$A$ and $B$ alternately throw a pair of dice. $A$ wins if he throws a sum of 5 before $B$ throws a sum of 8 …

$A$ 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. $\frac{8}{17}$
  2. $\frac{9}{19}$
  3. $\frac{9}{17}$
  4. $\frac{8}{19}$

Solution

For sum ' 5 ' $\rightarrow(1,4),(2,3),(3,2)$
$(4,1) \Rightarrow P(A)=\frac{4}{36}$
For sum ' 8 ' $\rightarrow(2,6),(3,5),(4,4)$
For sum ' 5 ' $\rightarrow(1,4),(2,3),(3,2)$
$(4,1) \Rightarrow P(A)=\frac{4}{36}$
For sum ' 8 ' $\rightarrow(2,6),(3,5),(4,4)$
$\begin{aligned} & (5,3),(6,2) \Rightarrow P(B)=\frac{5}{36} \\ & \begin{aligned} P(\bar{A})=\frac{32}{36} & , P(\bar{B})=\frac{31}{36}\end{aligned} \\ & \begin{aligned} P(A \text { wins })= & P(A)+P(\bar{A}) P(\bar{B}) P(A)+ \\ & +P(\bar{A}) P(\bar{B}) P(\bar{A}) P(\bar{B}) P(A)+\ldots \\ = & \frac{P(A)}{1-P(\bar{A}) P(\bar{B})}=\frac{9}{19}\end{aligned}\end{aligned}$ /

Asked in: JEE Main 2025 (24 Jan Shift 1)

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