Mathematics - Probability Question with Solution | TestHub

We know that $P\left(E/A\right)=\frac{P\left(E\cap A\right)}{P\left(A\right)}$

$\Rightarrow P\left(E\cap A\right)=P\left(E/A\right)·P\left(A\right) ...\left(i\right)$

Also, $P\left(A/E\right)=\frac{P\left(E\cap A\right)}{P\left(E\right)}$

$\Rightarrow P\left(E\cap A\right)=P\left(A/E\right)·P\left(E\right) \left(ii\right)$

From the equations $\left(i\right)$ and $\left(ii\right),$ we get

$P\left(E/A\right)·P\left(A\right)=P\left(A/E\right)·P\left(E\right)$

Now, since $0<P\left(A\right)\le 1$

$\Rightarrow P\left(E/A\right)\ge P\left(A/E\right)·P\left(E\right)$

Again, we have $P\left(A/E\right)=\frac{P\left(E\cap A\right)}{P\left(E\right)}$

$\Rightarrow P\left(A/E\right)·P\left(E\right)=P\left(E\cap A\right)$

Now, since $0<P\left(E\right)\le 1$

$P\left(A/E\right)\ge P\left(A\cap E\right)$

$\Rightarrow$Both the statements are true.