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If A and B are two events such that \mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{1}{4} , the P(A′ ∩ B′) equals
A. 1|12
B. 3|4
C. 1|4
D. 3|16

Answers (1)

Given-

\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{1}{4}

As we know,
P(A|B) × P(B) = P(A ∩ B) [Property of Conditional Probability]

\\ \Rightarrow \frac{1}{4} \times \frac{1}{3}=P(A \cap B) \\ \Rightarrow P(A \cap B)=\frac{1}{12} \\ P\left(A^{\prime} \cap B^{\prime}\right)=1-P(A \cup B) \\ =1-[P(A)+P(B)-P(A \cap B)] \\ {[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)]} \\ =1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right] \\ =1-\left[\frac{6+4-1}{12}\right] \\ =1-\frac{9}{12} \\ =\frac{12-9}{12} \\ =\frac{3}{12} \\ =\frac{1}{4}

Hence, the correct option is C

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