# Intro to Probability

two construction contracts are to be randomly assigned to one or more of 3 firms- I,II,III. A firm may receive more than one contract. if each contract has a potential profit of 90,000 dollars. find the expected potential profit for firm I. then find the expected potential profit for firms I and II together.
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Can i start by using the expected value formula?

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3. 👁 1,189
1. Yes, this looks like an expected value problem. The probability of one or the other getting a contract both times (2/3)^2 = 4/9. The probability of I or II getting one only is 2*1/3*2/3= 4/9. The probability of missing out both times is 1/9.
E.V. = (4/9)*180,000 + (4/9)*90,000) = ?

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2. thank you drwls for your help I greatly appreciate it.

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3. For firm 1
Probability of getting contract a only = probability of getting a - probability of getting both = 1/3 - 1/9 = 2/9

Probability of getting contract b only = 2/9

Probability of getting a and b both = 1/9
so
E(\$) = (4/9)90 k + (1/9) 180 k
= 40 + 20
= \$60 K

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4. I get 120K. With a total of 180K awarded, it seems to make sense that two of the three firms would, on the average, get 2/3 of it

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5. for 2 firms together:
There are still 9 possible configurations
probability of 1 getting A and 2 getting B = probability of 1 getting B and 2 getting A = 1/9
so (2/9) * 180 k = 40k so far

Then 1 gets 0, 2 gets A 1/9
1 gets 0, 2 gets B 1/9
1 gets A, 2 gets 0 1/9
1 gets B, 2 gets 0 1/9
so 4/9 * 90k = 40k

Then 1 gets AB 1/9
2 gets AB 1/9
so again
(2/9)*180 k = 40 k
so in the end

3 * 40 k = 120 k

Which is just what drwls said, but done in a much more difficult if simple way.

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6. drwls
I did the one firm problem first.

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7. I think you missed the first of 2 parts of the question.

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8. thanks damon

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9. Yes, I see that now. I just did the two firms together case. I should not skim through questions

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10. 1/9

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