Let E₁ and E₂ be two events such that the conditional probabilities P(E₁|E₂) = 1/2, P(E₂|E₁) = 3/4 and P(E₁ ∩ E₂) = 1/8. Then :

Solution

Maths sir · 2 steps

Get P(E₁) and P(E₂) first from the two conditionals, then test each option with actual numbers.

1. Find the singles. Use P(A ∩ B) = P(A|B) P(B) both ways. P(E₂) = (1/8)/(1/2) = 1/4 and P(E₁) = (1/8)/(3/4) = 1/6.

   P(E₁) P(E₂) = 1/6 × 1/4 = 1/24

   Option	Left side	Right side	Equal?
   (a)	P(E₁ ∩ E₂) = 1/8	1/24	No
   (b)	1 - P(E₁ ∪ E₂) = 1 - 7/24 = 17/24	5/6 × 1/4 = 5/24	No
   (c)	P(E₁) - P(E₁ ∩ E₂) = 1/6 - 1/8 = 1/24	1/24	Yes
   (d)	P(E₂) - P(E₁ ∩ E₂) = 1/4 - 1/8 = 1/8	1/24	No

2. So. Only P(E₁ ∩ E'₂) = 1/24 matches P(E₁) P(E₂), so the answer is (c).

The idea to keep. Two conditionals plus the intersection give you both single probabilities. Once you have those, every region of the Venn diagram is just subtraction, so check each option with numbers.

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