Question

In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and the total of scores on the two dice, in each throw is noted.

$A$ wins the game if he throws a total of $6$ before $B$ throws a total of $7$ and $B$ wins the game if he throws a total of $7$ before A throws a total of six

The game stops as soon as either of the players wins.

The probability of $A$ winning the game is :

• A. $\frac{5}{31}$
• B. $\frac{31}{61}$
• C. $\frac{30}{61}$
• D. $\frac{5}{6}$

Answer 1

The correct option is C

$\frac{30}{61}$

Explanation for the correct answer:

Find the probability of A winning the game is:

Let, $n\left(s\right)$ be the total number of possible outcomes if two dices are thrown at a time

So,

$n\left(s\right)=6^2=36$

Let, $n\left(E_1\right)$ be the favorable outcome of getting a sum is six

So,

$\begin{array}{rcl}n\left(E_1\right)& =& \left[(1,5)(2,4)(3,3)(4,2)(5,1)\right]\\ & =& 5\end{array}$

Let, $n\left(E_2\right)$ be the favorable outcome of getting a sum is seven.

So,

$\begin{array}{rcl}n\left(E_2\right)& =& \left[(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)\right]\\ & =& 6\end{array}$

Thus, the probability of getting a sum of $6$ is $p\left(E_1\right)=\frac{5}{36}$

The probability of getting a sum of $7$ is $p\left(E_2\right)=\frac{6}{36}$

Now,

The probability of A winning the game is $p\left(A_{win}\right)$

$\begin{array}{rcl}p\left(A_{win}\right)& =& p\left(E_1\right)+p\left(\overline{E_1}\right)p\left(\overline{E_2}\right)p\left(E_1\right)+p\left(\overline{E_1}\right)p\left(\overline{E_2}\right)p\left(\overline{E_1}\right)p\left(\overline{E_2}\right)p\left(E_1\right)+..........\\ & =& \frac{5}{36}+\frac{31}{36}\frac{30}{36}\frac{5}{36}+\frac{31}{36}\frac{30}{36}\frac{31}{36}\frac{30}{36}\frac{5}{36}+...............\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{p}\mathbf{\left(}\overline{\mathbf{X}}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{p}\mathbf{\left(}\mathbf{X}\mathbf{\right)}\mathbf{]}\\ & =& \frac{\frac{5}{36}}{1-\frac{31}{36}\frac{30}{36}}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{S}}_{\mathbf{\infty }}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}\mathbf{]}\\ & =& \frac{30}{61}\end{array}$

Hence, the correct option is C.