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

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exactly equal to $\frac{1}{2}$

$g r e a t e r t h a n \frac{1}{2}$

$g r e a t e r t h a n \frac{1}{4} b u t l e s s t h a n \frac{1}{2}$

$g r e a t e r t h a n \frac{1}{8} b u t l e s s t h a n \frac{1}{4}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} P (A) + P (B) - 2 P (A \cap B) & = & 1 - k - (1) \\ P (B) + P (C) - 2 P (B \cap C) & = & 1 - 2 k - (2) \\ P (C) + P (A) - 2 P (A \cap C) & = & 1 - k - (3) \\ P (A \cap B \cap C) & = & k^{2} \\ eq (1) + eq (2) + eq (3) \\ \Rightarrow & P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) = \frac{3 - 4 k}{2} \end{array}$

$\begin{array}{rcl} P (A \cup B \cup C) & = & \frac{3 - 4 k}{2} + k^{2} \\ = & k^{2} - 2 k + \frac{3}{2} \\ = & {(k - 1)}^{2} + \frac{3}{2} - 1 \\ = & {(k - 1)}^{2} + \frac{1}{2} \end{array}$

$0 < K < 1 \Rightarrow - 1 < K - 1 < 0 \Rightarrow 0 < {(K - 1)}^{2} < 1 \Rightarrow \frac{1}{2} < {(K - 1)}^{2} + \frac{1}{2} < \frac{3}{2}$

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