Among 17 metal parts produced in a machine shop, 6 are defective. If a random sample of four of these metal parts is selected, find:
(a) The probability that this sample will contain at least three defectives. (Do not round the intermediate answers. Round your answer to 4 decimal places.)
(b) The probability that this sample will contain at most one defective. (Round your answer to 4 decimal places.)
To solve this problem, we can use the concept of probability.
(a) To find the probability that the sample contains at least three defectives, we need to consider two cases: the sample contains exactly 3 defectives and the sample contains all 4 defectives.
1. The probability of selecting exactly 3 defectives out of 4 can be calculated using the binomial probability formula:
P(3 defectives) = (C(6,3) * C(11,1)) / C(17,4)
C(n,r) represents the combination function, which calculates the number of ways to choose r objects from a set of n objects.
2. The probability of selecting all 4 defectives can be calculated using the same formula:
P(4 defectives) = (C(6,4) * C(11,0)) / C(17,4)
3. Finally, we can find the probability of at least 3 defectives by adding the probabilities of the two cases:
P(at least 3 defectives) = P(3 defectives) + P(4 defectives)
Let's calculate these probabilities now:
P(3 defectives) = (C(6,3) * C(11,1)) / C(17,4) = (20 * 11) / 2380 = 220 / 2380 = 0.0924 (rounded to 4 decimal places)
P(4 defectives) = (C(6,4) * C(11,0)) / C(17,4) = (15 * 1) / 2380 = 15 / 2380 = 0.0063 (rounded to 4 decimal places)
P(at least 3 defectives) = P(3 defectives) + P(4 defectives) = 0.0924 + 0.0063 = 0.0987 (rounded to 4 decimal places)
Therefore, the probability that the sample will contain at least three defectives is approximately 0.0987.
(b) To find the probability that the sample will contain at most one defective, we need to consider two cases: the sample contains no defectives (0 defective parts) and the sample contains one defective part (1 defective part).
1. The probability of selecting no defectives out of 4 can be calculated using the binomial probability formula:
P(no defectives) = (C(6,0) * C(11,4)) / C(17,4)
2. The probability of selecting one defective part can be calculated using the same formula:
P(1 defective) = (C(6,1) * C(11,3)) / C(17,4)
3. Finally, we can find the probability of at most one defective by adding the probabilities of the two cases:
P(at most 1 defective) = P(no defectives) + P(1 defective)
Let's calculate these probabilities now:
P(no defectives) = (C(6,0) * C(11,4)) / C(17,4) = (1 * 330) / 2380 = 330 / 2380 = 0.1387 (rounded to 4 decimal places)
P(1 defective) = (C(6,1) * C(11,3)) / C(17,4) = (6 * 165) / 2380 = 990 / 2380 = 0.4151 (rounded to 4 decimal places)
P(at most 1 defective) = P(no defectives) + P(1 defective) = 0.1387 + 0.4151 = 0.5538 (rounded to 4 decimal places)
Therefore, the probability that the sample will contain at most one defective is approximately 0.5538.
To find the probabilities in this problem, we need to use the concept of combinations and apply the binomial probability formula. Let's solve each part step by step.
(a) The probability that the sample contains at least three defectives:
To find this probability, we need to consider two scenarios: when exactly 3 defectives are selected and when all 4 selected parts are defective.
1. Probability of selecting exactly 3 defectives:
The number of ways to select 3 defectives out of 6 defective parts is given by the combination formula: C(6,3) = 6! / (3! * (6 - 3)!) = 20.
The number of ways to select 1 non-defective part out of 11 non-defective parts is given by the combination formula: C(11,1) = 11.
The total number of ways to select 4 parts out of 17 is given by the combination formula: C(17,4) = 17! / (4! * (17 - 4)!) = 2380.
The probability of selecting exactly 3 defectives is: (20 * 11) / 2380.
2. Probability of selecting all 4 defective parts:
The number of ways to select 4 defectives out of 6 defective parts is given by the combination formula: C(6,4) = 6! / (4! * (6 - 4)!) = 15.
The number of ways to select 0 non-defective part out of 11 non-defective parts is given by the combination formula: C(11,0) = 1.
The total number of ways to select 4 parts out of 17 is given by the combination formula: C(17,4) = 2380.
The probability of selecting all 4 defective parts is: (15 * 1) / 2380.
To find the probability of at least three defectives, we add the probabilities of the two scenarios calculated above:
Probability = [(20 * 11) / 2380] + [(15 * 1) / 2380] = ? (Calculate the answer)
(b) The probability that the sample contains at most one defective:
To find this probability, we need to consider two scenarios: when 0 defective parts are selected and when exactly 1 defective part is selected.
1. Probability of selecting 0 defective parts:
The number of ways to select 0 defectives out of 6 defective parts is given by the combination formula: C(6,0) = 1.
The number of ways to select 4 non-defective parts out of 11 non-defective parts is given by the combination formula: C(11,4) = 11! / (4! * (11 - 4)!) = 330.
The total number of ways to select 4 parts out of 17 is given by the combination formula: C(17,4) = 2380.
The probability of selecting 0 defective parts is: (1 * 330) / 2380.
2. Probability of selecting exactly 1 defective part:
The number of ways to select 1 defective part out of 6 defective parts is given by the combination formula: C(6,1) = 6.
The number of ways to select 3 non-defective parts out of 11 non-defective parts is given by the combination formula: C(11,3) = 11! / (3! * (11 - 3)!) = 165.
The total number of ways to select 4 parts out of 17 is given by the combination formula: C(17,4) = 2380.
The probability of selecting exactly 1 defective part is: (6 * 165) / 2380.
To find the probability of at most one defective, we add the probabilities of the two scenarios calculated above:
Probability = [(1 * 330) / 2380] + [(6 * 165) / 2380] = ? (Calculate the answer)
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