RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
This chapter builds on basic probability by asking how new information updates the odds of an event — the idea of conditional probability. From there, the multiplication rule and the total probability theorem let you compute the probability of compound events built from partitioned scenarios, and Bayes' theorem reverses the flow, computing the probability of a cause given an observed effect. The chapter closes with random variables, their probability distributions, and the binomial distribution for repeated independent trials.
If P(A)=0.4, P(B)=0.3 and P(A∩B)=0.12, then events A and B are:
Answer
A and B are independent. Events are independent exactly when P(A∩B) = P(A)×P(B). Here P(A)×P(B) = 0.4×0.3 = 0.12, which exactly matches the given P(A∩B) = 0.12, confirming independence.
A card is drawn from a well-shuffled deck of 52 cards. The probability that it is a king, given that it is a face card, is:
Answer
The probability is 1/3. There are 12 face cards in total (Jack, Queen, King in each of 4 suits), and 4 of them are kings. P(king | face card) = (number of kings among face cards)/(total face cards) = 4/12 = 1/3.
Two fair dice are thrown together. The probability of getting a sum of 7 is:
Answer
The probability is 1/6. There are 36 equally likely outcomes in total. The pairs summing to 7 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) — 6 outcomes. P(sum=7) = 6/36 = 1/6.
For a binomial distribution with n=10 and p=0.4, the mean is:
Answer
The mean is 4. For a binomial distribution, mean = np. Here mean = 10 × 0.4 = 4.
Assertion (A): If two events A and B are mutually exclusive with P(A) > 0 and P(B) > 0, then A and B cannot be independent. Reason (R): For mutually exclusive events, P(A∩B) = 0, but independence requires P(A∩B) = P(A)·P(B), which is nonzero whenever both P(A) > 0 and P(B) > 0.
Answer
Both A and R are true and R is the correct explanation of A. Reason R correctly notes that mutual exclusivity forces P(A∩B)=0, while independence would require P(A∩B)=P(A)P(B), a strictly positive number when both P(A) and P(B) are positive. Since 0 cannot equal a positive number, the two conditions are incompatible, proving Assertion A exactly as R explains.
A bag contains 5 red and 3 black balls. Two balls are drawn one after another without replacement. Find the probability that both are red.
Answer
P(first ball red) = 5/8. After removing one red ball, 4 red balls remain out of 7 total, so P(second ball red | first red) = 4/7. P(both red) = (5/8)×(4/7) = 20/56 = 5/14.
If P(A)=0.5, P(B)=0.6 and P(A∩B)=0.3, find P(A|B).
Answer
By the definition of conditional probability, P(A|B) = P(A∩B)/P(B). Substituting the given values: P(A|B) = 0.3/0.6 = 0.5.
A random variable X has the probability distribution P(X=0)=0.2, P(X=1)=0.5, P(X=2)=0.3. Find the mean E(X).
Answer
First check the probabilities sum to 1: 0.2+0.5+0.3 = 1.0, confirmed valid. The mean is E(X) = Σxi·P(xi) = 0(0.2) + 1(0.5) + 2(0.3). = 0 + 0.5 + 0.6 = 1.1.
In a factory, machines A, B and C produce 25%, 35% and 40% of the total output respectively. Of their outputs, 5%, 4% and 2% are defective. An item is selected at random and found defective. Find the probability that it was produced by machine A, using Bayes' theorem.
Answer
Let A, B, C denote the events that the item came from each machine, with P(A)=0.25, P(B)=0.35, P(C)=0.40. Let D denote the event that the item is defective, with P(D|A)=0.05, P(D|B)=0.04, P(D|C)=0.02. By the Theorem of Total Probability: P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) = 0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.014 + 0.008 = 0.0345. By Bayes' theorem: P(A|D) = P(A)P(D|A)/P(D) = 0.0125/0.0345 = 125/345 = 25/69 ≈ 0.362. So the probability the defective item came from machine A is 25/69.
A fair coin is tossed 5 times. Using the binomial distribution, find the probability of getting exactly 3 heads.
Answer
This is a binomial setting with n=5 trials, success probability p=1/2 (getting a head), and failure probability q=1/2. P(X=r) = C(n,r) pʳ qⁿ⁻ʳ. For r=3: P(X=3) = C(5,3) (1/2)³ (1/2)². C(5,3) = 5!/(3!2!) = 10. P(X=3) = 10 × (1/8) × (1/4) = 10/32 = 5/16. So the probability of exactly 3 heads in 5 tosses is 5/16.
Read the following and answer the questions that follow: In a group of students, 60% study Mathematics (M) and 40% study Biology (B), with 20% studying both subjects. (a) Find P(M∪B), the probability a randomly chosen student studies at least one of the two subjects. (b) Find P(M|B), the probability a student studies Mathematics given they study Biology. (c) Find P(B|M), the probability a student studies Biology given they study Mathematics. (d) Are the events M and B independent? Justify your answer.
Answer
Given P(M)=0.6, P(B)=0.4, P(M∩B)=0.2. (a) P(M∪B) = P(M) + P(B) - P(M∩B) = 0.6 + 0.4 - 0.2 = 0.8. (b) P(M|B) = P(M∩B)/P(B) = 0.2/0.4 = 0.5. (c) P(B|M) = P(M∩B)/P(M) = 0.2/0.6 = 1/3 ≈ 0.333. (d) M and B are not independent, because P(M)×P(B) = 0.6×0.4 = 0.24, which does not equal the given P(M∩B) = 0.2. Since independence requires these to be equal, they are not independent.
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