Final answer:
To find the probability of selecting two non-defective and one defective item from a lot, we use combinations to determine the number of favorable outcomes and divide by the total number of outcomes. After calculating, we find that the probability is approximately 46.32%, which is closest to option D) 0.5.
Step-by-step explanation:
To find the probability that two items are non-defective and one item is defective when three items are selected, we need to calculate the number of favorable outcomes and divide it by the number of all possible outcomes. We can use the combination formula, which is C(n, k) = n! / (k! * (n-k)!), where 'n' is the total number of items and 'k' is the number of items to choose. In this case, 'n' would be the number of non-defective or defective items, and 'k' would represent the number of items being chosen that are non-defective or defective.
To find the probability of selecting two non-defective items from twelve, we calculate C(12, 2), and to find the probability of selecting one defective item from eight, we calculate C(8, 1). Then, to find the total number of ways to select three items from twenty, we calculate C(20, 3).
The probability of this combination occurring is the product of the probabilities of each individual selection, which equals to the product of the number of ways to select the non-defective and defective items divided by the number of ways to select any three items. Therefore, the probability is:
P(2 non-defective, 1 defective) = [C(12, 2) * C(8, 1)] / C(20, 3)
Which simplifies to:
P(2 non-defective, 1 defective) = [(12! / (2! * 10!)) * (8! / (1! * 7!))] / (20! / (3! * 17!)) = [66 * 8] / 1140 = 528 / 1140
After simplifying this, we find the probability:
P(2 non-defective, 1 defective) = 0.4632 or 46.32%
The closest answer would be option D) 0.5.