Let A, B and C be three events such that the probability that exactly one of A and B occur…

Question

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 :

Accepted Answer

$$P(\overline A  \cap B) + P(A \cap \overline B ) = 1 - k$$

$$P(\overline A  \cap C) + P(A \cap \overline C ) = 1 - 2k$$

$$P(\overline B  \cap C) + P(B \cap \overline C ) = 1 - k$$

$$P(A \cap B \cap C) = {k^2}$$

$$P(A) + P(B) - 2P(A \cap B) = 1 - k$$ .....(i)

$$P(B) + P(C) - 2P(B \cap C) = 1 - k$$ ..... (ii)

$$P(C) + P(A) - 2P(A \cap C) = 1 - 2k$$ ..... (iii)

$(i) + (ii) + (iii)$

$$P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) = {{ - 4k + 3} \over 2}$$

So,

$$P(A \cup B \cup C) = {{ - 4k + 3} \over 2} + {k^2}$$

$$P(A \cup B \cup C) = {{2{k^2} - 4k + 3} \over 2}$$

$$ = {{2{{(k - 1)}^2} + 1} \over 2}$$

$$P(A \cup B \cup C) > {1 \over 2}$$

Solutions · 1