Question

You have just arrived in a foreign country and you have to wait in line at immigration. There are three different desks, staffed by workers A B and C. Worker A is fast, and can process arrivals in 2 minutes, on average. Worker B takes 5 minutes, on average. And worker C takes 6 minutes, on average. You don't know who is who, so you pick someone at random and walk up to their desk. Assume processing times are exponential random variables.

What's the chance your processing time is less than 5 minutes?

Suppose your processing time actually ended up being less than 5 minutes. What's the chance you were helped by worker C?

Answer 1

By CDF of exponential distribution, P(X < x) = 1 - exp(-x)

where mean = 1/

For Worker A, probability that processing time is less than 5 minutes = 1 - exp(-5/2) = 0.917915

For Worker B, probability that processing time is less than 5 minutes = 1 - exp(-5/5) = 0.6321206

For Worker C, probability that processing time is less than 5 minutes = 1 - exp(-5/6) = 0.5654018

As, we pick someone at random, the probability of choosing ay desk worker is 1/3.

So, probability that processing time is less than 5 minutes

= (1/3) * 0.917915 + (1/3) * 0.6321206 + (1/3) * 0.5654018

= 0.7051458

Let F be the event that processing time actually ended up being less than 5 minutes.

Then P(F) = 0.7051458

Let C be the event that you were helped by worker C.

Given, that processing time actually ended up being less than 5 minutes, probability that you were helped by worker C = P(C | F)

By Bayes theorem,

P(C | F) = P(F | C) * P(C) / P(F)

We know that P(C) = 1/3 (All desks are randomly chosen)

P(F | C) = Probability that processing time is less than 5 minutes by worker C = 1 - exp(-5/6) = 0.5654018

Then,

P(C | F) = 0.5654018 * (1/3) / 0.7051458

= 0.2672742