Question

Suppose a bus arrives at a bus stop every 30 minutes. If you arrive at the bus stop at a random time what is the probability that you will have to wait at least 5 min for the bus?
(a) 1/6
(b) 1/5
(c) 2/3
(d) 5/6

Answer 1

C

Answer 2

Let X be the waiting time random variable. Then we can say that X has a continuous uniform distribution, \(X\sim U(0,30)\).

Then the question asks to find \(P(X\ge 5)\)
\(P(X\ge 5)= 1-P(X \le 5)\)

Now recall that the CDF for a conitnuous random variable with distribution \(U(a,b)\) is:
\[P(X \ge x)=\frac{x-a}{b-a}, \text{for }x\in[a,b) \]
So
\[1 - P(X \le 5) =1-\frac{5-0}{30-0}=\frac{5}{6}\]

Answer 3

thank you

Answer 4

Also, if you think of it intuitively, since at least 5 mins means 5 to 30 mins... that time interval is 25 mins. And 25 min divided by the whole 30 mins = 25/30 = 5/6

Answer 5

yw :)