JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Probability

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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:

(A) $\frac{5}{6}$
(B) $\frac{31}{61}$
(C) $\frac{30}{61}$ ✓ Correct
(D) $\frac{5}{31}$

Answer: (C)

Solution

A total of

6

comes up with probability

\frac{5}{36}

and a total of

7

with probability

\frac{6}{36} = \frac{1}{6}

. Player

A

throws first, so

A

wins on the opening throw, or after both players miss and the cycle repeats; the chance that one full cycle passes with neither winning is

\frac{31}{36} \cdot \frac{30}{36}

. Summing the geometric series, the probability that

A

wins is

\frac{5/36}{1 - \frac{31}{36} \cdot \frac{30}{36}} = \frac{30}{61}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Probability

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