Suppose that 2% of cereal boxes contain a prize and the other 98% contain the message, "sorry, try again." consider the random variable x, where x = number of boxes purchased until a prize is found. (round all answers to four decimal places.) (a) what is the probability that at most three boxes must be purchased? p (at most three boxes) = (b) what is the probability that exactly three boxes must be purchased? p (exactly three boxes) = (c) what is the probability that more than three boxes must be purchased? p (more than three boxes) =

Question

Diana Waller

Mathematics

27 December, 19:38

Suppose that 2% of cereal boxes contain a prize and the other 98% contain the message, "sorry, try again." consider the random variable x, where x = number of boxes purchased until a prize is found. (round all answers to four decimal places.) (a) what is the probability that at most three boxes must be purchased? p (at most three boxes) = (b) what is the probability that exactly three boxes must be purchased? p (exactly three boxes) = (c) what is the probability that more than three boxes must be purchased? p (more than three boxes) =

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Amaya Gonzalez · Answer 1

a) This is the sum of the probabilities of winning in 1, 2 or 3 boxes. 0.02 + (0.98*0.02) + (0.98*0.98*0.02) = 0.0588 b) Exactly three boxes requires the first two to lose. (0.98*0.98*0.02) = 0.0192 c) This is the complement of answer for a). 1-.0588 =.9412