Question

Two persons A and B take turns in throwing a pair of dice. The first person to throw 9 from both dice will be awarded the prize. If A throws first, then the probability that B wins the game is
(a) 9/17
(b) 8/17
(c) 8/9
(d) 1/9

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)\frac{8}{17} \\ 9\text{can be obtained from throw of two dice in only 4 cases as given below:} \\ \left[\left(3,6\right)\text{,}\left(4,5\right)\text{,}\left(5,4\right)\text{,}\left(6,3\right)\right] \\ \text{⇒P}\left(\text{getting 9}\right)\text{=}\frac{4}{36}\text{=}\frac{1}{9} \\ \text{P}\left(\text{not getting 9}\right)\text{=}\frac{32}{36}\text{=}\frac{8}{9} \\ \text{Now,} \\ \text{P}\left(\mathit{\text{B}}\text{is winning}\right)\text{= P}\left({\text{getting 9 in 2}}_{nd}\text{throw}\right)\text{+}\mathrm{P}\left({\text{getting 9 in 4}}_{th}\text{throw}\right)+P\left({\text{getting 9 in 6}}_{th}\text{throw}\right)+... \\ =\frac{8}{9}\times \frac{1}{9}+\frac{8}{9}\times \frac{8}{9}\times \frac{8}{9}\times \frac{1}{9}+... \\ =\frac{8}{81}\left[1+\frac{64}{81}+{\left(\frac{64}{81}\right)}^2+...\right] \\ =\frac{8}{81}\times \frac{1}{1-{\displaystyle \frac{64}{81}}} \\ =\frac{8}{81}\times \frac{81}{17} \\ =\frac{8}{17} \end{gathered}$$