NCERT MCQ’s : Class 12 Maths with Answers MCQ Chapter 13 Probability

NCERT Solutions for Class 12 Maths Chapter 13 Probability MCQ’s with Answers

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NCERT Maths MCQs for Class 12 Question and Answers Chapter 13 Probability PDF

Q1. You are given that A and B are two events such that P(b) = 3/5 , P(A|B) = 4/5, then P(A) equals

(a)3/10
(b)1/5
(c)1/2
(d)3/5

Option c – 1/2

Q2. A and B are events such that P(A) = 0.4, P(b) = 0.3 and P(A ∪ B) = 0.5. Then P(B’ ∩ A) equals

(a)3/6
(b)2/6
(c)1/10
(d) 1/5

Option d – 1/5

Q3. If P(A) = 0.4, P(b) = 0.8 and P(B|A) = 0.6, then P(A ∪ B) equal to

(a) 0.24
(b) 0.3
(c) 0.48
(d) 0.96

Option c – 0.48

Q4. If A and B are two events sue that P(A) =1/2 , P(b) =1/3 , P(A|B) = 1/4 then (A’ ∩ B’) equals

(a)1/2
(b) 3/4
(c) 1/4
(d) 3/12

Option c – 1/4

Q5. If P(A) = 2/5, P(B) = 3/10 and P(A ∩ B) = 1/5 , then P(A’|B’) . (P(B’|A’) is equal to

(a)5/6
(b)5/7
(c)6/7
(d) 1

Option b – 5/7

Q6. If P(A) = 3/10 , P(b) = 2/5 and P(A ∪ B) = 3/5, then P(B|A) + P(A|B) equals

(a)5/7
(b)2/7
(c)7/2
(d) 7/12

Option d – 7/12

Q7. If P(A ∩ B) = 5/10 and P(b) = ,17/20 P(A|B) equals

(a)14/17
(b)17/20
(c)12/7
(d)5/7

Option a – 14/17

Q8.The mean and the variance of a binomial distribution are 4 and 2 respectively. Find the probability of atleast 6 successes.

(a)37/256
(b)35/256
(c)32/256
(d)38/256

Option a – 37/256

Q9. In a binomial distribution, the sum of its mean and variance is 1.8. Find the probability of two successes, if the event was conducted times.

(a) 0.2623
(b) 0.2048
(c) 0.302
(d) 0.305

Option b – 0.2048

Q10. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both clubs. Find the probability of the lost card being a club.

(a)11/50
(b)17/50
(c)13/50
(d)22/50

Option a -11/50

Q11. A bag contains 3 green and 7 white balls. Two balls are drawn one by one at random without replacement. If the second ball drawn is green, what is the probability that the first ball was drawn in also green?

(a)5/9
(b)9/2
(c)2/9
(d)8/9

Option c – 2/9

Q12. A bag contains 4 balls. Two balls are drawn at random and are found to be white. What is the probability that all balls are white?

(a)2/5
(b)3/5
(c)4/5
(d)1/5

Option b – 3/5

Q13. If A and B are two independent events, then the probability of occurrence of at least of A and B is given by

(a) 1 – P(A) P(b)
(b) 1 – P(A) P(B’)
(c) 1 – P(A’) P(B’)
(d) 1 – P(A’) P(b)

Option c – 1 – P(A’) P(B’)

Q14. Two events A and B will be independent, if

(a) A and B are mutually exclusive
(b) P(A’ ∩ B’) = [1 – P(A)] [1 – P(B)]
(c) P(A) = P(B)
(d) P(A) + P(B) = 1

Option c – 1 – P(A) = P(B)

Q15. If three events of a sample space are E, F and G, then P(E ∩ F ∩ G) is equal to

(a) P(E) P(F|E) P(G|(E ∩ F))
(b) P(E) P(F|E) P(G|EF)
(c) Both (a) and (b)
(d) None of these

Option c – Both (a) and (b)

Q16. P(E ∩ F) is equal to

(a) P(E) . P(F|E)
(b) P(F) . P(E|F)
(c) Both (a) and (b)
(d) None of these

Option c – Both (a) and (b)

Q17. The probability of a man hitting a target is 1/4. How many times must he fire so that the probability of his hitting the target at least once is greater than 2/3 ?

(a) 4
(b) 3
(c) 2
(d) 1

Option a – 4

Q18. The variance of random variable X i.e. or var (X) is equal to

(a) E(X2) + [E(X2)2]2
(b) E(X) – [E(X2)]
(c) E(X2) – [E(X)]2
(d) None of these

Option c – E(X2) – [E(X)]2

Q19. Three balls are drawn from a bag containing 2 red and 5 black balls, if the random variable X represents the number of red balls drawn, then X can take values

(a) 0, 1, 2
(b) 0, 1, 2, 3
(c) 0
(d) 1, 2

Option a – 0,1, 2

Q20. Let A and B be two given independent events such that P(A) =p and P(B) = q and P(exactly one of A, B) = 2/3 , then value of 3p + 3q – 6pq is

(a) 2
(b) -2
(c) 4
(d) -4

Option a – 2

Q21. If A and B are independent events, then P(A and B) = P(A) + P(B). State true or false.

Option – False, as P(A and B) = P(A) P(B).

Q22. Given P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6. Find P(A ∪ B).

Option – 0.86

Q23. Prove that if E and F are independent events, then the events E and F are also independent.

Option – As E and F are independent events ∴ P(E ∩ F) = P(E)P(F) …(i) Consider, P(E)P(F) = P(E)[1 – P(F)] = P(E) – P(E)P(F) = P(E) – P(E ∩ F) P(E)P(F’) = P(E ∩ F’) ⇒ E and F’ are independent events.

Q24. A coin is biased so that the head is 3 times likely to occur as a tail. If the coin is tossed twice, then find the probability distribution of the number of tails.

Option – d

Q25. A pair of the die is thrown 4 times. If getting a doubled is considered a success, then find the probability distribution of a number of successes.

Option – a

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