Final answer:
To calculate the probability of at least two defectives or at most one defective in a random sample of 4 metal parts from 15, with 3 defectives among them, use the hypergeometric distribution approach and sum the relevant probabilities.
Step-by-step explanation:
To calculate the probability that a random sample of 4 out of 15 metal parts will contain at least two defectives, we must consider all possible ways to choose defective and non-defective parts. Since hypergeometric distribution is best suited for this scenario, we'll use it for our calculations.
(a) To find the probability of at least two defectives, we need to calculate the probabilities for exactly two, exactly three, and exactly four defectives and sum them up.
Probability of exactly 2 defectives P(X=2) is given by (Combination of 3 defectives taken 2 at a time * Combination of 12 non-defectives taken 2 at a time) / Combination of 15 parts taken 4 at a time.
Probability of exactly 3 defectives P(X=3) uses a similar approach.
Probability of exactly 4 defectives P(X=4) is actually zero since there are only 3 defectives available.
Sum these probabilities to get the total probability of at least two defectives.
(b) The probability of at most one defective can be found by calculating the probability for 0 or 1 defective parts and summing them.
Probability of exactly 0 defectives, P(X=0), is given by (Combination of 12 non-defectives taken 4 at a time) / Combination of 15 parts taken 4 at a time.
Probability of exactly 1 defective, P(X=1), uses a combination of 3 defectives taken 1 at a time and 12 non-defectives taken 3 at a time.
Combine these probabilities to find the total probability of at most one defective. Remember to round your final answers to 4 decimal places as instructed.