Question

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

(1) \(\frac{9}{17}\)  

(2) \(\frac{9}{19}\) 

(3) \(\frac{8}{17}\) 

(4) \(\frac{8}{19}\)

Answer 1

Correct option is (2) \(\frac{9}{19}\) 

For sum '5' \(\rightarrow (1, 4) (2, 3), (3, 2) \)

(4, 1) \(\Rightarrow P(A) = \frac{4}{36}\) 

For sum ‘8’ \(\rightarrow (2, 6), (3, 5), (4, 4) \) 

\( (5,3),(6,2) \Rightarrow P(B)=\frac{5}{36} \)

\(\begin{aligned} P(\bar{A})=\frac{32}{36}, P(\bar{B})=\frac{31}{36}\end{aligned} \)

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

\(=\frac{P(A)}{1-P(\bar{A}) P(\bar{B})}=\frac{9}{19}\)