Mathematics - Probability Question with Solution | TestHub

Statement : I If $P\left(H_i\cap E\right)=0$ for some $i$, then $$P\left(\frac{H_i}{E}\right)=P\left(\frac{E}{H_i}\right)=0$$ If $P\left(H_i\cap E\right)\ne 0$ for $\forall i=1,2,\dots .n$, then $$\begin{array}{rl}P\left(\frac{H_i}{E}\right)& =\frac{P\left(H_i\cap E\right)}{P\left(H_i\right)}\times \frac{P\left(H_i\right)}{P(E)}\\ & =\frac{P\left(\frac{E}{H_i}\right)\times P\left(H_i\right)}{P(E)}>P\left(\frac{E}{H_i}\right)\cdot P\left(H_i\right)\text{ [as }0<P(E)<1]\end{array}$$ Hence, Statement I may not always be true. Statement II Clearly, $H_1\cup H_2\cup \dots .\cup H_n=S$ (sample space) $$\Rightarrow P\left(H_1\right)+P\left(H_2\right)+\dots +P\left(H_n\right)=1.$$ Hence, Statement II is true.