Question

A men and a women decided to meet at a certain location. If each of them independently arrives at a time uniformly between 12pm and 1pm, find the probability that the first to arrive has to wait longer than 10 mins?

Answer 1

the probability that the first to arrive has to wait longer than 10 mins is 35/36.

Explanation:

First you need to denote by X and Y the time past 12 noon that the man and woman arrive.

Then you have to compute P(X+10<Y)+P(Y+10<X), which by symmetry equals 2P(X+10<Y).

So basically you have to compute P(X+10<Y).

check the attachment for the rest of the step by step.

Answer should be 25/72

But since we want 2× of the above probability 2× 25/72 = 25/36 final answer.