A and B throw a pair of dice alternately and they note the sum of the numbers appearing on the dice. A wins…

A and B throw a pair of dice alternately and they note the sum of the numbers appearing on the dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before $A$ throws 6 . If $A$ begins, the probability of his winning is
  1. $\frac{15}{61}$
  2. $\frac{21}{61}$
  3. $\frac{30}{61}$
  4. $\frac{36}{61}$

Solution

For sum 6 possible combinations are $(1,5),(5,1)$, $\begin{aligned} & (2,4),(4,2),(3,3) \\ & P(\text { sum } 6)=\frac{5}{36} \end{aligned}$
For sum 7 possible combinations are $(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$ $\begin{aligned} & P(\operatorname{sum} 7)=\frac{6}{36}=\frac{1}{6} \\ & \therefore P(A \text { wins })=\frac{5}{36}+\frac{31}{36} \times \frac{5}{6} \times \frac{5}{36}+\ldots=\frac{\frac{5}{36}}{1-\frac{31}{36} \times \frac{5}{6}}=\frac{30}{61}\end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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