To solve this problem, we will use the concept of hypergeometric distribution. The hypergeometric distribution models the probability of k successes (random draws for which the object drawn has a specified feature) in n draws, without replacement, from a finite population of size N that contains exactly K objects with that feature.

Here, N = 20 (total number of wafers), K = 7 (total number of defective wafers), n = 3 (number of wafers selected), and we want to find the probability that k (number of defective wafers selected) is 2 or 3.

The formula for hypergeometric distribution is:

P(X = k) = [C(K, k) * C(N-K, n-k)] / C(N, n)

where C(a, b) is the number of combinations of a items taken b at a time.

1. First, calculate the probability that exactly 2 wafers are defective:

P(X = 2) = [C(7, 2) * C(20-7, 3-2)] / C(20, 3)

2. Then, calculate the probability that all 3 wafers are defective:

P(X = 3) = [C(7, 3) * C(20-7, 3-3)] / C(20, 3)

3. Finally, add these two probabilities together to get the probability that 2 or more of the wafers selected are defective:

P(X = 2) = P(X = 2) + P(X = 3)

Question

To solve this problem, we will use the concept of hypergeometric distribution. The hypergeometric distribution models the probability of k successes (random draws for which the object drawn has a specified feature) in n draws, without replacement, from a finite population of size N that contains exactly K objects with that feature.

Here, N = 20 (total number of wafers), K = 7 (total number of defective wafers), n = 3 (number of wafers selected), and we want to find the probability that k (number of defective wafers selected) is 2 or 3.

The formula for hypergeometric distribution is:

P(X = k) = [C(K, k) * C(N-K, n-k)] / C(N, n)

where C(a, b) is the number of combinations of a items taken b at a time.

1. First, calculate the probability that exactly 2 wafers are defective:

P(X = 2) = [C(7, 2) * C(20-7, 3-2)] / C(20, 3)

2. Then, calculate the probability that all 3 wafers are defective:

P(X = 3) = [C(7, 3) * C(20-7, 3-3)] / C(20, 3)

3. Finally, add these two probabilities together to get the probability that 2 or more of the wafers selected are defective:

P(X >= 2) = P(X = 2) + P(X = 3)

Knowee AI · Accepted Answer

To solve this problem, we will use the concept of hypergeometric distribution. The hypergeometric distribution models the probability of k successes (random draws for which the object drawn has a specified feature) in n draws, without replacement, from a finite population of size N that contains exactly K objects with that feature.

Here, N = 20 (total number of wafers), K = 7 (total number of defective wafers), n = 3 (number of wafers selected), and we want to find the probability that k (number of defective wafers selected) is 2 or 3.

The formula for hypergeometric distribution is:

P(X = k) = [C(K, k) * C(N-K, n-k)] / C(N, n)

where C(a, b) is the number of combinations of a items taken b at a time.

1. First, calculate the probability that exactly 2 wafers are defective:

P(X = 2) = [C(7, 2) * C(20-7, 3-2)] / C(20, 3)

2. Then, calculate the probability that all 3 wafers are defective:

P(X = 3) = [C(7, 3) * C(20-7, 3-3)] / C(20, 3)

3. Finally, add these two probabilities together to get the probability that 2 or more of the wafers selected are defective:

P(X >= 2) = P(X = 2) + P(X = 3)

A tray of 20 silicon wafers contains 7 defective wafers. A sample of 3 wafers is randomlyselected from the tray and tested to destruction.Calculate the probability that 2 or more of the wafers selected are defective

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