# Solved: If the number of minutes that a doctor spends with a patient is a random variable having an exponential distribution

#### ByDr. Raju Chaudhari

Feb 24, 2021

If the number of minutes that a doctor spends with a patient is a random variable having an exponential distribution with the parameter $\theta = 9$, what are the probabilities that it will take the doctor at least 20 minutes to treat

(a) one patient;
(b) two patients;
(c) three patients?

#### Solution

We use the relation between Exponential distribution and Poisson distribution to solve this question.

Let $N$ denote the number of patients that have been take care of during 20 minutes.

Then $N$ has a Poisson distribution with parameter $\lambda = \theta \frac{20}{60} = 9\times \frac{20}{60} = 3$.

The pmf of $N$ is

 \begin{aligned} P(N=n) &= \frac{e^{-\lambda}\lambda^n}{n!};\; n=0,1,2,\cdots\\ &= \frac{e^{-3}3^n}{n!};\; n=0,1,2,\cdots \end{aligned}

(a) The probability that the first patient is still with the doctor after 20 minutes is the probabilities that it will take the doctor at least 20 minutes to treat.

Thus required probability is

 \begin{aligned} P(N=0) &= \frac{e^{-3}3^0}{0!}\\ &= 0.0497871 \end{aligned}

(b) The probability that it will take the doctor at least 20 minutes to treat two patient is

 \begin{aligned} P(N=1) &= \frac{e^{-3}3^1}{1!}\\ &= 0.1493612 \end{aligned}

(c) The probability that it will take the doctor at least 20 minutes to treat three patient is

 \begin{aligned} P(N=2) &= \frac{e^{-3}3^2}{2!}\\ &= 0.2240418 \end{aligned}