JEE Main 2022 · 28 July · evening · Single correct · +4 / −1

Let A and B be two events such that P(B | A)=2/5, P(A | B)=1/7 and P(A ∩ B)=1/9 · Consider
(S1) P(A′ ∪ B)=5/6,
(S2) P(A′ ∩ B′)=1/18
Then :

  1. (a) Both (S1) and (S2) are true
  2. (b) Both (S1) and (S2) are false
  3. (c) Only (S1) is true
  4. (d) Only (S2) is true
Answer and solution

Answer (a) Both (S1) and (S2) are true

Solution

Maths sir · 4 steps

Use each conditional to recover P(A) and P(B), then turn both statements into complements.

  1. Find P(A) and P(B). Since P(B | A) = P(A ∩ B)/P(A), we get P(A) = (1/9)/(2/5) = 5/18. In the same way, P(B) = (1/9)/(1/7) = 7/9 = 14/18.
  2. Check S2. By De Morgan, A' ∩ B' is the complement of A ∪ B. Here P(A ∪ B) = 5/18 + 14/18 - 2/18 = 17/18, so P(A' ∩ B') = 1 - 17/18 = 1/18. S2 is true.

    P(A' ∪ B) = 1 - P(A ∩ B') = 1 - [P(A) - P(A ∩ B)]

  3. Check S1. The complement of A' ∪ B is A ∩ B', which is the part of A outside B. So P(A ∩ B') = 5/18 - 2/18 = 3/18, and P(A' ∪ B) = 1 - 3/18 = 15/18 = 5/6. S1 is true.
  4. So. Both statements hold, so the answer is (a).

The idea to keep. Given P(A ∩ B) and a conditional, divide to get the single-event probability. For any union or intersection with complements, use De Morgan and find its complement first.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Probability — Advanced

Asked about 100 times in JEE Main since 2020. This question is from 2022.