Answer to Question #152242 in Statistics and Probability for Kelly

Question

The probability that a randomly selected box of a certain type of cereal has a particular prize is 0.1. Suppose you purchase box after box until you have obtained four of these prizes.

(a)
What is the probability that you purchase x boxes that do not have the desired prize?
h(x; 4, 0.1)
b(x; 4, 1, 10)
h(x; 4, 1, 10)
b(x; 4, 0.1)
nb(x; 4, 1, 10)
nb(x; 4, 0.1)


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

(c)
What is the probability that you purchase at most six 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

(a) Let $X$ be the number of purchase boxes that do not have the desired prize.

The random variable $X$ follows negative binomial distribution with parameters $p=0.1, r=4$

$nb(x; 4, 0.1)$

(b) In 6 boxes we obtain 4 successes (prizes) and 2 failures

$$nb(2; 4, 0.1)=\dbinom{x+r-1}{r-1}p^r(1-p)^x$$

$$=\dbinom{2+4-1}{4-1}(0.1)^4(1-0.1)^2$$

$$=0.00081\approx0.0008$$

(c)

$$nb(0; 4, 0.1)+nb(1; 4, 0.1)+nb(2; 4, 0.1)$$

$$=\dbinom{0+4-1}{4-1}(0.1)^4(1-0.1)^0$$

$$+\dbinom{1+4-1}{4-1}(0.1)^4(1-0.1)^1$$

$$+\dbinom{2+4-1}{4-1}(0.1)^4(1-0.1)^2$$

$$=0.0001+0.00036+0.00081=0.00127$$

$$\approx0.0013$$

(d) The mean (the expected value) is

$$\mu=\dfrac{r(1-p)}{p}=\dfrac{4(1-0.1)}{0.1}=36$$

This means that we expect to purchase 36 boxes without the prize (failure) until we have obtain the four prizes (success)

In total we expected to need $36+4=40$ boxes.

Boxes without the desired prize: $36.$ do you expect to purchase?

Boxes to purchase: $40.$