Final answer:
The problem is about finding the probability of selecting exactly 2 defective transistors from a selection of 4 out of 10 transistors, 4 of which are defective. This can be solved by calculating the combinations for the number of ways to choose the defective and non-defective transistors, then finding their ratio within the overall possible combinations.
Step-by-step explanation:
The student is asking about the probability of selecting exactly 2 defective transistors from a box of 10 transistors, 4 of which are defective, when 4 transistors are selected at random. This is a problem that can be solved using the concept of combinations and the hypergeometric probability distribution.
We want to find the probability of getting exactly 2 defective transistors. The probability formula for exactly x successes in a hypergeometric distribution is:
P(X = x) = [(good combinations) * (bad combinations)] / (total combinations)
To calculate the total number of ways to choose 4 transistors from 10, we use the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items and k is the number of items to choose:
Total combinations = C(10, 4)
For the good combinations (selecting 2 defective transistors out of 4), since there are 4 defective, we choose 2 from these:
Good combinations = C(4, 2)
For the bad combinations (selecting 2 non-defective transistors out of the remaining 6), since there are 6 good transistors, we choose 2 from these:
Bad combinations = C(6, 2)
Plugging these into our formula:
P(X = 2) = [C(4, 2) * C(6, 2)] / C(10, 4)
Calculating the combinations we get:
P(X = 2) = [(4! / (2! * 2!)) * (6! / (2! * 4!))] / (10! / (4! * 6!)) = [(6 * 15) / 210] = 90 / 210
Simplify the fraction:
P(X = 2) = 3 / 7
Converting to a decimal:
P(X = 2) = 0.4286
This number is not in the passage, which means either the choices given are incorrect, or there has been an error in calculation. However, since we must make a choice from the given options, the answer closest to our calculation is D) 0.286.