Question

If $A$ and $B$ are mutually exclusive events such that $P\left(A\right)=\frac{3}{5}$ and $P\left(B\right)=\frac{2}{7}$, then find $P(A\cup B)$

Answer 1

We know that if $A$ and $B$ are mutually exclusive events then $P(A\cup B)=P\left(A\right)+P\left(B\right)$.

It is given that $P\left(A\right)=\frac{3}{5}$ and $P\left(B\right)=\frac{2}{7}$, therefore,

$P(A\cup B)=P\left(A\right)+P\left(B\right)=\frac{3}{5}+\frac{2}{7}=\frac{3\times 7}{5\times 7}+\frac{2\times 5}{7\times 5}=\frac{21+10}{35}=\frac{31}{35}$

Hence, $P(A\cup B)=\frac{31}{35}$.