Question

A computer software company has submitted bids on two separate government contracts, A and B. The company estimates that it has a 36 percent chance of winning contract A and a 53 percent chance of winning contract B. The company believes that winning contract A is independent of winning contract B. Answer the following questions, rounding your answers to two decimal places where appropriate.
a) What is the probability that the company will win both contracts?
b)What is the probability that the company will win at least one of the two contracts?

Now suppose that the company believes that, given that it wins contract B, it has a 40 percent chance of winning contract A.
c)What is the probability that the company will win both contracts?
d) What is the probability that the company will win at least one of the two contracts?
e) If the company wins contract B, what is the probability that it will not win contract A?

Answer 1

Using probability concepts, it is found that there is a:

a) 0.1908 = 19.08% probability that the company will win both contracts.

b) 0.6992 = 69.92% probability that the company will win at least one of the two contracts.

c) 0.212 = 21.2% probability that the company will win both contracts.

d) 0.6992 = 69.92% probability that the company will win at least one of the two contracts.

e) 60% probability that it will not win contract A.

Conditional probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

• P(B|A) is the probability of event B happening, given that A happened.
• $P(A \cap B)$ is the probability of both A and B happening.
• P(A) is the probability of A happening.

Item a:

• 0.36 probability of winning contract A.
• 0.53 probability of winning contract B.

The probability of both is the multiplication of each probability, thus:

$p = 0.36(0.53) = 0.1908$

0.1908 = 19.08% probability that the company will win both contracts.

Item b:

• 1 - 0.36 = 0.64 probability of not winning contract A.
• 1 - 0.53 = 0.47 probability of not winning contract B.

The probability of winning none is:

$p_N = 0.64(0.47) = 0.3008$

The probability of winning at least one is:

$p = 1 - p_N = 1 - 0.3008 = 0.6992$

0.6992 = 69.92% probability that the company will win at least one of the two contracts.

Item c:

Now, conditional probability is used, considering $P(B) = 0.53, P(A|B) = 0.4$. Hence:

$P(A \cap B) = P(B)P(A|B) = 0.53(0.4) = 0.212$

0.212 = 21.2% probability that the company will win both contracts.

Item d:

If it doesn't wins contract A, the probability of not winning B is the same as in item B, so the probability of winning none is the same, which means that the probability of winning at least one is the same, that is:

0.6992 = 69.92% probability that the company will win at least one of the two contracts.

Item e:

• If it wins contract B, it has a 40% percent change of winning contract A, thus 100% - 40% = 60% probability that it will not win contract A.

A similar problem is given at https://brainly.com/question/25229135