The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.20.

Suppose that you purchase box after box until you have obtained 2 of these prizes.

(a) What is the probability that you purchase exactly 8 boxes?

(b) What is the probability that you purchase at least 9 boxes?

(c) How many boxes would you expect to purchase, on average?

a) the probability is 0.0734 (7.34%)

b) the probability is 0.7382 (73.82%)

c) 10 boxes

Step-by-step explanation:

Since each box is independent of the others and defining the random variable X = number of boxes of cereal purchased without prices until having 2 prices, then X has a negative binomial distribution. Thus

P (X=k) = C (k+r-1, k) * p^k * (1-p) ^r

where

p = probability to not obtain a price when a box is purchased = 1 - 0.2 = 0.8

C () = combinations

k = number of boxes without prices

r = number of boxes with prices=2

P (X=k) = probability of purchasing k boxes without prices until obtaining r boxes with prices

a) for k = 8 - r = 6

P (X=6) = C (7,6) * 0.8^6*0.2^2 = 7*0.8^6*0.2^2 = 0.0734 (7.34%)

b) for k≥9-2=7

P (X≥7) = 1 - P (X<7) = 1 - Fn (X=7)

where Fn (X) is the cumulative negative binomial distribution. We can calculate it through its relationship with the cumulative binomial distribution Fb (X) that is easily found in tables:

Fn (k=7, r=2, p=0.8) = Fb (k=7, n=k+r=9, p=0.8) = 0.2618

thus

P (X≥7) = 1 - P (X<7) = 1 - Fn (X=7) = 1 - 0.2618 = 0.7382 (73.82%)

c) the expected value E (X) for a negative binomial distribution is

E (X) = p*r / (1-p) = 0.8*2/0.2 = 8 boxes that do not contain prices

thus

n=k+r = 8 + 2 = 10 boxes