Question

Suppose a man has ordered twelve 1-gallon paint cans of a particular color (lilac) from the local paint store in order to paint his mother's house. Unknown to the man, three of these cans contains an incorrect mix of paint. For this weekend's big project, the man randomly selects four of these 1-gallon cans to paint his mother's living room. Let x = the number of the paint cans selected that are defective. Unknown to the man, x follows a hypergeometric distribution. Find the probability that at least one of the four cans selected contains an incorrect mix of paint.A) 0.49091 B) 0.74545 C) 0.78182 D) 0.50909

Answer 1

Answer:

B) 0.74545

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The hypergeometric distribution can be simplified using the combinations formula.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem:

12 cans

3 contain an incorrect mix.

4 are selected

Probability that at least one of the four cans selected contains an incorrect mix of paint.

Either none contain an incorrect mix, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$

So

$P(X \geq 1) = 1 - P(X = 0)$

Finding P(X = 0):

None with an incorrect mix

Desired outcomes:

4 from a set of 9(without an incorrect mix). So

$D = C_{9,4} = \frac{9!}{4!(9-4)!} = 126$

Total outcomes:

4 from a set of 12. So

$T = C_{12,4} = \frac{12!}{4!(12-4)!} = 495$

Probability:

$P(X = 0) = p = \frac{D}{T} = \frac{126}{495} = 0.25455$

Probability that at least one of the four cans selected contains an incorrect mix of paint:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.25455 = 0.74545$

So the correct answer is:

B) 0.74545