Question

Let $A$ and $B$ be events such that $p\left(A\right)=\frac{5}{11},P\left(B\right)=\frac{6}{11}$, and $P\left(A\cup B\right)=\frac{7}{11}$ and $P\left(B/A\right).$

Answer 1

As we know,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

On substituting the values, we get

$\frac{7}{11}=\frac{5}{11}+\frac{6}{11}-P(A\cap B)$

$P(A\cap B)=\frac{11}{11}-\frac{7}{11}$

$P(A\cap B)=\frac{4}{11}.$

Now,

$P(A\cap B)=P\left(A\right)P\left(B/A\right).$

On substituting the values, we get

$\frac{4}{11}=\frac{5}{11}\times P\left(B/A\right)$

$P\left(B/A\right)=\frac{4}{5.}$