1. ## probability help..poisson process

During normal working hours, the help desk of a car repair company(also repairs small amount of motorbikes) receives telephone calls according to a Poisson process at the rate of 25 calls per hour.
it has been noted that 5% of calls relate to the company’s motorbike repairs.

Calculate the probability that exactly four calls are received between 10.20am and 10.30 am, exactly one of which relates to motorbike repairs.

Exactly eight calls are received between 10.20am and 10.30am one morning. Calculate the probability that exactly two of them relate to motorbike repairs
can any one show me how this is done please , thanks.

2. Originally Posted by frank567

During normal working hours, the help desk of a car repair company(also repairs small amount of motorbikes) receives telephone calls according to a Poisson process at the rate of 25 calls per hour.
it has been noted that 5% of calls relate to the company’s motorbike repairs.

Calculate the probability that exactly four calls are received between 10.20am and 10.30 am, exactly one of which relates to motorbike repairs.

Exactly eight calls are received between 10.20am and 10.30am one morning. Calculate the probability that exactly two of them relate to motorbike repairs
can any one show me how this is done please , thanks.
First to calculate $\lambda$ we need to find the average number of calls in a ten minute interval this gives

$\displaystyle \frac{25 \text{calls}}{1 \text{hr}}=\frac{25 \text{calls}}{1 \text{hr}}\frac{1 \text{ hr}}{60 \text{min}}=\frac{2.5 \text{ calls}}{10 \text{min}} \implies \lambda =2.5$

This gives the probability function

$f(k)=\frac{(2.5)^ke^{-2.5}}{k!}$

This gives $f(4) \approx .1336$

The 2nd part of this is a binomial distribution with

$g(k)=\binom{4}{k}(.05)^k(.95)^{4-k}$

This gives $g(1) \approx .1715$

So the probability of them both happening is

$f(4)\cdot g(1) \approx 0.0229$

Now you try the other one!