Aman and a woman agree to meet at a cafe about noon. if the man arrives at a time uniformly distributed between 11: 40 and 12: 10 and if the woman independently arrives at a time uniformly distributed between 11: 45 and 12: 20, what is the probability that the first to arrive waits no longer than 5 minutes?

0.3888

Step-by-step explanation:

To solve the exercise we need a neutral time, so we define at 12:00 as time 0.

If the man arrives in a uniformly distributed time between 11:40 and 12:10, the time is distributed from -20 to 10.

If the woman arrives in a time evenly distributed between 11:45 and 12:20, the time is distributed from -15 to 20.

Assuming that each one arrives at 12: X and 12: Y for men and women, then the space of our time is defined as

Man [-20,10]

Woman [-15,20]

So,

Then,

The probability of finding an arrival less than 5 minutes is

on plase n

we have been given a piecewise function which has three parts

we can graph each part one by one.

first part is x+1 which is just like equation y=x+1

restriction with first part is x< 0 so that means we can use any number for x which is less than 0 for example x=-1 and x=-2

now plug x=-1 into y=x+1 to get y

y=x+1=-1+1=0

so the point will be (-1,0)

similarly we get another point (-2,-1)

now we graph a line joining both points but keep in mind that line must be on left side of x=0 because of restriction x< 0

same way we can graph 2nd and third parts

if we have < or > then we use open circle.

if we have < = or > = then we use close circle.

so using above information, final graph can be drawn as: