JEE Mathematics 2024 Question with Solution

Given: A fair die is tossed until the first six appears, so $X$ follows a geometric distribution with success probability $p = \frac{1}{6}$ and failure probability $q = \frac{5}{6}$.

Find: $\frac{b+c}{a}$ where $a = P(X=3)$, $b = P(X \geq 3)$, and $c = P(X \geq 6 \mid X > 3)$.

First calculate $a$:

Now calculate $b$. For $X \geq 3$, the first two tosses must not be six:

Next calculate $c = P(X \geq 6 \mid X > 3)$. Given $X > 3$, no six has appeared in the first three tosses. To have $X \geq 6$, tosses $4$ and $5$ must also not be six. By the memoryless property of the geometric distribution:

Substitute these values:

Therefore, the correct option is A.

Given: $X$ is geometric with success probability $\frac{1}{6}$.

Find: $\frac{b+c}{a}$.

For a geometric random variable,

• $P(X=3) = q^2 p$
• $P(X \geq 3) = q^2$
• $P(X \geq 6 \mid X > 3) = q^2$ because after conditioning on $X > 3$, the process restarts from toss $4$.

So,

Therefore, the correct option is A.