Question

URGENT QUICK PROBABILITY QUESTIONS

Problem #2

Part a: A data scientist is is building a model for a business to evaluate various metrics about the products and outlook. In looking at the data, she decides that a product should be continued if it sold 8,000 over the previous year. In addition, the product is considered “popular” if it receives 200 mentions by the local press over the past year.

In selecting a product at random from the company’s web page, let C be the likelihood that this product sold 8000 products the past year. Let P be the likelihood that the product received the 200 or more mentions by the local press.

The analyst determines that P(C) = 0.307, P(P) = 0.194, and the probability that a product has sold 8000 items, and was ‘popular’ is 0.083. What is the probability that a randomly selected product either sold the requisite 8000 items, or that it is ‘popular’?

Part b: Where would the analyst have come up with the probaility values for P(C) andP(P)?

Problem #3

At a insurance sales conference, exactly 100 people are selected at random from conference registration records. Under “specialty”, 25 people indicated that they are specialize in health insurance, 36 said they focused on life insurance, and 24 said “Other”. The remaining individuals specified either Car insurance or Home Insurance. The rest of the individuals did not specify any specialty.

a)     If you were to pick 3 individuals at random, what is the probability that the first one specializes in life insurance, and the second one also specializes in life insurance?

b)      If you were to pick 3 individuals at random, what is the probability that the first one specializes in life insurance, and the second one specializes in car insurance?

c)       If you were to pick 3 individuals at random, what is the probability that at least one specializes in health insurance?

Answer 1

Problem 3:

a ) required probability = 1st from life insurance × 2nd from life insurance × 3rd from the remaining 98 = {(36C1)(35C1)(98C1)} / (100C3) = 0.7636

b ) required probability = 1st from life insurance × 2nd from car insurance × 3rd from the remaining 98 = {(36C1)(15C1)(98C1)} / (100C3) = 0.3273

c ) required probability = there can be three cases = 1 specializes in health insurance + 2 specialises in health insurance + all the 3 specialises in health insurance = {(25C1)(74C2)+(25C2)(73C1)+(25C3)} / 100C3 = 0.5673