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:
[tex]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[/tex]
The probability that none of the four cans selected contains an incorrect mix of paint is P=0.2545.
