Al Cascade ,president of the litre coporation is studying hiscompany's chances of being awarded an important water purificationsystem contract for the Tennessee Valley Authority.Accordingly twoevents are intrest to him.First ,litres major competitor,WTR isconducting purification research,which it hopes to complete beforecontract award deadline.Second,TWA is investigating all recentcontrcators, of which litres is one and WTR is not.If WTRfinishes its research and there is no investigation,then litresprobability of being awarded the contract is 0.67.If there is aninvestigation but WTR doesn't finish its research,the probabilityis 0.72.If both event occur , the probability is 0.58, if neitheroccurs,the probability is 0.85. The occurence of an investigationand WTR's completion of research in time are independentevents.

1.Suppose that AL knows that the probability of WTR'scompleting its research in time is 0.80.How low must theprobability of an investigation be so that the probability ofLitre's being awarded the contract is at least 0.65 ?
2.Suppose Al knows that the probability of an investigation is0.70.How low must the probability of WTR's completing its researchon time be so that the probability of litres being awarded thecontract is at least 0.65?
Suppose that the probability of an investigation is 0.75 andthe probability of WTR's completiong its research in time is0.85.What is the probability of litre's being awarded thecontract?

That is YOUR assignment. No one here will do your assignment for you. However, if you have questions about the assignment's instructions, be sure to ask.

To solve these questions, we can use the concept of conditional probability.

1. Let's denote the probability of an investigation as P(I) and the probability of WTR completing its research on time as P(W). We are given that P(W) = 0.80. We need to find the minimum value of P(I) such that the probability of Litre's being awarded the contract is at least 0.65.

From the information given, we can create the following probabilities:
P(A | ¬I, ¬W) = 0.85
P(A | I, ¬W) = 0.72
P(A | ¬I, W) = 0.67
P(A | I, W) = 0.58

Using the law of total probability, we have:
P(A) = P(A | ¬I, ¬W) * P(¬I, ¬W) + P(A | I, ¬W) * P(I, ¬W) + P(A | ¬I, W) * P(¬I, W) + P(A | I, W) * P(I, W)

Since P(¬I, ¬W) = (1 - P(I)) * (1 - P(W)), P(I, ¬W) = P(I) * (1 - P(W)), P(¬I, W) = (1 - P(I)) * P(W), and P(I, W) = P(I) * P(W), we can rewrite the equation as:
P(A) = 0.85 * (1 - P(I)) * (1 - P(W)) + 0.72 * P(I) * (1 - P(W)) + 0.67 * (1 - P(I)) * P(W) + 0.58 * P(I) * P(W)

We want to find the minimum value of P(I) such that P(A) ≥ 0.65. This can be formulated as an inequality:
0.85 * (1 - P(I)) * (1 - 0.80) + 0.72 * P(I) * (1 - 0.80) + 0.67 * (1 - P(I)) * 0.80 + 0.58 * P(I) * 0.80 ≥ 0.65

Simplifying the inequality will give us the answer.

2. Let's denote the probability of an investigation as P(I) and the probability of WTR completing its research on time as P(W). We are given that P(I) = 0.70. We need to find the minimum value of P(W) such that the probability of Litre's being awarded the contract is at least 0.65.

Using the same probabilities as in question 1, we can write the equation as:
P(A) = 0.85 * (1 - 0.70) * (1 - P(W)) + 0.72 * 0.70 * (1 - P(W)) + 0.67 * (1 - 0.70) * P(W) + 0.58 * 0.70 * P(W) ≥ 0.65

Again, we simplify the inequality to find the minimum value of P(W) that satisfies the condition.

3. Given P(I) = 0.75 and P(W) = 0.85, we can use the same equation as above to calculate P(A):
P(A) = 0.85 * (1 - 0.75) * (1 - 0.85) + 0.72 * 0.75 * (1 - 0.85) + 0.67 * (1 - 0.75) * 0.85 + 0.58 * 0.75 * 0.85

Evaluating this expression will give us the probability of Litre's being awarded the contract.