A quality control inspector has drawn a sample of 1616 light bulbs from a recent production lot. Suppose 30%30% of the bulbs in the lot are defective. What is the probability that exactly 44 bulbs from the sample are defective? Round your answer to four decimal places.

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

Step-by-step explanation:

For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent from other bulbs, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

A quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.

This means that

What is the probability that exactly 4 bulbs from the sample are defective?

This is P(X = 4).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]

[tex]P(X = 4) = C_{16,4}.(0.3)^{4}.(0.7)^{12} = 0.2040[\tex]

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

sorry not sure

step-by-step explanation: