We are given that a batch of products is produced by three different factories. The proportion of products from Factory 1 is 30%, from Factory 2 is 50%, and from Factory 3 is 20%. The defect rates for products from these three factories are 2%, 1%, and 1%, respectively. We want to find the probability that a randomly selected product from this batch is defective.

Probability and StatisticsProbabilityConditional ProbabilityLaw of Total Probability
2025/4/3

1. Problem Description

We are given that a batch of products is produced by three different factories. The proportion of products from Factory 1 is 30%, from Factory 2 is 50%, and from Factory 3 is 20%. The defect rates for products from these three factories are 2%, 1%, and 1%, respectively. We want to find the probability that a randomly selected product from this batch is defective.

2. Solution Steps

Let F1F_1, F2F_2, and F3F_3 represent the events that a product is from Factory 1, Factory 2, and Factory 3, respectively. Let DD be the event that a product is defective.
We are given the following probabilities:
P(F1)=0.30P(F_1) = 0.30
P(F2)=0.50P(F_2) = 0.50
P(F3)=0.20P(F_3) = 0.20
P(DF1)=0.02P(D|F_1) = 0.02
P(DF2)=0.01P(D|F_2) = 0.01
P(DF3)=0.01P(D|F_3) = 0.01
We want to find P(D)P(D), the probability that a randomly selected product is defective. We can use the law of total probability:
P(D)=P(DF1)P(F1)+P(DF2)P(F2)+P(DF3)P(F3)P(D) = P(D|F_1)P(F_1) + P(D|F_2)P(F_2) + P(D|F_3)P(F_3)
Substituting the given values:
P(D)=(0.02)(0.30)+(0.01)(0.50)+(0.01)(0.20)P(D) = (0.02)(0.30) + (0.01)(0.50) + (0.01)(0.20)
P(D)=0.006+0.005+0.002P(D) = 0.006 + 0.005 + 0.002
P(D)=0.013P(D) = 0.013

3. Final Answer

The probability that a randomly selected product from this batch is defective is 0.013.

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