JEE PYQ: Probability Question 20
Question 20 - 2020 (04 Sep Shift 2)
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) $\frac{5}{31}$
(2) $\frac{31}{61}$
(3) $\frac{5}{6}$
(4) $\frac{30}{61}$
Show Answer
Answer: (4)
Solution
$P(A \text{ gets } 6) = \frac{5}{36}$, $P(B \text{ gets } 7) = \frac{6}{36} = \frac{1}{6}$. $P(A \text{ wins}) = \frac{5}{36} + \frac{31}{36} \cdot \frac{30}{36} \cdot \frac{5}{36} + \cdots = \frac{5/36}{1 - \frac{31 \times 30}{36^2}} = \frac{5}{36} \cdot \frac{36^2}{36^2 - 930} = \frac{5 \times 36}{1296 - 930} = \frac{180}{366} = \frac{30}{61}$.
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