Question

Pierre is waiting to be seated at a popular restaurant where the waiting time is a random variable with an exponential PDF, and the mean waiting time is 75 minutes. Pierre has already been waiting for 40 minutes. What is the probability that Pierre will have to wait more than 30 more minutes, given that he has already waited 40 minutes? Compute your answer rounded to 4 decimal places.

Answer 1

The PDF for the wait time (denoted by the random variable X) is

`f_X(x) = \begin{cases}\lambda e^(-\lambda x) & \text{if }x \ge 0 \\ 0 &\text{otherwise}\end{cases}`

where λ = 1/75. We want to find Pr[X > 70 | X ≥ 40]. Pierre has already been waiting for 40 min, so if he waits another 30 min he will have waited for a total of 70 min.

By definition of conditional probability,

Pr[X > 70 | X ≥ 40] = Pr[X > 70 and X ≥ 40] / Pr[X ≥ 40]

If X > 70, then automatically X ≥ 40 is satisified, so the right side reduces to

Pr[X > 70 | X ≥ 40] = Pr[X > 70] / Pr[X ≥ 40]

Use the PDF or CDF to find the remaining probabilities. For instance, using the PDF,

`\mathrm{Pr}[X > 70] = \displaystyle \int_(-\infty)^(70) f_X(x) \, dx = \int_0^(70) f_X(x) \, dx \approx 0.3932`

Or, using the CDF,

`F_X(x) = \displaystyle \int_(-\infty)^x f_X(t) \, dt = \begin{cases}0&\text{if }x<0 \\ 1-e^(-\lambda x) & \text{if }x \ge 0\end{cases}`

`\implies \mathrm{Pr}[X > 70] = 1 - \mathrm{Pr}[X \le 70] = 1 - F_X(70) \approx 0.3932`

Similarly, you'll find that Pr[X ≥ 40] ≈ 0.5866.

It follows that

Pr[X > 70 | X ≥ 40] ≈ 0.3932 / 0.5866 ≈ 0.6703