# Exponential Probability

The number of arrivals at a car wash follows a Poisson distribution with an average of 7.50 cars per hour during business hours. We start to observe and record the arrival cars at 1pm of a particular day. Given that no car arrives in the first four minutes, what is the probability that the first car will arrive before 1.10pm?

Let Y be the time (in min) between 2 consecutive arrival cars. Mean no of arrivals is 0.125 per minute Variance = 64 Standard deviation = 8

I understand that this is a conditional probability. But I am not sure how to continue. Any hints please?

Let A be the event that 1 car arrives between $1.04\ \textrm{pm}$ and $1.10\ \textrm{pm}$. And B is the event that 1 car arrives in 10 minutes. Then it is asked for

$$P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}$$

The probability that one car arrives in 10 minutes given that 1 car arrives between $1.04\ \textrm{pm}$ and $1.10\ \textrm{pm}$ is $1$.

The probability that 1 car arrives between $1.04\ \textrm{pm}$ and $1.10\ \textrm{pm}$ is

$$P(A)=e^{-0.75}\cdot \frac{0.75^{ 1}}{1!}$$

Here the parameter is $\lambda\cdot \Delta t=0.125\cdot 6=0.75$

The probability that the first car will arrive before 1.10pm

$$P(B)=e^{-1.25}\cdot \frac{1.25^{ 1}}{1!}$$

The parameter is $\lambda\cdot \Delta t=0.125\cdot 10=1.25$

Thus $$P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}=\frac{e^{-0.75}\cdot \frac{0.75^{ 1}}{1!}}{e^{-1.25}\cdot \frac{1.25^{ 1}}{1!}}=e^{0.5}\cdot \frac{3/4}{5/4}=0.6\cdot e^{0.5}\approx 98.92\%$$

• thank u for the clear explanation! Commented Feb 21, 2017 at 10:37