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solmaris [256]
3 years ago
15

A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h

as a 10% chance of winning a given bid and that winning bids are independent of one another. Let X-the number of bids won by the company.
a. Write out the PMF of X.
b. How many bids can the company expect to win?
c. What is the standard deviation of the number of bids won?
It costs the company $10,000 to prepare and submit all three bids. If the company wins the bid, it will produce S50,000 of income to the company. What is the expected profit for the company?
Mathematics
1 answer:
julsineya [31]3 years ago
5 0

Answer:

a.P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

b. E(x) = 0.3

c. S(x)=0.5196

d. E=5,000

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\

For binomial distribution:

E(x)=np\\S(x)=\sqrt{np(1-p)}

Therefore, the company can expect to win 0.3 bids and it is calculated as:

E(x) = np = 3*0.1 = 0.3

Additionally, the standard deviation of the number of bids won is:

S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196

Finally, the probability to won 1, 2 or 3 bids is equal to:

P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001

So, the expected profit for the company is equal to:

E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

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