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A soft drink company has 4 production lines (I, II, III, IV) that represent 20%, 35%, 25% and 20% of total production. If 0.4% of the bottles of line I, 0.6% of line II, 0.5% of line III and 0.65% of line IV, come out with some defect and quality control takes a bottle at random that is defective. Calculate the probability that it comes from line IV.

A. 0.3853% B.23.85% C.38.53% D.0.2385%

User NKSM
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1 Answer

5 votes

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

User Patrickmdnet
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