Exponential/Poisson distribution

Question

Tom enters the post office where 5 people are being served, each by a different sales clerk. He will be called up as soon as any one of the 5 people currently being attended to are finished. The service time for each individual by each cleark has an exponential distribution with an average service time of 5 minutes, and is independent of all other service times. Find the probability Tom has to wait for more than 2 minutes before he is called up.

I'm struggling with determining how to set this up, mainly with the fact that there are 5 people being served.

Answer 1

For Tom to wait more than 2 minutes, each of the 5 clerks will have to take more than 2 minutes on their respective customers. So if x is the probability that a single clerk will take longer than 2 minutes (I'll let you compute x), then the final answer would just be x to the power 5. This is a joint probability distribution. P(tom waits longer than 2 minutes) = P(clerk 1 takes longer than 2 minutes,clerk 2 takes longer than 2 minutes,etc.etc.) = P(a single clerk takes longer than 2 minutes)^5.

Answer 2

Here is how you can solve the problem using theory (with memory-less property of the exponential distribution, with the fact that the random variables are i.i.d.) and also with simulation using R:

# P(/\(X_i > 2)) = Prod_i(P((X_i > 2))), i=1,..,5, X_i ~ Exp(1/5) i.i.d., where /\ denotes intersection
# P((X_i > 2)) = F_X_i(2) = exp(-(1/5)*2), F is th CDF function

# with theory
(exp(-(1/5)*2))^5
# [1] 0.1353353

(1-pexp(2, rate=1/5))^5 
# [1] 0.1353353

# with simulation
set.seed(1)
res <- replicate(10^6,{rexp(5, rate=1/5)})
probs <- table(colSums(res > 2)) / ncol(res)

probs # prob that exactly i clerks will have service time > 2, i=1,..,5
# we are interested in the event that i = 5

#      0        1        2        3        4        5 
#0.003900 0.039546 0.161347 0.327012 0.332583 0.135612 

barplot(probs)