Question

Please answer the questions clearly

4. A company currently has two active projects, one in Asia and the other in Europe. Let A be the event that the Asian project is successful and B be the event that the European project is successful. Suppose that A and B are independent events with P(A)-0.4 and P(B)-0.17 (a) If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning using "common sense." Then, list which theorems are applicable to this problem (b) What is the probability that at least one of the two projects will be successful? (c) Given that at least one of the projects is successful, what is the probability that only the Asian project is successful? 5. Explain why there must be a mistake in each of the following statements. For each statement, provide at least one theorem or rule of probability from the text has been violated. (a) The probability that my cat, Sugar Bean, will scratch and then bite my dog, Ernie, on any given day is 0.34. The probability that Sugar Bean only scratches Ernie is 0.45. Therefore, the probability that Sugar Bean bites Ernie is 0.34-0.45.11 b) Ernie has 3 different colored balls and each day he can choose one of the three to play with. The probability that he chooses the blue ba is 0.7, the green ball is 0.3, and the orange ball is 0.1

Answer 1

4. (a) Bayes' Theorem is applicable to this problem

Probability that European project is not successful = 1 - 0.7 = 0.3

So, he answer is 0.3 because the events are independent.

(b) P(at least one of the project will be successful) = 1 - P(both projects wont be succssful)

= 1 - P(A') x P(B')

= 1 - (1 - 0.4)x(1 - 0.7)

= 1 - 0.6x0.3

= 1 - 0.18

= 0.82

(c) Bayes' theorem can be used to answer this

P(only Asian project is successful | at least 1 project is successful) = P((A and B') | (A or B))

= P(A and B') / P(A or B)

= (0.4x0.3) / (0.4 + 0.6x0.7)

= 0.12/0.82

= 0.146

5. (a) Probability of an event cannot be negative.

The rule of probability states that for any event A, 0 P(A) 1

(b) Probability rules says that, sum of probabilities within a sample space must be 1

Here, this rule is violated because, P(blue)+P(green)+P(orange)

= 0.7+0.3+.01

= 1.1 > 1, which is a violation of the above stated rule.