A particular product that you consider for marketing will either sell well or be

Question

A particular product that you consider for marketing will either sell well or be a bust. You assess probability 1/3 that it will sell well and 2/3 that it will be a bust. Consumers will be selected at random and asked whether they like the product. If the product will sell well, there is probability 6 that any consumer will say that he or she likes the product. If the product will be a bust, there is probability 06 that any consumer will say that he or she likes the product. (a) What is the probability that a given consumer will say that he or she likes the product? (b) Suppose that the first consumer you ask says that he or she likes the product. After hearing the first consumers response, what do you now assess is the probability that the product will sell well? (c) If you now ask a second consumer, what do you assess is the probability that this second consumer will say that he or she likes the product? (d) Suppose five consumers are chosen at random and asked whether they like the product. What is the probability that three of the five will say that they like it? Can you use the binomial distribution to compute that probability? Why or why not?

Explanation / Answer

P(sell well) = 1/3

P(bust) = 2/3

P(likes the product | sell well) = 0.6

P(likes the product | bust) = 0.06

a) P(likes the product) = P(likes the product | sell well) * P(sell well) + P(likes the product | bust) * P(bust)

                                    = 0.6 * 1/3 + 0.06 * 2/3

                                    = 0.24

b) P(sell well | likes the product) = P(likes the product | sell well) * P(sell well) / P(likes the product)

                                                    = 0.6 * (1/3) / 0.24

                                                    = 0.8333

c) P(second consumer also likes the product) = 0.24

d) Yes this is a bnomial distribution. Because the event is independent

p = 0.24

n = 5

P(X = x) = 5Cx * 0.24x * (1 - 0.24)5-x

P(X = 3) = 5C3 * 0.243 * 0.762 = 0.0798

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