The problem involves finding the probability $p = P(X > \max \{ Y_1, Y_2, \dots, Y_n \})$ where $X$ and $Y_i$'s are independent exponential random variables with rates $\lambda$ and $\mu$, respectively. It suggests that you use a certain identity for probability, namely:

$p = P(A_1 A_2 \cdots A_n) = P(A_1)P(A_2|A_1)P(A_3|A_1 A_2) \cdots P(A_n | A_1 A_2 \dots A_{n-1})$

Here's a breakdown:

Steps:

1. Define Events $A_j$: The event $A_j$ is that the $j$-th smallest among the $n + 1$ random variables (the $n$ $Y_i$'s and $X$) is one of the $Y_i$'s. The goal is to calculate the probability that $X$ is larger than all the $Y_i$'s.

2. First Step: $P(A_1)$: The first event, $A_1$, means that the smallest variable is one of the $Y_i$'s. Since there are $n+1$ total variables, the probability that the smallest is one of the $Y_i$'s is $\frac{n}{n+1}$.

3. Second Step: $P(A_2|A_1)$: Conditional on $A_1$, the second smallest must again be one of the $Y_i$'s. After removing the smallest, we are left with $n$ variables, so the probability is $\frac{n-1}{n}$.

4. Continuing this Pattern: For the $j$-th step, the conditional probability becomes $\frac{n-j+1}{n-j+2}$.

5. Final Step: $P(A_n | A_1, A_2, \dots, A_{n-1})$: The last probability is simply $\frac{1}{2}$, as after all other variables are removed, we are left with only one $Y_i$ and $X$.

Overall Formula:

Multiplying all the conditional probabilities, we get:

$p = \frac{n}{n+1} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdot \dots \cdot \frac{1}{2} = \frac{1}{n+1}$

Verification for $n = 2$:

For $n = 2$, we are looking for $p = P(X > \max \{Y_1, Y_2\})$. Using the identity:

$p = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$

To verify, consider conditioning on $X$:

• The maximum of $Y_1$ and $Y_2$, denoted by $Z = \max\{Y_1, Y_2\}$, follows the exponential distribution with rate $2\mu$ because the maximum of two independent exponentials with rate $\mu$ has rate $2\mu$.
• Thus, $P(X > Z) = \frac{\lambda}{\lambda + 2\mu}$.
• When $\lambda = \mu$, $P(X > Z) = \frac{1}{3}$, matching our result.

Would you like further details on any part of this solution? Here are five related questions that could extend your understanding:

1. How does the memoryless property of the exponential distribution play into this solution?
2. What happens to $p$ as $n$ increases indefinitely?
3. Can this method be applied to other distributions besides exponential? Why or why not?
4. What is the intuition behind the product of conditional probabilities in this context?
5. How would the calculation change if the $Y_i$'s were not identically distributed?

Tip: The exponential distribution is memoryless, meaning that for any $t$, $P(X > s + t | X > s) = P(X > t)$. This is often key in simplifying problems involving sums and maximums of exponentials.