Mathematics · Probability

JEE Main 2025 — 24 January, Morning Shift — Question 11

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

Option A:
$\frac{9}{17}$
Option B:
$\frac{9}{19}$
Correct
Option C:
$\frac{8}{17}$
Option D:
$\frac{8}{19}$

Answer: B

Step-by-step solution

p(S_8) = \frac{1}{9}

p(S_5) = \frac{5}{36}

\text{required\; prob.\; } = \frac{1}{9} + \frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9} + \left( \frac{8}{9} \cdot \frac{31}{36} \right)^2 \cdot \frac{1}{9} + ... \infty

= \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{9}{19}

\text{Option\; (2)}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Probability
Topic: Total Probability and Baye's Theorem

For a statistical data x 1, x 2, ldots, x 10 of 10 values, a student obtained the mean as 5.5 and sum i=1^10 x i^2=371 . He later found…

Answer: B

Step-by-step solution

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