Question

In a single throw of two dice, find the probability that neither a doublet nor a total of $9$ will appear.

Answer 1

We know that,

$n\left(S\right)=36$

$\begin{array}{c}A=\left\{\left(1,1\right)\left(2,2\right)\left(3,3\right)\left(4,4\right)\left(5,5\right)\left(6,6\right)\right\}\\ B=\left\{\left(2,6\right)\left(3,5\right)\left(4,4\right)\left(5,3\right)\left(6,2\right)\right\}\\ n\left(A\right)=6,n\left(B\right)=5,n\left(A\cap B\right)=1\end{array}$

$\therefore$Required probability$=P\left(A\cap B\right)$

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

$=\frac{6}{36}+\frac{5}{36}-\frac{1}{36}=\frac{5}{18}$

Then,

We get $\frac{5}{18}$