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

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