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I am working on a question for class where there are two patients waiting on kidneys. The arrival of transplant kidneys is a Poisson process with rate λ. A will die after an exponential time with rate $μ_A$, and B after an exponential time with rate $μ_B$. A is ahead of B on the waiting list, so the first kidney goes to A as long as they are still alive. We are asked to compute the probability that A and B each survive to receive their kidneys.

For A, I defined $P( A < T_1 )$ and used the total probability law to expand that out into $P( A > T_1 | T_1 = t)P( T_1 = t)$, which simplified pretty easily into $μ_A \exp( - μ_At - λt )$.

For B, I tried a similar approach and got $P( A < T_1)P( B > T_1 ) + P( A > T_1 )P( B > T_1 + T_2)$.

What's the easiest way to approach $P( B > T_1 + T_2)$? Expanding that out with total probability just confused me more. I got this: $P( B > T_1 + T_2 | T_1 + T_2 = t)P( T_1 + T_2 = t | T_1 = s)P( T_1 = s)$ and I just don't know what to do with the variable $s$. Are those supposed to go away? Should I expect the expression for B to depend on two variables? I'm not even sure that my answer for A should be in terms of t.

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Think about the relationship between Poisson process and exponential distribution. You will find you work with 4 random variables following the exponential distributions. Let $X_1$ be the time that first kidney arrives, $X_2$ the time between two kidneys arrive, $Y_1$ is A's survival time without transplantation, and $Y_2$ is A's survival time without transplantation. From the description of the question, $X_1, X_2, Y_1,Y_2$ are independent and follow exponential distribution. So their joint pdf is the product of 4 pdf.

The event "A and B each survive to receive their kidneys" is $(Y_1 > X_1) \cap (Y_2 > X_1 + X_2)$.

Integrating the joint pdf on the region defined by the event is the probability of event.

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