**Question:** The probability of a car passing a certain intersection in a 30 minute window is 0.95. What is the probability of a car passing the intersection in a 10 minute window? Assume that the cars are randomly distributed.

Try solving the problem yourself or keep reading for the solution.

### Solution

We can start by saying the probability of a car passing the intersection in 10 minutes is equal to P. Accordingly, the probability of not seeing a car in 10 minutes is equal to 1-P:

P(seeing 10) = P

P(not seeing 10) = 1-P

Consider coin flips – the probability of flipping heads 3 times in a row is given by 0.5^{3}. Similarly, to not see a car for 30 minutes, we would need to not see a car for 10 minutes three times in a row. This is the only real insight we need in order to solve the problem.

P(not seeing 30) = P(not seeing 10)

^{3}= (1-P)^{3}

Since the probability of seeing a car in 30 minutes is already given (0.95), we know that the probability of not seeing a car in 30 minutes is equal to 0.05. We can use this to solve for the probability of not seeing a car in 10 minutes by taking the cube root.

P(not seeing 10) = 0.05

^{1/3}= 0.3684

We originally wanted to know the probability of seeing a car in 10 minutes so we just need to subtract the probability of not seeing a car in 10 minutes from 1.

P(seeing 10) = 1 – 0.3684 = 0.6316

Alternatively, we could have done all the algebra in one go:

0.95 = 1 – (1 – P)

^{3 }0.05 = (1 – P)^{3}

0.05^{1/3}= 1 – P

0.3684 = 1 – P

1 – 0.3684 = P

0.6316 = P

If there is a 95% chance of seeing a car in 30 minutes, there is a 63% chance of seeing a car in 10 minutes.