Answer:
P(A obtains a new kidney) = P(T1 < TA) = λ /(λ + µA)
P(B obtains a new kidney)
=[λ/(λ + µB)]+ [(λ + µA)/ (λ + µA + µB)]
Step-by-step explanation:
P(A obtains a new kidney) = P(T1 < TA) = λ /(λ + µA)
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]
