Question

Given two mutually exclusive events A and B such that $P\left(A\right)=\frac{1}{2}$ and $P\left(B\right)=\frac{1}{3}$ find P(A or B).

Answer 1

Given,

$P\left(A\right)=\frac{1}{2}$

$P\left(B\right)=\frac{1}{3}$

$\therefore$ A and B are mutually exclusive events, then $P(A\cap B)$= 0

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

= $\frac{1}{2}+\frac{1}{3}$

= $\frac{3+2}{6}=\frac{5}{6}$

$\therefore P(A\cup B)=\frac{5}{6}$