Answer:
a.![P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D%2Ap%5E%7Bx%7D%2A%281-p%29%5E%7Bn-x%7D%5C%5C)
b. ![E(x) = 0.3](https://tex.z-dn.net/?f=E%28x%29%20%3D%200.3)
c. ![S(x)=0.5196](https://tex.z-dn.net/?f=S%28x%29%3D0.5196)
d. ![E=5,000](https://tex.z-dn.net/?f=E%3D5%2C000)
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}\\](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D%2Ap%5E%7Bx%7D%2A%281-p%29%5E%7Bn-x%7D%5C%5C)
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}\\](https://tex.z-dn.net/?f=P%28x%29%3D%5Cfrac%7B3%21%7D%7Bx%21%283-x%29%21%7D%2A0.1%5E%7Bx%7D%2A%281-0.1%29%5E%7Bn-x%7D%5C%5C)
For binomial distribution:
![E(x)=np\\S(x)=\sqrt{np(1-p)}](https://tex.z-dn.net/?f=E%28x%29%3Dnp%5C%5CS%28x%29%3D%5Csqrt%7Bnp%281-p%29%7D)
Therefore, the company can expect to win 0.3 bids and it is calculated as:
![E(x) = np = 3*0.1 = 0.3](https://tex.z-dn.net/?f=E%28x%29%20%3D%20np%20%3D%203%2A0.1%20%3D%200.3)
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](https://tex.z-dn.net/?f=S%28x%29%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B3%280.1%29%281-0.1%29%7D%3D0.5196)
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](https://tex.z-dn.net/?f=P%281%29%3D%5Cfrac%7B3%21%7D%7B1%21%283-1%29%21%7D%2A0.1%5E%7B1%7D%2A%281-0.1%29%5E%7B3-1%7D%3D0.243%5C%5CP%282%29%3D%5Cfrac%7B3%21%7D%7B2%21%283-2%29%21%7D%2A0.1%5E%7B2%7D%2A%281-0.1%29%5E%7B3-2%7D%3D0.027%5C%5CP%283%29%3D%5Cfrac%7B3%21%7D%7B3%21%283-3%29%21%7D%2A0.1%5E%7B3%7D%2A%281-0.1%29%5E%7B3-3%7D%3D0.001)
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](https://tex.z-dn.net/?f=E%3D-10%2C000%2B50%2C000%280.243%29%2B100%2C000%280.027%29%2B150%2C000%280.001%29%5C%5CE%3D5%2C000)
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.