Question

A worker leaves for work between 9:00am and 9:30am and takes between 45 and 55 minutes to arrive. Let the random variable Y denote this worker’s time of departure, and the random variable X the travel time. Assuming that Y and X are independent and uniformly distributed, find the probability that the worker arrives at work before 10:00am.

Answer 1

Answer:

Given,

Let us consider,

X = U(0 , 30)

Y = U(45,55)

Now we have to determine P(X+Y < 60)

P(X+Y < 60) = fx(x)*fy(y) dx dy

= 1/30 * 1/(55-45) dx dy

= 1/30*1/10 dx dy

= 1/300 * (60-y) dy

= 1/300 * (60y - y^2/2) | interval limits (45 to 55)

= 1/300 * [(60*55 - 55^2/2) - (60*45 - 45^2/2)]

= 1/300 * 100

= 0.333