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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 $(1 - 2 k),$ 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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Greater than $\frac{1}{2}$

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

Greater than $\frac{1}{8}$ but less than $\frac{1}{4}$

Greater than $\frac{1}{4}$ but less than $\frac{1}{2}$

answer is A.

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Detailed Solution

$P (\bar{A} \cap B) + P (A \cap \bar{B}) = 1 - k$

$P (\bar{A} \cap C) + P (A \cap \bar{C}) = 1 - k$

$P (\bar{B} \cap C) + P (B \cap \bar{C}) = 1 - 2 k$

$p (A \cap B \cap C) = K^{2}$

$p (A) + P (B) - 2 P (A \cap B) = 1 - K ........ (1)$

$p (B) + P (C) - 2 P (B \cap C) = 1 - 2 K ......... (2)$

$p (C) + P (A) - 2 P (A \cap C) = 1 - K ........ (3)$

$(1) + (2) + (3)$

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

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

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

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

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