Question

an aerospace company has submitted bids on two separate federal government defense contracts. the company president believes that there is a 42% probability of winning the first contract. if they win the first contract, the probability of winning the second is 71%. however, if they lose the first contract, the president thinks that the probability of winning the second decreases to 47%. a. what is the probability that they win both contracts? probability = b. what is the probability that they lose both contracts? probability = c. what is the probability that they win only one contract? probability =

Explanation:

Step1: Define probabilities

Let $P(A)$ be the probability of winning the first - contract and $P(B)$ be the probability of winning the second contract. We know that $P(A) = 0.42$, if they win the first contract, $P(B|A)=0.71$, if they lose the first contract, $P(B|\overline{A}) = 0.47$.

Step2: Calculate $P(A\cap B)$ (probability of winning both contracts)

By the formula for conditional probability $P(B|A)=\frac{P(A\cap B)}{P(A)}$, so $P(A\cap B)=P(A)\times P(B|A)=0.42\times0.71 = 0.2982$.

Step3: Calculate $P(\overline{A}\cap\overline{B})$ (probability of losing both contracts)

First, $P(\overline{A})=1 - P(A)=1 - 0.42 = 0.58$. And $P(\overline{B}|\overline{A})=1 - P(B|\overline{A})=1 - 0.47 = 0.53$. Then $P(\overline{A}\cap\overline{B})=P(\overline{A})\times P(\overline{B}|\overline{A})=0.58\times0.53 = 0.3074$.

Step4: Calculate $P((A\cap\overline{B})\cup(\overline{A}\cap B))$ (probability of winning only one contract)

We know that $P(A\cap\overline{B})=P(A)\times(1 - P(B|A))=0.42\times(1 - 0.71)=0.42\times0.29 = 0.1218$. And $P(\overline{A}\cap B)=P(\overline{A})\times P(B|\overline{A})=0.58\times0.47 = 0.2726$. Then $P((A\cap\overline{B})\cup(\overline{A}\cap B))=P(A\cap\overline{B})+P(\overline{A}\cap B)=0.1218 + 0.2726=0.3944$.

Answer:

A. Probability = 0.2982
B. Probability = 0.3074
C. Probability = 0.3944