1. ## Poisson Probability

The number of bugs in a computer is a Poisson random variable with a rate parameter of 3 bugs a year. When a bug occurs, a technician A is called to fix it with probability 0.5, a technician B is called to fix it with probability 0.2 and a technician C with probability 0.1. The probability that C won't be called to fix any bug during a year is:

a. e^-2.7 b. e^-0.3 c. 1-e^-2.7 d. 1-e^-0.3

cheers !!

2. Start by finding the probability that c is not called, given that there are n bugs.

P(C' | n) = 0.9^n

Now, the required probability is:

P(n=0) * 1
+P(n=1)*0.9
+P(n=2)*0.9^2
...

We are told the distribution of N. $\displaystyle P(N=n) = \frac{e^{-3} 3^n}{n!}$

So the sum you want to do is
$\displaystyle \sum \frac{e^{-3} 3^n *0.9^n}{n!}$

$\displaystyle \sum \frac{e^{-3} (3*0.9)^n }{n!}$

$\displaystyle \sum e^{-0.3} \frac{e^{-2.7} (2.7)^n }{n!}$

$\displaystyle e^{-0.3} \sum \frac{e^{-2.7} (2.7)^n }{n!}$

You are supposed to recognise the above as
$\displaystyle e^{-0.3} * P(Y< \infty)$
where Y is a poisson(2.7) variable.

So the answer is $\displaystyle e^{-0.3}$

3. Talking about Simeon-Denis Poisson , today is the birthday of him (229 years old ) .

4. thats a different kettle of fish entirely ^^