Question

On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is

• A. $\frac{2}{45}$
• B. $\frac{2}{5}$
• C. $\frac{1}{81}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{2}{5}$

let P(A) be the probability of throwing total of 5 and P(B) the probability of throwing total 7.
$\therefore P\left(A\right)=\frac{4}{36}=\frac{1}{9},P\left(B\right)=\frac{6}{36}=\frac{1}{6},P\left(AorB\right)=\frac{5}{18}andP\left(AandB\right)=0P\left(neitherAnorB\right)=\frac{13}{18},\therefore P\left(5before7\right)=\frac{1}{9}+\frac{13}{18}.\frac{1}{9}+\frac{13}{18}.\frac{13}{18}.\frac{1}{9}+....\infty =\frac{1}{9}\left[\frac{1}{1-\frac{13}{18}}\right]=\frac{2}{5}.$