Question

The time between customers is exponentially distributed with a mean of 0.5 minutes. Given that 60 minutes have elapsed since the last customer, find the probability that we must wait an additional 5 mins for the arrival of the next customer

Answer 1

Answer:

The probability that we must wait 5 more minutes is 0.0000454

Step-by-step explanation:

Let X be the amount of minutes that passed between the two customers. X has emponential distribution with parameter $\lambda = 1/0.5 = 2$ and its density is

$f_X(x) = 2 \, e^{-2 \, x}$

We want to compute is P(X> 65 | X > 60), which is equal to P(X>65)/P(X>60) (Because the conditional event is contained in the other event). Lets do some computations

$P(X>60) = \int\limits_{60}^{+\,\infty} {2e^{-2 \, x} } \, dx = - e^{-2x} \, |^{+\infty}_{\, 60} =e^{-120}$

$P(X>65) = \int\limits_{65}^{+\,\infty} {2e^{-2 \, x} } \, dx = - e^{-2x} \, |^{+\infty}_{\, 65} = e^{-130}$

As a consequence

$P(X > 65 | X > 60) = \frac{P(X > 65)}{P(X>60)} = \frac{e^{-130}}{e^{-120}} = e^{-10} = 0.0000454$

Note that this is the same result than just calculate P(X > 5), and 5 is, by the way, the difference between 60 and 65. This is commonly referred as the memoryless property: The past doesnt have influence in the future.

We conclude that the probability that we must wait 5 more minutes is 0.0000454.