Mathematics - Probability Question with Solution | TestHub

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

Using the addition theorem of probability, 

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

$\Rightarrow P\left(A\right)=P\left(B\right)$

$P\left(A\cap B^{\mathrm{\text{'}}}\right)=0$

$P\left(A^{\mathrm{\text{'}}}\cap B\right)=0$

Here, $A$ and $B$ are equally likely.

But we can't say that $P\left(A\right)=P\left(B\right)=\frac{1}{2}$

Therefore, $P\left(A\right)+P\left(B\right)\ne 1$