Poisson distribution - Number of Failures The number of machine failures per day in a certain factory has a Poisson Distribution with parameter $\lambda=3$.
At present maintenance facilities can repair 3 machines per day. Failures in excess of 3 are repair by a contractor.
a) On a given day, what is the probability of having machines repaired by the contractor?
I did:
$P(X>3)=1-P(X\le3)$  which is
$1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)]$
Using the Poisson distribution formula $P(X=x)\frac{e^-\lambda.\lambda^x}{x!}$ 
I got that:
$P(X>3)=1-P(X\le3)$=
$1-0.6423$=0.3577
Can anyone let me know if I am thinking correctly?
b) What is the most probable number of machine failures per day?
I was thinking of doing:
Using the Poisson Distribution table to calculate:
P(X=0)
P(X=1)=P(X<=1) - P(X=0)
P(X=2)=P(X<=2) - P(X=1)
etc. until X=11
Is this correct?
c) Find the expected number of machines repaired in the factory each day, that is those not repaired by the contractor.
I did:
I know that $\mu=E(X)=\lambda$ in this case the expected number is 3 since $\lambda = 3$  
d) What is the expected number of machines repaired by the contractor each day?
I am not sure how to do this one. Since in this case $\lambda =3$
Can anyone help me on this one?
 A: a)  The thinking is right. I have not checked the arithmetic.
b) You need to find out which one (?) of $\Pr(X=0)$, $\Pr(X=1)$, $\Pr(X=2)$, $\Pr(X=3)$, $\Pr(X=4)$, and so on is largest. You could calculate (best done without ever touching a calculator). The probabilities are 
$$e^{-3}, \quad e^{-3}\frac{3^1}{1!},\quad e^{-3}\frac{3^2}{2!},\quad e^{-3}\frac{3^3}{3!},\quad e^{-3}\frac{3^4}{4!}, \quad e^{-3}\frac{3^5}{5!},$$
and so on. Note that $\Pr(X=n)$ reaches a maximum at two places, $n=2$ and $n=3$. To get the "$n+1$" term from the "$n$" term, we multiply by $\frac{\lambda}{n+1}$. So probabilities typically increase and then start to decrease. If $\lambda$ is an integer, we get a two way tie for largest probability.
c)  For in-house repairs, the number of failures must be $0$, $1$, $2$, or $3$. The expected number of such repairs is $1\cdot\Pr(X=1)+2\cdot\Pr(X=2)+3\cdot\Pr(X=3)$.
d) The total expected number of repairs is $3$, by a standard property of the Poisson. So the expected number of contractor repairs is $3$ minus the answer to Question c).
