Question

A man and a woman agree to meet at a certain location about 12:30 PM. Suppose the man arrives at a time uniformly distributed between 12:15 and 12:45. Suppose the woman independently arrives at a time uniformly distributed between 12:00 and 1 PM. Find the probability that the man arrives first.

Answer 1

Answer:

The probability that the man arrives first is 0.50.

Step-by-step explanation:

Let X denote the arrival time of the man and Y denote the arrival time of the woman.

It is provided that:

$X\sim U[15,45]\\\\Y\sim U[0,60]$

The pdf of X an Y are:

$f_{X}(x)=\frac{1}{45-15}=\frac{1}{30};\ 15<X<45\\\\f_{Y}(y)=\frac{1}{60-0}=\frac{1}{60};\ 0<Y<60$

It is provided that X and Y are independent of each other.

Then the joint pdf of X and Y is:

$f_{XY}(x,y)=f_{X}(x)\times f_{Y}(y)\\\\\Rightarrow f_{XY}(x,y)=\frac{1}{30}\times \frac{1}{60}\\\\\Rightarrow f_{XY}(x,y)=\frac{1}{1800};\ 15<X<45,\ 0<Y<60$

Compute the probability that the man arrives first as follows:

$P(X<Y)=\int\limits^{45}_{15} {\int\limits^{60}_{x} {\frac{1}{1800}} \, dy } \,dx$

                 $=\frac{1}{1800}\times \int\limits^{45}_{15} y|^{60}_{x} \,dx\\\\=\frac{1}{1800}\times \int\limits^{45}_{15} {60-x} \,dx\\\\=\frac{1}{1800}\times |60x-\frac{x^{2}}{2}|^{45}_{15}\\\\=\frac{(1687.5-787.5)}{1800}\\\\=\frac{900}{1800}\\\\=\frac{1}{2}\\\\=0.50$

Thus, the probability that the man arrives first is 0.50.