Question

A man and a woman agree to meet at a certain location about 12:30 p.m. If the man arrives at a time uniformly distributed between 12:15 and 12:45, and if the woman independently arrives at a time uniformly distributed between 12:00 and 1 p.m., find the probability that the first to arrive waits no longer than 5 minutes. What is the probability that the man arrives first?

Short answer

The probability that the first to arrive wants no longer that 5 minute is $0.1667$

And the probability that man arrives first is$0.5$

Step-by-step solution

Content Introduction

The probability that the first to arrive waits no longer than 5 minutes. also probability that the first to arrive .

Probability of arrival within 5 minutes

$$\begin{gathered} P\left(\right|X-Y|\le 5)=P(X-5<Y<X+5) \\ =P(x-5<Y<x+5) \\ ={\int }_{-\infty }^{\infty }{\int }_{x-5}^{\pi +5}{f}_{X,Y}(x,y)dydx \\ ={\int }_{15x-5}^{45}{\int }_{1800}^{15}\frac{1}{1800}dydx \\ =\frac{1}{1800}{\int }_{15}^{45}\left\{(y{)}_{x-5}^{x+5}\right\}dx \\ =\frac{1}{1800}{\int }_{15}^{45}10dx \\ =\frac{1}{1800}\times 10\times (x{)}_{15}^{45} \end{gathered}$$

$$\begin{gathered} =\frac{1}{1800}\times 10\times 30 \\ =\frac{1}{6} \\ \approx 0.1667 \end{gathered}$$

Therefore, the probability that the first to arrive wants no longer that 5 minute is $0.1667$

Compute probability that man arrives first.

$$\begin{gathered} P(X<Y)=P(x<Y)={\int }_{-\infty }^{\infty }{\int }_x{f}_{x,y}(x,y)dydx \\ ={\int }_{15}^{45}{\int }_x^{60}\frac{1}{1800}dydx \\ =\frac{1}{1800}{\int }_{15}^{45}\left\{(y{)}_x^{60}\right\}dx \\ =\frac{1}{1800}{\int }_{i5}^{45}(60-x)dx \\ =\frac{1}{1800}\times {\left(60x-\frac{x^2}{2}\right)}_{15}^{45} \\ =\frac{1}{1800}\times 900 \\ =\frac{1}{2} \end{gathered}$$