Question

If $P\left(A\right)=\frac{6}{11},P\left(B\right)=\frac{5}{11}andP(A\cup B)=\frac{7}{11},$ find
$P(A\cap B)$

$P\left(\frac{A}{B}\right)$

$P\left(\frac{B}{A}\right)$

Answer 1

Given, $P(A\cup B)=\frac{7}{11},P\left(A\right)=\frac{6}{11},P\left(B\right)=\frac{5}{11}\therefore WeknowthatP(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A\cap B)\Rightarrow P(A\cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}$

We know that $P\left(\frac{A}{B}\right)=\frac{P(B\cap A)}{P\left(A\right)}\Rightarrow P\left(\frac{B}{A}\right)=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{11}\times \frac{11}{6}=\frac{4}{6}=\frac{2}{3}$

We know that $P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P\left(A\right)}\Rightarrow P\left(\frac{B}{A}\right)=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{11}\times \frac{4}{6}=\frac{2}{3}[\because P(A\cap B)=P(B\cap A\left)\right]$