Answer:
0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.
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.
![P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)[\tex]</p><p>In which [tex]C_(n,x)[\tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.</p><p>[tex]C_(n,x) = (n!)/(x!(n-x)!)[\tex]</p><p>And p is the probability of X happening.</p><p><strong>Suppose 30% of the bulbs in the lot are defective.</strong></p><p>This means that [tex]p = 0.3]()
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.