# Solved: A general contractor has submitted two bids for two projects; A and B. The probability of getting project A

#### ByDr. Raju Chaudhari

Feb 25, 2021

A general contractor has submitted two bids for two projects; A and B. The probability of getting project A is 0.60. The probability of getting project B is 0.75. The probability of getting at least one of the projects is 0.85.

a. What is the probability that the contractor will get both projects?
b. Are the events of getting the two projects mutually exclusive? Explain using probabilities.
c. Are the events of getting the two projects independent? Explain using probabilities.
d. If the contractor gets project A, what is the probability that he will get project B?
e. If the contractor gets project B, what is the probability that he will get project A?

#### Solution

Let $A$ be the event of getting project $A$ and $B$ be the event of getting project $B$.

Given that $P(A) = 0.60$, $P(B) = 0.75$ and $P(A\cup B) =0.85$.

a. The probability that the contractor will get both projects is

 \begin{aligned} P(A\cap B) &= P(A) + P(B) - P(A\cup B)\\ &= 0.60+0.75 -0.85\\ &= 0.5 \end{aligned}

b. $P(A\cap B)\neq 0$, the events of getting the two projects are not mutually exclusive.

c. $P(A)\times P(B) = 0.60\times 0.75 = 0.45\neq P(A\cap B)$, the events of getting the two projects are not independent.
d. If the contractor gets project A, the probability that he will get project B is

 \begin{aligned} P(B|A) &=\frac{P(A\cap B)}{P(A)}\\ &= \frac{0.5}{0.60}\\ &= 0.8333 \end{aligned}
e. If the contractor gets project B, the probability that he will get project A is

 \begin{aligned} P(A|B) &=\frac{P(A\cap B)}{P(B)}\\ &= \frac{0.5}{0.75}\\ &= 0.6667 \end{aligned}