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?

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

Step-by-step 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 probability2× 25/72 = 25/36 final answer.

The side opposite a 30° angle is the same as the side adjacent to a 60° angle in a right triangle. On a unit circle, the y (sin) distance of a 30° angle is the same as the x (cos) distance of a 60° angle.

:)) heheheheheheh