Question

Assume you are waiting for a cab and that the probability that a cab will come in any one minute is 15%, and the probability of a cab in any one minute is an independent event. In this instance, your waiting time could be modeled as a geometric random variable. Given this situation, what is the probability you have to wait more than 20 minutes for a cab?

Answer 1

Let X be the waiting time for the first cab.

Then X~Geometric (p=0.15)

Then P(X=k) = (1-p)^k-1 * p...... where k=1,2,3,....

To get the first cab in more than 20 minutes, considering the time taken is in minutes, we should get our cab in 21,22,23,.... and so on minutes.

i.e we calculate P(X>20) = 1- P(X<=20) = 1-0.967054 = 0.032946

We can calculate this value using online calculators as well as other software.

Hence the probability that we have to wait more than 20 minutes for a cab is 3.29% .