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 none of the four cans selected contains an incorrect mix of paint.

Answer 1

Answer:

The probability that none of the four cans selected contains an incorrect mix of paint is P=0.2545.

Step-by-step explanation:

We have 12 cans, out of which 3 are defective (incorrect mix of paint).

The man will choose 4 cans to paint his mother's house living room.

Let x = the number of the paint cans selected that are defective.

The variable x is known to follow a hypergeometric distribution.

The probability of getting k=0 defectives in a selected sample of K=4 cans, where there are n=3 defectives in the population of N=12 cans is:

$P(X=k)=\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\\\\\\\\ P(X=0)=\dfrac{\binom{4}{0}\binom{12-4}{3-0}}{\binom{12}{3}}=\dfrac{\binom{4}{0}\binom{8}{3}}{\binom{12}{3}}=\dfrfac{1*56}{220}=\dfrac{56}{220}=0.2545$

The probability that none of the four cans selected contains an incorrect mix of paint is P=0.2545.