Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2309 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{12}$
We have, $P(A \cup B)=\frac{3}{4}, P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{2}{3}$
$\begin{aligned} & \text { We know that, } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(B)-P(A \cap B)=P(A \cup B)-P(A) \\ & \Rightarrow P(\bar{A} \cap B)=P(A \cup B)-P(A) \\ & =\frac{3}{4}-\left(1-\frac{2}{3}\right)=\frac{3}{4}-\frac{1}{3}=\frac{5}{12} \\ & \end{aligned}$
$\begin{aligned} & \text { We know that, } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ & \Rightarrow P(B)-P(A \cap B)=P(A \cup B)-P(A) \\ & \Rightarrow P(\bar{A} \cap B)=P(A \cup B)-P(A) \\ & =\frac{3}{4}-\left(1-\frac{2}{3}\right)=\frac{3}{4}-\frac{1}{3}=\frac{5}{12} \\ & \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.