Question

In a race, the probabilities of A and B winning the race are $\frac{1}{3}$ and $\frac{1}{6}$ respectively. find the probability of neither of them winning the race.

Answer 1

Given probabilities of winning of $A$ and $B:$ $P\left(A\right)=\frac{1}{3}$ and $P\left(B\right)=\frac{1}{6}$

Probability of neither of them winning the race $:1-P(A\cup B)$

Now since the event is mutually exclusive, therefore

$P(A\cup B)=P\left(A\right)+P\left(B\right)=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

Hence, $1-P(A\cup B)=1-\frac{1}{2}=\frac{1}{2}$

$\therefore$ Probability of neither of them winning the race $:\frac{1}{2}$