# Using probability to calculate profit

Q) A building contractor has submitted bids for three jobs. If the contractor obtains the jobs, they will yield respective profits of £20000, £25000 and £40000. On the other hand, for each job that the contractor does not win, due to time and money already spent on making the bid he will incur a loss (interpreted as a negative profit) of £2,000. If the probabilities that the contractor will get these jobs are, respectively, 0.3, 0.6, and 0.2, what is the expected total profit?

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The total expected profit is the sum of expected profits from each bid. For the first bid the expected profit is $$20000*0.3 + (-2000)*(1-0.3) = 4600$$ Similarly find expected profits from other bids, and add them up.

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@AndréNicolas Yes, thank you. I realized that as well. – Sasha May 10 '12 at 1:47

Here's a straightforward approach:

The first thing you should do is define the random variable of interest, list its values, and list their respective probabilities. Not only will this help you in the computations that follow, but it it will relieve your reader from the task of surmising your intent.

Here, we are interested in the profit the contractor makes, so naturally we will let $X$ be his profit. I will use units of thousands of pounds. The values $X$ could possibly take can be easily enumerated if we divide the possibilities up into how many jobs he wins:

One job only:

$X$ takes the values:

$\ \ \ \ \ 20-4=16$ (he gets the first job only, note then he "loses" 4k pounds since he didn't get the $\qquad \qquad \ \ \ \ \ \ \ \ \ \ \ \ \$other two jobs),

$\ \ \ \ \ 25-4=21$ (he gets the second job only),

and

$\ \ \ \ \ 40-4=36$ (he gets the third job only).

Let's go ahead and calculate probabilities here. (It turns out that the values of $X$ for the other cases are distinct from the ones found so far; so what follows is valid)

I'll do one example and leave the others for you. The probability that $X$ takes the value $16$ is the probability that the contractor obtained the first (with the $20$ profit) job only. He did not get the other two. Assuming independence, the probability that he got the first, and did not get the second, and did not get the third, is

$\ \ \ \ \ P[X=16] = (0.3)(1-0.6)(1-0.2)$.

Two jobs only:

$X$ takes the values

$\ \ \ \ \ 20+25-2=43$,

$\ \ \ \ \ 20+40-2=58$,

and

$\ \ \ \ \ 25+40-2=63$.

As an example calculating probabilities here, let's find $P[X=63]$. Here, the contractor got the second and third jobs but not the first. So

$\ \ \ \ \ P[X=63]=(1-.3)(.6)(.2)$.

Three jobs:

$X$ takes only one value:

$\ \ \ \ \ 20+25+40=85$.

With probability

$\ \ \ \ \ P[X=85]=(.3)(.6)(.2)$

And lest we forget, $X$ can take the value $-6$ if the contractor gets none of the jobs. This has probability $(1-.3)(1-.6)(1-.2)$ of occurring.

Now that we have all the information needed, we can find the expected profit. The expected value of a discrete random variable $X$ is $$\Bbb E(X)=\sum_x x\cdot P[X=x],$$ where the sum is taken over all distinct values that $X$ takes.

So to find the expected profit, compute the sum

$$16\cdot p_{16}+21\cdot p_{21}+36\cdot p_{36}+43\cdot p_{43}+58\cdot p_{58}+63\cdot p_{63}+85\cdot p_{85} +(-6)\cdot p_{-6},$$ where $p_x=P[X=x]$.

That sum is the expected profit in thousands of pounds.

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