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$\frac{9}{17}$
B$\frac{8}{17}$
C$\frac{8}{9}$
D$\frac{1}{9}$
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B$\frac{8}{17}$
$9$ can be obtained from throw of two dice in only $4$ cases as given below:
$\{[(3, 6), (4, 5), (5, 4), (6, 3)]\}$
$\Rightarrow\ \text{P}($getting $9)=\frac{4}{36}=\frac{1}{9}$
$\text{P}($not getting $9)=\frac{32}{36}=\frac{8}{9}$
Now,
$P(B$ is winning$) = P($getting $9$ in $2^{nd}$ throw$) + P($getting $p$ in $4^{th}$ throw$) + P($getting $9$ in $6^{th}$ throw$) + .....$
$=\frac{8}{9}\times\frac{1}{9}+\frac{8}{9}\times\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}+\ .....$
$=\frac{8}{81}\Big[1+\frac{64}{81}+\Big(\frac{64}{81}\Big)^2+\ ......\Big]$
$=\frac{8}{81}\times\frac{1}{1-\frac{64}{81}}$
$=\frac{8}{81}\times\frac{81}{17}$
$=\frac{8}{17}$
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