Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 - k), the probability that exactly one of B and C occurs is (1 - 2k), the probability that exactly one of C and A occurs is (1 - k) and the probability of all A, B and C occur simultaneously is k², where 0 < k < 1. Then the probability that at least one of A, B and C occur is :

Solution

Maths sir · 3 steps

Add the three 'exactly one' probabilities, and the inclusion-exclusion pieces fall out ready to use.

1. Exactly one formula. The probability that exactly one of A and B occurs is P(A) + P(B) - 2P(A ∩ B). Write the same for the pairs B, C and C, A.
2. Add all three. Each single probability appears twice and each pair term appears with -2. So 2[P(A)+P(B)+P(C)] - 2[P(A∩ B)+P(B∩ C)+P(C∩ A)] = (1-k)+(1-2k)+(1-k) = 3-4k. Halving gives Σ P(A) - Σ P(A∩ B) = 3/2 - 2k.

   P(A∪ B∪ C) = Σ P(A) - Σ P(A∩ B) + P(A∩ B∩ C) = 3/2 - 2k + k² = (k-1)² + 1/2

3. So. Since 0 < k < 1, k ≠ 1, so (k-1)² > 0. Hence P(A∪ B∪ C) > 1/2, which is option (b).

The idea to keep. When a question gives 'exactly one of each pair', add all of them: you get twice the single sum minus twice the pair sum, which plugs straight into inclusion-exclusion.

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