Two persons $A$ and $B$ throw a pair of dice alternately until one of them gets the sum of the numbers…

Two persons $A$ and $B$ throw a pair of dice alternately until one of them gets the sum of the numbers appeared on the dice as 4 and the person who gets this result first is declared as the winner. If A starts the game: then the probability that $B$ wins the game is
  1. $\frac{11}{23}$
  2. $\frac{1}{2}$
  3. $\frac{5}{11}$
  4. $\frac{8}{17}$

Solution

Outcomes for sum 4 are $(1,3),(3,1),(2,2)$
Probability of sum $4=\frac{3}{36}=\frac{1}{12}$ $\begin{aligned} & P(B \text { wins })=\left(\frac{11}{12} \times \frac{1}{12}\right)+\left(\frac{11}{12} \times \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}\right)+\ldots \\ & =\frac{\frac{11}{12} \times \frac{1}{12}}{1-\left(\frac{11}{12}\right)^2}=\frac{\frac{11}{144}}{\frac{144-121}{144}}=\frac{11}{23}\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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