Maximizing Expected Profit: A Contractor's Dilemma and How to Calculate It

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A contractor is evaluating a potential sale with a profit of $38,000 at a 70% probability and a loss of $16,000 at a 30% probability. The expected profit is calculated using the formula E = E[x*P(x)], resulting in a loss contribution of -$4,800 and a gain contribution of $26,600. The total expected profit is $21,800, which confirms the contractor's calculations are accurate. This approach effectively illustrates how to maximize expected profit in uncertain conditions.
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Expected profit, help please

1. A contractor is considering a sale that promises a profit of 38,000 with a probability of 0.7 or a loss (due to bad weather, strikes, and such) of 16,00 with a probability of .3. What is the expected profit?

2. I used the expected value formula, E= E[x*P(x)]

3. Then I computed it: Lose= -16000 * 0.3 -4800
Gain = 38,000 * 0.7 26600
Total= 21800
Therefore his expected profit is 21800. does this answer make sense? thank you
 
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Yes, it makes perfect sense.
 


Not only does it make sense, it is the correct answer!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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