It is B.
Sure, I can help you solve this problem quickly! This is a conditional probability problem that can be solved using Bayes' theorem. Let's denote the events as follows:
- $I$, $II$, $III$, and $IV$ represent the events that a randomly selected bottle comes from production line I, II, III, and IV, respectively.
- $D$ represents the event that a randomly selected bottle is defective.
From the problem statement, we know that:
- $P(I) = 0.20$, $P(II) = 0.35$, $P(III) = 0.25$, and $P(IV) = 0.20$.
- $P(D|I) = 0.004$, $P(D|II) = 0.006$, $P(D|III) = 0.005$, and $P(D|IV) = 0.0065$.
We want to find the probability that a randomly selected defective bottle comes from line IV, which is given by $P(IV|D)$. Using Bayes' theorem, we can write:
$$
P(IV|D) = \fracP(D{P(D)}.
$$
To calculate the denominator, we can use the law of total probability:
$$
P(D) = P(D|I)P(I) + P(D|II)P(II) + P(D|III)P(III) + P(D|IV)P(IV).
$$
Substituting the known values, we get:
$$
P(D) = (0.004)(0.20) + (0.006)(0.35) + (0.005)(0.25) + (0.0065)(0.20) = 0.005425.
$$
Now we can calculate the desired probability:
$$
P(IV|D) = \fracIV)P(IV){P(D)} = \frac{(0.0065)(0.20)}{0.005425} \approx 0.2392.
$$
So, the probability that a randomly selected defective bottle comes from line IV is approximately 23.92%, which corresponds to answer choice B.
I hope this helps! Let me know if you have any further questions