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

i'm not entirely sure

step-by-step explanation:

if you ask home depot or lowe's they might be able to you out