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.
[tex]f(x) = \displaystyle\frac{1}{b-a} = \frac{1}{1-0} = 1\\\\f(y) = \displaystyle\frac{1}{b-a} = \frac{1}{1-0} = 1[/tex]
Since X and Y are independent events:
[tex]f(X,Y) = f(x).f(y) = 1.1 =1[/tex]
Now, we have to evaluate:
[tex]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[/tex]
0.25 is the probability that the passenger will catch the bus.
