Answer to Question #175141 in Statistics and Probability for Bindhu

Question

A passenger arrives at a bus stop at 10 am knowing that bus will arrive at random time between 10 am and 10:30 am. What is the probability that he will have to wait lower than 10 minutes?If the bus has not arrived until 10 : 15 am, what is the probability that he will have to wait another 10 minutes?

Expert's answer

Answer 1

Solution:

Given that Bus arrives uniformly at the bus stop between 10 am and 10 :

30 am. Let X denote the arrival of the bus at the bus stop. Then, X is a uniform random variable having density as

$f(x)=\frac{1}{30} \quad ; x \in[0,30]$

(a): Probability that a passenger arriving at the bus stop between 10 and 10: 30 will have to wait less than 10 minutes.

$P[X<10]=1-P[X\ge10]=1-\int_{10}^{30} f(x) d x$

$=1-\int_{10}^{30} \frac{1}{30} d x=1-\frac{1}{30}[x]_{10}^{30}=1-\frac{1}{30}[30-10]=1-\frac{20}{30}=\frac{1}{3}$

(b): The bus hasn't arrived till 10: 15 am. The probability that the passenger will have to wait for an additional 10 minutes is given as

$\begin{aligned} P[X=25 \mid X>15]=\frac{P[X=25 \cap X>15]}{P[X>15]} &=\frac{P[X\ge25]}{P[X>15]} \\ &=\frac{\int_{25}^{30} \frac{1}{30} d x}{\int_{15}^{30} \frac{1}{30} d x} \\ &=\frac{\frac{1}{30}[30-25]}{\frac{1}{30}[30-15]} \\ &=\frac{5}{15} \\ &=\frac{1}{3} \end{aligned}$