Question

A bus arrives at a bus stop at a uniformly distributed time over the interval 0 to 1 hour. A passenger also arrives at the bus stop at a uniformly distributed time over the interval 0 to 1 hour. Assume that the arrival times of the bus and passenger are independent of one another and that the passenger will wait for up to 1/4 hour for the bus to arrive.What is the probability that the passenger will catch the bus?

Answer 1

Answer:

0.25 is the probability that the passenger will catch the bus.          

Step-by-step explanation:

We are given the following information in the question:

Let X denote the bus arrival times and Y denote the passenger arrival time.

Here we will evaluate the joint density of X and Y.

$f(x) = \displaystyle\frac{1}{b-a} = \frac{1}{1-0} = 1\\\\f(y) = \displaystyle\frac{1}{b-a} = \frac{1}{1-0} = 1$

Since X and Y are independent events:

$f(X,Y) = f(x).f(y) = 1.1 =1$

Now, we have to evaluate:

$P( y < x < y+\frac{1}{4})\\\\=\displaystyle\int_0^1\int_{y}^{y+\frac{1}{4}} f(x,y) dxdy\\\\=\displaystyle\int_{0}^{1} [y]_{y}^{y+\frac{1}{4}} dx= \int_0^1 \frac{1}{4}dx = \frac{1}{4} = 0.25$

0.25 is the probability that the passenger will catch the bus.