Question

The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.2. Suppose you purchase box after box until you have obtained three of these prizes.

(b)

What is the probability that you purchase five boxes? (Round your answer to four decimal places.)

(c)

What is the probability that you purchase at most five boxes? (Round your answer to four decimal places.)

(d)

How many boxes without the desired prize do you expect to purchase?

How many boxes do you expect to purchase?

Answer 1

Let X denotes the number of boxes one will purchase to obtain three of these prizes.

X ~ Negative Binomial( r = 3, p = 0.2)

The probability mass function of X is

$P(X=x) = \binom{x-1}{3-1}*0.2^{3}*(1-0.2)^{x-3},x=3,4,...$

b) The probability that you purchase five boxes

$=P(X=5) = \binom{5-1}{3-1}*0.2^{3}*(1-0.2)^{5-3} = 0.03072$

c) The probability that you purchase at most five boxes

$=P(X\leq 5) = \binom{5-1}{3-1}*0.2^{3}*(1-0.2)^{5-3} = 0.05792$

d) The expected number of boxes to purchase = $E(X) = \frac{r}{p} = \frac{3}{0.2} = 15$

The expected number of boxes to purchase without the desired prize $\frac{r*(1-p)}{p} = \frac{3*(1-0.2)}{0.2} = 12$