Let A-B C be 3 arbitary events defined on a sample space S and if- PA - PB - PC -p1- PA - B - P B - C - P C - A- p2 PA - B - C- p3- then the probability that exactly one of the three events occurs is given by-

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Let A,B C b...

Question

Let $A, B$ & $C$ be $3$ arbitary events defined on a sample space $S$ and if, $P (A) + P (B) + P (C) = p_{1}, P (A \cap B) + P (B \cap C) + P (C \cap A) = p_{2}$ & $P (A \cap B \cap C) = p_{3}$ , then the probability that exactly one of the three events occurs is given by:

p_{1} - p_{2} + p_{3}

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p_{1} - p_{2} + 2 p_{3}

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p_{1} - 2 p_{2} + p_{3}

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p_{1} - 2 p_{2} + 3 p_{3}

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Solution

The correct option is D $p_{1} - 2 p_{2} + 3 p_{3}$
Probability that exactly one of event occurs
$= P (only A) + P (only B) + P (only C)$

$P (only A) = P (A) - P (A \cap B) - P (A \cap C) + P (A \cap B \cap C)$
$P (only B) = P (B) - P (A \cap B) - P (B \cap C) + P (A \cap B \cap C)$
$P (only C) = P (C) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C)$
$= (P (A) - P (A \cap B) - P (A \cap C) + P (A \cap B \cap C)) + (P (B) - P (A \cap B) - P (B \cap C) + P (A \cap B \cap C)) + (P (C) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C))$
$= p_{1} - 2 p_{2} + 3 p_{3}$

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Let A,B & C be 3 arbitary events defined on a sample space S and if, P(A)+P(B)+P(C)=p1,P(A∩B)+P(B∩C)+P(C∩A)=p2 & P(A∩B∩C)=p3, then the probability that exactly one of the three events occurs is given by:

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