Question

Two persons $A$ and $B$ take turns throwing a pair of dice. The first person to throw $9$ from both dice will be awarded the prize. If $A$throws first then probability that $B$ wins the game is

• A. $\frac{9}{17}$
• B. $\frac{8}{17}$
• C. $\frac{8}{9}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B

$\frac{8}{17}$

Explanation for the correct answer.

Find the probability?

Given: Two dice are rolled

$9$ can be obtained from two dice in these ways $\left\{\left(4,5\right),\left(5,4\right),\left(3,6\right),\left(6,3\right)\right\}=4$

Probability of getting $9=\frac{4}{36}=\frac{1}{9}$

Probability of not getting $9=1-\frac{1}{9}=\frac{8}{9}$

Required probability $=$ Probability of getting $9$ in first-row $+$Probability of getting $9$ in the fourth row $+$……..

$$\begin{gathered} =\frac{8}{9}\times \frac{1}{9}+\frac{8}{9}\times \frac{8}{9}\times \frac{8}{9}\times \frac{1}{9}+\frac{8}{9}\times \frac{8}{9}\times \frac{8}{9}\times \frac{8}{9}\times \frac{8}{9}\times \frac{1}{9}...... \\ =\frac{8}{81}\left[1+\frac{64}{89}+{\left(\frac{64}{81}\right)}^2+.....\right] \\ =\frac{8}{81}\times \frac{1}{1-{\displaystyle \frac{64}{81}}} \\ =\frac{8}{81}\times \frac{1}{{\displaystyle \frac{81-64}{81}}} \\ =\frac{8}{81}\times \frac{81}{17} \\ =\frac{8}{17} \end{gathered}$$

Hence option (b) is correct.