# How to solve this conditional probability problem?

A bus arrives at a station every day at a random time between 1 PM and 1:30 PM. A person arrives at this station at 1 and waits for the bus. If at 1:15 the bus has not yet arrived, what is the probability that the person will have to wait at least an additional 5 minutes?

-

Assuming the probability distribution is uniform, the probability is $2/3$. The bus didn't come until 1:15, therefore there's now an uniform distribution over the remaining quarter hour. Due to the uniform distribution, for each $5$ minute interval the probability is the same. Thus the bus has a probability of $1/3$ to arrive in the first $5$ minutes of the probability, and $2/3$ in the remaining $10$ minutes. Only in the latter case the person will have to wait at least another $5$ minutes.

-

P(the person will have to wait at least an additional 5 minutes|bus hasn't arrived at 1:15)

=P(bus arrive at 1:20-1:30|bus hasn't arrived at 1:15)

=P(bus arrive at 1:20-1:30 & bus hasn't arrived at 1:15)/P(bus hasn't arrived at 1:15)

=P(bus arrive at 1:20-1:30)/P(bus hasn't arrived at 1:15)

=(1/3)/(1/2)

=2/3

-

"Random" is imprecise, the probability distribution of the arrival times has not been specified. We are presumably expected to assume, unreasonably, that number $T$ of minutes elapsed after $1$ o'clock until the arrival of the bus has uniform distribution over the interval $[0,30]$.

Let $A$ be the event $T\gt 20$ and $B$ be the event $T\gt 15$. We want $\Pr(A|B)$. By the usual formula, we have $$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$ We have $\Pr(A\cap B)=\Pr(A)=\frac{10}{30}$. Similarly, $\Pr(B)=\frac{15}{30}$.

Remark: There is a better informal way of arriving at the answer. However, the more general machinery we used above can be of great help in more complicated situations.

-