Question

Suppose that the number of minutes that you need to wait for a bus is uniformly distributed on the interval [0, 15]. If you take the bus seven times, what is the probability that your longest wait is less than 12 minutes

Answer 1

0.2097 = 20.97% probability that your longest wait is less than 12 minutes

Explanation:

To solve this question, we need to understand the uniform and the binomial distribution.

Uniform distribution:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distribution has two bounds, a and b, and the probability of finding a value lower than x is given by:

`P(X < x) = (x - a)/(b - a)`

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Probability of a single bus having a waiting time of less than 12 times:

Uniformly distributed on the interval [0, 15] means that
`a = 0, b = 15`

`P(X < 12) = (12 - 0)/(15 - 0) = 0.8`

What is the probability that your longest wait is less than 12 minutes?

This is the probability that all 7 buses have waiting time less than 12 minutes, which is
`P(X = 7)` when
`n = 7`, with
`p = 0.8`. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 7) = C_(7,7).(0.8)^(7).(0.2)^(0) = 0.2097`

0.2097 = 20.97% probability that your longest wait is less than 12 minutes