Two individuals, A and B, both require kidney transplants. If she does not receive a new kidney, then A will die after an exponential time with rate µA, and B after an exponential time with rate µB. New kidneys arrive in accordance with a Poisson process having rate λ. It has been decided that the first kidney will go to A (or to B if B is alive and A is not at that time) and the next one to B (if still living). a) What is the probability that A obtains a new kidney? b) What is the probability that B obtains a new kidney?

To obtain the probability of B obtains a new kidney, we will condition on the first event (and then unconditioned), i.e., which happens first: a kidney arrives, A dies or B dies. Thus

P(B obtains a new kidney)

=P(B obtains a new kidney|T1= min {T1, TA, TB})P(T1 = min {T1, TA, TB})+P(B obtains a new kidney|TA = min {T1, TA, TB})P(TA = min {T1, TA, TB})+P(B obtains a new kidney|TB = min {T1, TA, TB})P(TB = min {T1, TA, TB})

=P(T2 < TB)P(T1 = min {T1, TA, TB}) + P(T1 < TB)P(TA = min {T1, TA, TB}) + 0

=[λ/( λ + µB)][λ /(λ + µA + µB)]+ [λ/( λ + µB)] [µA/( λ + µA + µB)]

=[λ/(λ + µB)]+ [(λ + µA)/ (λ + µA + µB)]

Suppose that each person survives a kidney operation (independently) with probability p, 0 < p < 1, and, if so, has an exponentially distributed remaining life with mean 1/µ

We consider the time until either A dies or the kidney arrives, whichever is first. The mean time until that is 1/(µA + λ). A will of course live until that time. We now consider the expected remaining time A will live after that time. The conditional probability that A will live until the kidney arrives is λ/(λ + µA), independent of the time itself. Let LA be the lifetime of A. Given that the kidney operation is performed, the expected remaining life is p/µ. Thus

E[LA] = 1 /(λ + µA )+ (λ /λ + µA)( p/µ)= (µ + λp)/[(λ + µA)µ]