Question

Two construction contracts are to be randomly assigned to one or more of three firms: I, II, and III. Any firm may receive both contracts. Each contract will yield a profit of $90,000. Assume that firms I and II are actually owned by the same individual. Find the probability distribution of X, the owner's total profit.

Answer 1

Answer:

The owner's total profit is $120,000.

Step-by-step explanation:

Assume that:

X₁ = contract is assigned to firm 1

X₂ = contract is assigned to firm 2

The sample space for assigning the two contracts is:

S = {(I, I), (I, II), (I, III), (II, I), (II, II), (II, III), (III, I), (III, II) and (III, III)}

There are total of 9 possible combinations.

So, the probability of selecting any of the combination is, 1/9.

Compute the probability distribution of X₁ and X₂ as follows:

X₁      P (X₁)                      X₂        P (X₂)

0        4/9                        0           4/9

1         4/9                        1            4/9

2         1/9                        2           1/9

Compute the expected values of X₁ and X₂ as follows:

$E(X_{1})=\sum X_{1}\cdot P(X_{1})\\\\=(0\times\frac{4}{9})+(1\times\frac{4}{9})+(2\times\frac{1}{9})\\\\=\frac{6}{9}\\\\=\frac{2}{3}$               $E(X_{2})=\sum X_{2}\cdot P(X_{2})\\\\=(0\times\frac{4}{9})+(1\times\frac{4}{9})+(2\times\frac{1}{9})\\\\=\frac{6}{9}\\\\=\frac{2}{3}$

It is provided that each contract will yield a profit of $90,000.

Compute the owner's total profit as follows:

$\text{Total Profit}=\text{Profit}\times [E(X_{1})+E(X_{2})]$

                   $=90000\times[\frac{2}{3}+\frac{2}{3}]\\\\=120000$

Thus, the owner's total profit is $120,000.