In a game two players A and B take turns

Question:

In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and 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 :

  1. (1) $\frac{5}{31}$

  2. (2) $\frac{31}{61}$

  3. (3) $\frac{5}{6}$

  4. (4) $\frac{30}{61}$

Correct Option: 4,

Solution:

Probability of sum getting $6, P(A)=\frac{5}{36}$

Probability of sum getting $7, P(B)=\frac{6}{36}=\frac{1}{6}$

$P(A$ wins $)=P(A)+P(\bar{A}) P(\bar{B}) P(A)$ 

$+P(\bar{A}) \cdot P(\bar{B}) P(\bar{A}) P(\bar{B}) P(A)+\ldots . .$

$\Rightarrow \frac{5}{36}+\left(\frac{31}{36}\right)\left(\frac{30}{36}\right)\left(\frac{5}{36}\right)+\ldots . . \infty$

$\Rightarrow \frac{5}{36}\left(1+\frac{155}{216}+\left(\frac{155}{216}\right)^{2}+\ldots \ldots \ldots \infty\right)$

$\Rightarrow \frac{\frac{5}{36}}{\frac{61}{216}}=\frac{30}{61}$ $\left(\because S_{\infty}=\frac{a}{1-r}\right)$

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