JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Probability

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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

(A) $\frac{9}{17}$
(B) $\frac{9}{19}$ ✓ Correct
(C) $\frac{8}{17}$
(D) $\frac{8}{19}$

Answer: (B)

Solution

On a single throw of two dice,

P (sum = 5) = \frac{4}{36} = \frac{1}{9}

(pairs

(1, 4), (2, 3), (3, 2), (4, 1)

) and

P (sum = 8) = \frac{5}{36}

(pairs

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

).

A throws first. A wins on the first throw with probability

\frac{1}{9}

. Otherwise (probability

\frac{8}{9}

that A doesn't get 5), B throws; A still has a chance if B doesn't get 8 (probability

\frac{31}{36}

), after which A gets another attempt, and so on.

P (A wins) = \frac{1}{9} + \frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9} + (\frac{8}{9} \cdot \frac{31}{36})^{2} \cdot \frac{1}{9} + \dots

This is a geometric series with first term

a = \frac{1}{9}

and common ratio

r = \frac{8}{9} \cdot \frac{31}{36} = \frac{62}{81}

P (A wins) = \frac{1/9}{1 - 62/81} = \frac{1/9}{19/81} = \frac{81}{9 \times 19} = \frac{9}{19} .

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Probability

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