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) 0.74545
B) 0.50909
C) 0.78182
D) 0.49091

Question

Quizplus · Accepted Answer

The probability that at least one can is defective is 1 minus the probability that none are defective. The probability that none of the four selected cans are defective is calculated using the hypergeometric distribution: $\frac{{\binom{9}{4}}}{{\binom{12}{4}}} = \frac{126}{495} = 0.25455$. Therefore, the probability that at least one can is defective is $1 - 0.25455 = 0.74545$.

Suppose a Man Has Ordered Twelve 1-Gallon Paint Cans of a Particular