The given problem can be solved using the binomial distribution. Here, we need to determine the probability that the lot will fail inspection, which happens if at least one defective bulb is found in the sample.

Step-by-step solution:

1. Parameters of the binomial distribution:

   • The probability of a bulb being defective ($p$) is 20%, or $p = 0.20$.
   • The number of bulbs in the sample ($n$) is 20.
   • The complement of $p$, which is the probability of a bulb being non-defective, is $1 - p = 0.80$.

2. Restate the problem: We need to find the probability that at least one bulb in the sample is defective. This can be calculated as the complement of the probability that all bulbs in the sample are non-defective.

3. The probability of all bulbs being non-defective: The probability that all 20 bulbs are non-defective is: $P(\text{no defective bulbs}) = (0.80)^{20}$

4. The probability of at least one defective bulb: This is the complement of the probability that no bulbs are defective, i.e.: $P(\text{at least one defective bulb}) = 1 - (0.80)^{20}$

Now, let's calculate this probability.

$P(\text{at least one defective bulb}) = 1 - (0.80)^{20}$

Let me compute this for you.The probability that the lot will fail inspection (i.e., at least one bulb is defective) is approximately 0.9885.

Would you like a detailed explanation of any step? Here are a few related questions:

1. How do we determine the sample size in quality control?
2. What is a binomial distribution, and when is it used?
3. How can you calculate probabilities for different outcomes using the binomial formula?
4. What is the difference between independent and dependent events in probability?
5. How can quality control sampling be optimized to reduce inspection costs?

Tip: The complement rule in probability is a powerful tool—use it when calculating "at least one" probabilities!