A box contains 10 bulbs in total, of which exactly 3 are defective and the remaining 7 are good. If a random sample of 5 bulbs is drawn without replacement, what is the probability that the sample contains exactly one defective bulb?

Introduction / Context:
This question is a classic example of probability without replacement, which is modeled by the hypergeometric distribution. The bulb box contains both defective and good bulbs, and we draw several bulbs simultaneously. We are specifically interested in the case where the sample of five bulbs contains exactly one defective bulb, which requires counting combinations rather than simple single draw probabilities.

Given Data / Assumptions:

• Total number of bulbs in the box = 10.
• Defective bulbs = 3.
• Good bulbs = 7.
• We draw 5 bulbs at random without replacement.
• Every set of 5 bulbs is equally likely to be chosen.

Concept / Approach:
The sample space consists of all possible combinations of 5 bulbs that can be drawn from 10 bulbs. This is counted using combinations. For the favorable outcomes, we require exactly one defective bulb and the remaining four bulbs to be good. That means we choose 1 defective bulb out of 3 and 4 good bulbs out of 7. The probability is the ratio of favorable combinations to total combinations, which is a straightforward application of the hypergeometric formula.

Step-by-Step Solution:
Step 1: Compute the total number of ways to choose 5 bulbs from 10 bulbs: C(10,5) = 252.Step 2: To have exactly one defective bulb, choose 1 defective from 3, which gives C(3,1) = 3 ways.Step 3: For the remaining 4 bulbs, choose them from the 7 good bulbs, which gives C(7,4) = 35 ways.Step 4: The total number of favorable samples with exactly one defective bulb is 3 * 35 = 105.Step 5: The required probability is favorable combinations divided by total combinations, so probability = 105 / 252.Step 6: Simplify 105 / 252 by dividing numerator and denominator by 21 to obtain 5 / 12.
Verification / Alternative check:
A useful check is to verify that probabilities for all possible numbers of defective bulbs in a sample of five add up to 1. One can compute probabilities for zero defective, one defective, two defective and three defective bulbs using similar combination counts and confirm that the sum is 1. Once that is verified, the specific value 5/12 stands as the correct probability for exactly one defective bulb.

Why Other Options Are Wrong:
7/12 would be the complement if the other probabilities aligned in a particular way, but detailed calculation shows that it overestimates the chance of exactly one defective. 3/14 and 1/12 correspond to much smaller favorable counts, which do not match the computed value of 105 favorable samples. 11/42 does not arise from any consistent hypergeometric calculation for this setup and therefore cannot be correct.

Common Pitfalls:
Students often mistakenly multiply simple probabilities like 3/10 and 7/9 multiple times instead of using combinations, which does not correctly handle the requirement for exactly one defective bulb. Another frequent issue is forgetting that the order of selection does not matter in this context, leading to overcounting. It is essential to use combinations carefully and ensure that the favorable and total counts match the actual sampling process without replacement.

Final Answer:
The probability that the sample of five bulbs contains exactly one defective bulb is 5/12.