Probability – Independent Event – JEE Main 24 January 2025 Shift 1

Solution:

Let $S_A$ denote the event that A throws a sum of 5.
The possible outcomes for a sum of 5 are: $(1,4), (4,1), (2,3), (3,2)$.
$$ n(S_A) = 4 \Rightarrow P(S_A) = \frac{4}{36} = \frac{1}{9} $$
Let $p_1 = \frac{1}{9}$ and $q_1 = 1 – \frac{1}{9} = \frac{8}{9}$.

Let $S_B$ denote the event that B throws a sum of 8.
The possible outcomes for a sum of 8 are: $(2,6), (6,2), (3,5), (5,3), (4,4)$.
$$ n(S_B) = 5 \Rightarrow P(S_B) = \frac{5}{36} $$
Let $p_2 = \frac{5}{36}$ and $q_2 = 1 – \frac{5}{36} = \frac{31}{36}$.

A starts the game. A wins in the 1st throw, or 3rd throw (after A fails, B fails), or 5th throw, etc.
$$ P(A \text{ wins}) = p_1 + q_1 q_2 p_1 + (q_1 q_2)^2 p_1 + \dots $$
This is an infinite Geometric Progression with first term $a = p_1$ and common ratio $r = q_1 q_2$.

$$ r = \frac{8}{9} \times \frac{31}{36} = \frac{62}{81} $$
$$ P(A \text{ wins}) = \frac{a}{1-r} = \frac{1/9}{1 – 62/81} $$
$$ = \frac{1/9}{(81-62)/81} = \frac{1}{9} \times \frac{81}{19} = \frac{9}{19} $$

Ans. (2)