ball, then the value of n is :
(A) 18(B) 15(C) 10(D) 20
Show answer & solution
Answer: (B) 15
P(green) = n + 5 n , P(red) = n + 5 5 n + 5 n = 3 × n + 5 5 ⇒ n = 15
Two coins are tossed together. The probability of getting at least one tail is :
(A) 4 1 (B) 2 1 (C) 4 3 (D) 1
Show answer & solution
Answer: (C) 4 3
Outcomes: HH, HT, TH, TT (4). At least one tail: HT, TH, TT (3). P = 4 3 .
Which of the following numbers cannot be the probability of happening of an event ?
(A) 0(B) 0.01 7 (C) 0.07(D) 3 0.07
Show answer & solution
Answer: (B) 0.01 7
Probability lies between 0 and 1. 0.01 7 = 700 > 1 , so it cannot be a probability.
Two dice are rolled together. What is the probability of getting a sum greater than 10 ?
(A) 9 1 (B) 6 1 (C) 12 1 (D) 18 5
Show answer & solution
Answer: (C) 12 1
Total outcomes = 36. Sum greater than 10: (5, 6), (6, 5), (6, 6), i.e. 3 outcomes. P = 36 3 = 12 1 .
One card is drawn at random from a well shuffled pack of 52 playing cards. The probability that the drawn card is a queen, is :
(A) 13 4 (B) 52 4 (C) 13 2 (D) 26 1
Show answer & solution
Answer: (B) 52 4
A pack has 4 queens out of 52 cards. P(queen) = 52 4 ( = 13 1 ) .
A bag contains 5 pink, 8 blue and 7 yellow balls. One ball is drawn at random from the bag. What is the probability of getting neither a blue nor a pink ball ?
(A) 4 1 (B) 5 2 (C) 20 7 (D) 20 13
Show answer & solution
Answer: (C) 20 7
Total balls = 5 + 8 + 7 = 20. Neither blue nor pink means yellow: 7 balls. P = 20 7
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability of getting a face card is
(A) 2 1 (B) 13 3 (C) 13 4 (D) 13 1
Show answer & solution
Answer: (B) 13 3
There are 12 face cards (J, Q, K of 4 suits). P = 52 12 = 13 3
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 a prime number less than 23 is
(A) 90 7 (B) 9 1 (C) 45 4 (D) 89 9
Show answer & solution
Answer: (C) 45 4
Primes less than 23: 2, 3, 5, 7, 11, 13, 17, 19, i.e. 8 numbers. P = 90 8 = 45 4
Cards bearing numbers 3 to 20 are placed in a bag and mixed thoroughly. A card is taken out of the bag at random. What is the probability that the number on the card taken out is an even number ?
(A) 17 9 (B) 2 1 (C) 9 5 (D) 18 7
Show answer & solution
Answer: (B) 2 1
Numbers 3 to 20: 18 cards. Even numbers: 4, 6, ..., 20, i.e. 9 cards. P = 18 9 = 2 1
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