If A and B are events of a sample space such that P(A∪B)=34,P(A∩B)=14and P(A¯)=23,then P(A¯∩B) is

P(A∪B)=34,P(A∩B)=14and P(A)=23We know that, P(A∪B)=P(A)+P(B)−P(A∩B)⇒P(B)−P(A∩B)=P(A∪B)−P(A)⇒P(A¯∩B)=P(A∪B)−P(A)⇒34−(1−23)=34−13=512

If A and B are events of a sample space such that P(A∪B)=34,P(A∩B)=14and P(A¯)=23,then P(A¯∩B) is

P(A∪B)=34,P(A∩B)=14and P(A)=23We know that, P(A∪B)=P(A)+P(B)−P(A∩B)⇒P(B)−P(A∩B)=P(A∪B)−P(A)⇒P(A¯∩B)=P(A∪B)−P(A)⇒34−(1−23)=34−13=512

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If A and B are events of a sample space such that $P (A \cup B) = \frac{3}{4}, P (A \cap B) = \frac{1}{4}$ and $P (\bar{A}) = \frac{2}{3,}$ then $P (\bar{A} \cap B)$ is

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$\frac{5}{12}$

$\frac{3}{8}$

$\frac{4}{5}$

$\frac{5}{4}$

answer is A.

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Detailed Solution

$P (A \cup B) = \frac{3}{4}, P (A \cap B) = \frac{1}{4} a n d P (A) = \frac{2}{3}$
We know that,
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
$\begin{array}{l} \Rightarrow P (B) - P (A \cap B) = P (A \cup B) - P (A) \\ \Rightarrow P (\bar{A} \cap B) = P (A \cup B) - P (A) \\ \Rightarrow \frac{3}{4} - (1 - \frac{2}{3}) = \frac{3}{4} - \frac{1}{3} = \frac{5}{12} \end{array}$