Question

If A and B are two events such that $P(A\cup B)=\frac{3}{4},\mathrm{P}(A\cap B)=\frac{1}{4},P\left(\overline{\mathit{A}}\right)=\frac{2}{3},\mathrm{then}P(\overline{\mathit{A}}\cap B)$ is ______________.

Answer 1

For two events of A and B,

given, P(A⋃B) = $\frac{3}{4}$

P(A⋂B) = $\frac{1}{4}$

$P\left(\overline{A}\right)=\frac{2}{3}$

then $P\left(\overline{A}\cap B\right)=P\left(B\right)-P\left(A\cap B\right)$

since $P\left(\overline{A}\right)=\frac{2}{3}$
$P\left(A\right)=1-P\left(\overline{A}\right)=1-\frac{2}{3}=\frac{1}{3}$

Hence P(B) = P(A⋃B) – P(A) + P(A⋂B) [using P(A⋃B) = P(A) + P(B) – P(A⋂B)

$$\begin{gathered} \mathrm{i}.\mathrm{e}P\left(B\right)=\frac{3}{4}-\frac{1}{3}+\frac{1}{4}=1-\frac{1}{3}=\frac{2}{3} \\ \therefore P\left(\overline{A}\cap B\right)=\frac{2}{3}-\frac{1}{4}=\frac{8-3}{12} \\ \mathrm{i}.\mathrm{e}P\left(\overline{A}\cap B\right)=\frac{5}{12} \end{gathered}$$