A box contains 10 bulbs, 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 no defective bulbs?

Introduction / Context:
This question is a typical application of combinations in probability, specifically the hypergeometric distribution. We have a finite population of bulbs with some defective and some good bulbs. A sample of fixed size is drawn without replacement, and we are asked for the probability that no defective bulbs appear in the sample. This is a common model for quality control and acceptance sampling problems.

Given Data / Assumptions:

Total bulbs in the box = 10.

Defective bulbs = 3.

Good bulbs = 10 - 3 = 7.

A random sample of 5 bulbs is drawn without replacement.

We want the probability that all 5 bulbs in the sample are good (no defective bulbs).

Concept / Approach:
We treat every possible 5 bulb sample as equally likely. The total number of possible samples is the number of combinations of 10 bulbs taken 5 at a time. The favourable samples are those that contain only good bulbs, which means all 5 are chosen from the 7 good bulbs. The required probability is favourable combinations divided by total combinations.

Step-by-Step Solution:
Step 1: Total number of ways to choose any 5 bulbs from 10 = 10C5.Step 2: Compute 10C5 = (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1) = 252.Step 3: Favourable samples must contain only good bulbs. There are 7 good bulbs.Step 4: Number of ways to choose 5 good bulbs from 7 = 7C5.Step 5: Compute 7C5 = 7C2 = (7 * 6) / (2 * 1) = 21.Step 6: Probability that a random sample of 5 has no defective bulbs = favourable / total = 21 / 252.Step 7: Simplify 21 / 252 by dividing numerator and denominator by 21 to get 1 / 12.

Verification / Alternative check:
We can confirm by using the sequential probability approach. The probability that the first bulb is good is 7/10, the second is good given the first was good is 6/9, then 5/8, then 4/7 and finally 3/6. The product (7/10) * (6/9) * (5/8) * (4/7) * (3/6) simplifies to 21/252, which again equals 1/12. This matches the combination method and verifies the result.

Why Other Options Are Wrong:
The fractions 5/12 and 7/12 are much larger and would suggest that most samples are completely free of defectives, which is not consistent with having 3 defective bulbs among 10. The value 3/14 is also incorrect and arises from incomplete simplification or wrong counting. The value 1/4 is closer but still not equal to the exact fraction 1/12 that follows from both combinational and sequential reasoning.

Common Pitfalls:
One common error is to treat the draws as with replacement and calculate (7/10)^5, which ignores the decreasing pool of good bulbs. Another mistake is to accidentally choose some defective bulbs when counting favourable outcomes, or to miscompute 10C5 or 7C5. Always verify combination calculations and keep track of whether the sampling is with or without replacement.

Final Answer:
The probability that the five bulb sample contains no defective bulbs is 1/12.