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Let A, B and C be three events associated with sample space S. A, B and C are pair wise independent and $P (A) = P (B) = P (C) = p$ . If all of them can not occur simultaneously. Then $P (A \cup B \cup C)$

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$1 - {(1 - p)}^{3}$

$3 p (1 - p)$

$p^{3}$

answer is B.

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Detailed Solution

$P (A \cap B \cap C) = 0 P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (C \cap A) = P + P + P - P^{2} - P^{2} - P^{2} = 3 P - 3 P^{2} = 3 P (1 - P)$

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