The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.4. Suppose you purchase box after box until you have obtained two of these prizes. (a) What is the probability that you purchase x boxes that do not have the desired prize? nb(x; 2, 0.4) Oh(x; 2, 0.4) b(x; 2,0.4) Ob(x; 2, 4, 10) Onb(x; 2, 4, 10) Oh(x; 2, 4, 10) (b) What is the probability that you purchase four boxes? (Round your answer to four decimal places.) (c) What is the probabilit

1 answer
Question:

The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.4. Suppose you purchase box after box until you have obtained two of these prizes.
(a) What is the probability that you purchase x boxes that do not have the desired prize?
nb(x; 2, 0.4)
Oh(x; 2, 0.4)
b(x; 2,0.4)
Ob(x; 2, 4, 10)
Onb(x; 2, 4, 10)
Oh(x; 2, 4, 10)
(b) What is the probability that you purchase four boxes? (Round your answer to four decimal places.)
(c) What is the probability that you purchase at most four boxes? (Round your answer to four decimal places.)
(d) How many boxes without the desired prize do you expect to purchase?

Answers

Answer:a. 0.6^xb. 0.3376 c. 0.6624d. 2 boxesStep-by-step explanation:probability that the selected box of  cereal has a  prize (P)is 0.4probability that the selected box of  cereal has no  prize(p') is 1-0.4 = 0.6a. probability that you purchase x boxes that do not have the desired prize= 0.6 *0.6 *0.6 *....up to the x times=0.6^x(b) probability that four boxes are purchased = [tex]P^{4} + P^{3} *Q + P^{2} *Q ^{2} + P *Q^{3} + Q^{4} \\\\= 0.4x^{4} + 0.4^{3} *0.6 + 0.4^{2} *0.6^{2} + 0.4 *0.6^{3} + 0.6^4\\\\ =0.3376[/tex](c) probability that you purchase at most four boxes=1 - probability that four boxes are purchased= 1- 0.3376 = 0.6624(d) The desired prize do you expect to purchase:     [tex]=\frac{Q}{1-Q} \\=\frac{0.6}{1-0.6} \\\\=\frac{0.6}{0.4} \\\\= 1.5[/tex]≈2 boxes

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