Find $P(min(X_A, X_B,X_C,X_D) = X_A)$

Question

Problem

Four runners, A, B, C and D are organized in two teams, with runners A and B forming Team 1, and runners C and D forming Team 2. All four runners will run the same distance. Suppose that the running times of the 4 runners are all exponentially distributed, with respective means $µ_A$ = 9,$µ_B$ = 10, $µ_C$ = 8 and $µ_D$ = 12 minutes. Assume also that all four running times are independent.

What is the probability that the winner of the race (the fastest one of the four runners) will come from Team 1?

My Question

To solve, I need to find $P(min(X_A, X_B,X_C,X_D) = X_A) + P(min(X_A, X_B,X_C,X_D) = X_B)$. I can find the pdf of $Z = min(X_A, X_B,X_C,X_D)$, but that won't tell me if $min(X_A, X_B,X_C,X_D) = X_A$, so I'm not sure how to proceed

Answer 1

HINT

For independent, exponentially distributed random variables

$$P(\min(X_A,X_B,X_C,X_D)=X_A)=$$ $$=P(X_A\leq X_B\cap X_A\leq X_C\cap X_A\leq X_D)=$$ $$=E[P(X_A\leq X_B\cap _A\leq X_C\cap X_A\leq X_D\mid X_A)]=$$ $$=\int_0^{\infty}P(x\leq X_B\cap x\leq X_C\cap x\leq X_D\mid X_A=x)\lambda_Ae^{-\lambda_A x}dx=$$ $$=\int_0^{\infty}P(x\leq X_C)P(x\leq X_B)P(x\leq X_D)\lambda_Ae^{-\lambda_A x}\ dx=$$ $$=\int_0^{\infty}e^{-\lambda_B x}e^{-\lambda_C x}e^{-\lambda_D x}\lambda_Ae^{-\lambda_A x}\ dx=$$ $$=\int_0^{\infty}\lambda_Ae^{-(\lambda_A+\lambda_B+\lambda_C+\lambda_D)x}\ dx=$$ $$=\frac{\lambda_A}{\lambda_A+\lambda_B+\lambda_C+\lambda_D}.$$

Answer 2

If $X\sim\mathrm{Exp}(\lambda)$ and $Y\sim\mathrm{Exp}(\mu)$ are independent, then \begin{align} \mathbb P(X\leqslant Y) &= \int_0^\infty\int_0^y \lambda e^{-\lambda x}\mu e^{-\mu y}\ \mathsf dx\ \mathsf dy\\ &= \int_0^\infty \mu e^{-\lambda y}\left(1- e^{-\lambda y}\right)\ \mathsf dy\\ &= 1 - \mu\int_0^\infty e^{-(\lambda+\mu)y}\ \mathsf dy\\ &= \frac\lambda{\lambda+\mu}. \end{align} Moreover, for any $t>0$ we have \begin{align} \mathbb P(X\wedge Y>t) &= \mathbb P\left(\{X>t\}\cap\{Y>t\}\right)\\ &= \mathbb P(X>t)\mathbb P(Y>t)\\ &= e^{-\lambda t}e^{-\mu t}\\ &= e^{-(\lambda+\mu)t}, \end{align} so that $X\wedge Y\sim\mathrm{Exp}(\lambda + \mu)$. Now, $$ \{\min(X_A,X_B,X_C,X_D) = X_A \}\cup \{\min(X_A,X_B,X_C,X_D) = X_B \} $$ is equivalent to $$ \{X_A\wedge X_B \leqslant \min(X_A,X_B,X_C,X_D)\}, $$ and thus we compute \begin{align} \mathbb P\left(\{X_A\wedge X_B \leqslant \min(X_A,X_B,X_C,X_D)\}\right) &= \frac{\frac19 + \frac1{10}}{\frac19 + \frac1{10}+\frac18+\frac1{12}}\\ &= \frac{76}{151}. \end{align}