For any two events A, B if P(A)=13,P(B)=12 P(A∩B)=16 , then the probability that exactly one of them occurs is

P(A)=13,P(B)=12,P(A∩B)=16P( exactly one of A or B)=P(A−B)+P(B−A)=P(A)+P(B)−2P(A∩B)=13+12−2×16=12

For any two events A, B if P(A)=13,P(B)=12 P(A∩B)=16 , then the probability that exactly one of them occurs is

P(A)=13,P(B)=12,P(A∩B)=16P( exactly one of A or B)=P(A−B)+P(B−A)=P(A)+P(B)−2P(A∩B)=13+12−2×16=12

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For any two events A, B if $P (A) = \frac{1}{3}, P (B) = \frac{1}{2}$ $P (A \cap B) = \frac{1}{6}$ , then the probability that exactly one of them occurs is

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$\frac{1}{3}$

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{5}{6}$

answer is B.

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Detailed Solution

$P (A) = \frac{1}{3}, P (B) = \frac{1}{2}, P (A \cap B) = \frac{1}{6}$

$P (exactly one of A or B) = P (A - B) + P (B - A)$

$= P (A) + P (B) - 2 P (A \cap B) = \frac{1}{3} + \frac{1}{2} - 2 \times \frac{1}{6} = \frac{1}{2}$

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For any two events A, B if P(A)=13,P(B)=12 P(A∩B)=16 , then the probability that exactly one of them occurs is

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For any two events A, B if $P (A) = \frac{1}{3}, P (B) = \frac{1}{2}$ $P (A \cap B) = \frac{1}{6}$ , then the probability that exactly one of them occurs is