Consider a supermarket with three cash settling boxes and only one queue created from buyers. Four people, A, B, C and D enter the supermarket at almost the same time and in that order. A, B and C go directly into service while D waits for the first avaliable cash box. Suppose that the service time for each cash box is exponentially distributed with mean $5$.
How can I answer the following question: What is the probability D is the last one to leave?
Am I wrong if I try P(D last one) = P(D last | cbox 1 first)*P(cbox 1 first) + P(D last | cbox 2 first)*P(cbox 2 first) + P(D last | cbox 3 first)*P(cbox 3 first)
So,
P(D last one) = P(T1 > max{T2,T3})*P(T1 < min{T2,T3}) + P(T2 > max{T1,T3})*P(T2 < min{T1,T3}) + P(T3 > max{T1,T2})*P(T3 < min{T1,T2})
And for independence:
P(D last one) = P(T1 > T2)*P(T1 > T3)*P(T1 < T2)*P(T1 < T3) + P(T2 > T1)*P(T2 > T3)*P(T2 < T1)*P(T2 < T3) + P(T3 > T1)*P(T3 > T2)*P(T3 < T1)*P(T3 < T2)
$$(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4+(\frac{1/5}{1/5+1/5})^4 = = 3(\frac{1}{2})^4 = 3\frac{1}{16} = \frac{3}{16}$$
Setting $X_i$ the rv's of the service time of A,B,C and $Y$ the one for D you have to calculate
$$\mathbb{P}[\min(X_i)+Y>\max(X_i)] = \mathbb{P}[Y>\max(X_i)-\min(X_i)]$$