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  • A factory produces two types of products: Type A and Type B. It is known that 10% of Type A products are defective, while 5% of Type B products are defective. The factory produces Type A products 6

A factory produces two types of products: Type A and Type B. It is known that 10% of Type A products are defective, while 5% of Type B products are defective. The factory produces Type A products 60% of the time and Type B products 40% of the time. If a randomly selected product is defective, what is the probability that it is Type A?

 

Option: 1

5.88%


Option: 2

10%


Option: 3

16.67%


Option: 4

25%


Answers (1)

best_answer

To solve this problem, we need to use Bayes' theorem. Let's define the events:

A: Product is Type A

B: Product is defective

We are given:

\mathrm{P(A)=0.60} (probability of selecting a Type A product)
\mathrm{P\left ( B | A \right )=0.10} (probability of a Type A product being defective)
\mathrm{P\left ( B |\, not\, A \right )=0.05} (probability of a Type B product being defective)
We want to find \mathrm{P\left ( A|B \right )}, the probability that the product is Type A given that it is defective.

Using Bayes' theorem:
P(A \mid B)=P(B \mid A) \times P(A) / P(B) \text{To calculate} \mathrm{P}(\mathrm{B})$, \text{we can use the law of total probability}: $$ P(B)=P(B \mid A) \times P(A)+P(B \mid \text { not } A) \times P(\text { not } A)
Given:

\begin{aligned} & P(A)=0.60 \\ & P(B \mid A)=0.10 \\ & P(B \mid \text { not } A)=0.05 \\ & P(\text { not } A)=1-P(A)=1-0.60=0.40 \end{aligned}

Substituting the values into the equation, we can calculate \mathrm{P(A \mid B)} to be approximately 0.0588 or 5.88%.

Therefore, the probability that a randomly selected defective product is Type A is 5.88%.


 

 

Posted by

himanshu.meshram

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