JEE Main 2025 Jan 24 Shift 1, Mathematics Q11: Total Probability Theorem And Bayes' Theorem

JEE Main2025Jan 24, Shift 1Mathematics

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
$\frac{9}{19}$
C
$\frac{8}{17}$
D
$\frac{8}{19}$

Solution

Count outcomes for the two events on one throw of two dice (36 equally likely outcomes):

Sum 5: $(1, 4), (4, 1), (2, 3), (3, 2) \Rightarrow p_{A} = \frac{4}{36} = \frac{1}{9}$ .

Sum 8: $(2, 6), (6, 2), (3, 5), (5, 3), (4, 4) \Rightarrow p_{B} = \frac{5}{36}$ .

Probabilities of NOT scoring: $q_{A} = \frac{8}{9}$ , $q_{B} = \frac{31}{36}$ .

A wins if A scores on throw 1, OR both fail and then A scores on throw 3, etc. Each full round (A then B, both failing) multiplies by $q_{A} q_{B} = \frac{8}{9} \cdot \frac{31}{36} = \frac{248}{324} = \frac{62}{81}$ .

$P (A) = \frac{p _{A}}{1 - q _{A} q _{B}} = \frac{1/9}{1 - 62/81} = \frac{1/9}{19/81} = \frac{81}{9 \cdot 19} = \frac{9}{19}$ .

$∴$ Option (2).

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JEE Main 2025 Jan 24 Shift 1, Mathematics Q11: Total Probability Theorem And Bayes' Theorem

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