Question

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as "seconds."

(a) Among seven randomly selected goblets, how likely is it that only one is a second??____
(b) Among seven randomly selected goblets, what is the probability that at least two are seconds? ___
(c) If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?___

Answer 1

To find the probability that only one out of seven randomly selected goblets is a second, use the binomial probability formula. For at least two seconds, find the complement of the probability of fewer than two. To find the probability that at most five goblets must be selected to find four that are not seconds, use the hypergeometric probability formula.

Step-by-step explanation:

(a) To find the probability that only one out of seven randomly selected goblets is a second, we can use the binomial probability formula. The formula is:

P(X = k) = C(n, k) · p^k · (1-p)^(n-k)

where n is the number of trials, k is the number of successes, C(n, k) is the combination formula, p is the probability of success, and (1-p) is the probability of failure. In this case, n = 7, k = 1, p = 0.13, and (1-p) = 0.87. Plugging these values into the formula, we get:

P(X = 1) = C(7, 1) · 0.13^1 · 0.87^6

Calculating this expression gives us the answer to part (a).

(b) To find the probability that at least two out of seven randomly selected goblets are seconds, we need to find the complement of the probability that fewer than two are seconds. The complement probability is 1 minus the probability of fewer than two. By using the binomial probability formula with n = 7, k = 0 and k = 1, and summing those probabilities, we can find the probability of fewer than two. Subtracting this probability from 1 will give us the answer to part (b).

(c) To find the probability that at most five goblets must be selected to find four that are not seconds, we can use the hypergeometric probability formula. The formula is:

P(X ≤ k) = (∑(i=0)^(k) C(K, i) · C(N-K, n-i)) ÷ C(N, n)

where N is the population size, K is the number of success states in the population, n is the number of draws, k is the number of observed success states, and C(n,k) is the combination formula. In this case, N = total number of goblets, K = total number of non-seconds, n = number of draws, and k = number of observed non-seconds. Plugging these values into the formula, we get:

P(X ≤ 5) = (∑(i=0)^(5) C(K, i) · C(N-K, n-i)) ÷ C(N, n)

Calculating this expression gives us the answer to part (c).