A firm has $10$ cars for hire, the hire charge for each of which is $250$ dollars per day. The overheads are $25$ dollars per car per day, whether or not they hired. If the daily demand for cars has a Poisson distribution with mean $8$, what is the probability that the firm makes a profit on any given day?
How do I tackle this question? I know that $\lambda$ is $8$, and know that obviously profit = gain - cost. where $X$ is less than or equal to $10$ I'm just not sure how to plug it all in?
Perhaps we can start with the distribution. $ X\sim\mathrm{Poi}(\lambda)$ with
$f(x)$ = $Pr(X=x)$ = $\frac{\lambda^x e^{-\lambda}}{x!}$
To have a mean of 8, means that the $\sum x\cdot f(x)$ = 8 = $\lambda$. So the Expectation is $\lambda$ and a little calculating will give you the variance Var$(X)=\lambda$.
We know that hired car or not there is loss of \$250.00 per day. (10 $\times$ 25 per car)
And if we hires only 1 car he make nothing. So he must hire at least 2 cars
Perhaps, think about what it means to make a profit and how you can apply probabilities to it.