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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?

Chloe
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  • It is asking what is the probability that 2 or more cars are hired on any one day (The break even point is 1 hire as that brings in $$250$ and the overheads are $10 \times 25 = $250$) given that the number of hires has a Poisson distribution with mean 8. – Conrad Turner Mar 26 '15 at 11:59

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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.

Jonathan Miller
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