Among 15 metal parts produced in a machine shop, 4 are defective. If a random sample of five of these metal parts is selected, find
a) The probability that this sample will contain at least three defectives.
b) The probability that this sample will contain at most two defective.
To solve this problem, we can use the concept of probability and combinations.
a) The probability that the sample will contain at least three defectives can be found by summing the probabilities of selecting exactly three, four, or five defectives.
To calculate the probability of selecting exactly three defectives, we can use combinations. There are a total of 15 metal parts, and we want to choose 3 of them to be defectives. The number of ways to choose 3 defectives out of 4 defective parts is given by the combination formula: C(4, 3) = 4.
Similarly, to calculate the probability of selecting exactly four defectives, we use C(4, 4) = 1.
Finally, to calculate the probability of selecting exactly five defectives, we use C(4, 5) = 0, because we can't select 5 defectives when there are only 4 available.
Therefore, the probability of selecting at least three defectives is:
P(at least 3 defectives) = P(exactly 3 defectives) + P(exactly 4 defectives) + P(exactly 5 defectives)
= C(4, 3) / C(15, 5) + C(4, 4) / C(15, 5) + C(4, 5) / C(15, 5)
= (4/3003) + (1/3003) + 0
= 5/3003
≈ 0.0017
b) The probability that the sample will contain at most two defectives can be found by subtracting the probability of selecting three, four, or five defectives from 1 (the total probability).
P(at most 2 defectives) = 1 - P(at least 3 defectives)
= 1 - 5/3003
= 2998/3003
≈ 0.9983
Therefore, the probability that the sample will contain at most two defectives is approximately 0.9983.