Q. 5.13 - A First Course in Probability

Question

You arrive at a bus stop at $10$ a.m., knowing that the bus will arrive at some time uniformly distributed between $10$ and $10:30$.

(a) What is the probability that you will have to wait longer than   $10$ minutes?

(b) If, at $10:15$, the bus has not yet arrived, what is the probability that you will have to wait at least an additional $10$ minutes?

Step-by-Step Solution

Answer

(a) The probability that the waiting time will be longer than $10$minutes is $\frac{2}{3}$.

(b) The probability of waiting at least an additional $10$ minutes is $\frac{1}{3}$.

Part (a) Step 1. Given Information.

Here, it is given that the arrival time of bus is uniformly distributed between $10 - 10.30 \text{a.m.}$

A passenger arrives at the bus stop at $10 \text{a.m}$.

Part (a) Step 2. Determine the probability density function.

Let $X$ be the random variable that represents the person's waiting time in the bus stop. Then the probability density function is given by:

$f\left(x\right)=\frac{1}{b-a}=\frac{1}{30}$

Part (a) Step 3. Calculate the probability that the waiting time of passenger will be more than 10 minutes.

$$\begin{gathered} P(X>10) = {\int }_{10}^{30}f\left(x\right) dx \\ ={\int }_{10}^{30}\frac{1}{30} dx \\ =\frac{1}{30}{\left(30\right)}_{10}^{30} \\ = \frac{1}{30}\left(30-10\right) \\ =\frac{2}{3} \end{gathered}$$

Therefore, the probability that the waiting time of passenger will be more than$10$ minutes is $\frac{2}{3}$.

Part (b) Step 1. Calculate the probability that the waiting time of passenger will be at least additional 10 minutes.

The probability will be:

$P\left(X\ge 25 \overline{)X>15}\right) = \frac{P(25\le X\le 30)}{P(X>15)}$

$$\begin{gathered} =\frac{{\int }_{25}^{30}f\left(x\right) dx}{{\int }_{15}^{30}f\left(x\right) dx} \\ =\frac{{\int }_{25}^{30}{\displaystyle \frac{1}{30}} dx}{{\int }_{15}^{30}{\displaystyle \frac{1}{30}} dx} \\ =\frac{{\left[{\displaystyle \frac{1}{30}}\right]}_{25}^{30}}{{\left[{\displaystyle \frac{1}{30}}\right]}_{15}^{30}} \\ =\frac{{\displaystyle \frac{1}{30}}\left(30-25\right)}{{\displaystyle \frac{1}{30}}\left(30-15\right)} \\ =\frac{1}{3} \end{gathered}$$

Therefore, the probability that the waiting time of passenger will be at least additional $10$ minutes is $\frac{1}{3}$.