The probability that a randomly selected box of a certain type of cereal has a p

Question

The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.1. Suppose you purchase box after box until you have obtained three of these prizes. (a) What is the probability that you purchase x boxes that do not have the desired prize? b(x; 3, 0.1) b(x: 3, 1, 10) nb(x; 3, 1, 10) h(x; 3, 1, 10) b(x; 3, 0.1) h(x; 3, 0.1) (b) What is the probability that you purchase five boxes? (Round your answer to four decimal places.) (c) What is the probability that you purchase at most five boxes? (Round your answer to four decimal places.) (d) How many boxes without the desired prize do you expect to purchase? How many boxes do you expect to purchase?

Explanation / Answer

p = 0.1

This is negative binomial distribution.

a) Option-E) nb(x; 3, 0.1)

b) P(X = 5) = 4C2 * 0.12 * 0.92 * 0.1 = 0.0049

c) P(X < 5) = P(X = 3) + P(X = 4) + P(X = 5)

                  = 0.13 + 3C2 * 0.12 * 0.9 * 0.1 + 4C2 * 0.12 * 0.92 * 0.1

                  = 0.001 + 0.0027 + 0.0049

                  = 0.0086

d) Expected number of boxex without the desired prize to be purcahse = 1/0.1 = 10

Expected number of boxex to be purcahse = 3/0.1 = 30

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