Atul Kumar

Atul Kumar

Answered question

2022-08-14

Answer & Explanation

Eliza Beth13

Eliza Beth13

Skilled2023-05-31Added 130 answers

To solve the problem, let's denote the following events:
- Event T1: The computer is produced in plant T1.
- Event T2: The computer is produced in plant T2.
- Event D: The computer turns out to be defective.
We are given that plant T1 produces 20% of the total computers and plant T2 produces 80% of the total computers. Therefore, we have:
P(T1)=0.20 and P(T2)=0.80.
We are also given that 7% of the computers produced in the factory turn out to be defective. Therefore, we have:
P(D)=0.07.
Additionally, we are given that:
P(D|T1)=10P(D|T2).
We need to find the probability that a randomly selected non-defective computer is produced in plant T_2, denoted as P(T2|D), where D represents the event that the computer does not turn out to be defective.
To find this probability, we can apply Bayes' theorem:
P(T2|D)=P(D|T2)·P(T2)P(D).
First, let's calculate P(D|T2), the probability that a computer produced in plant T_2 is not defective:
P(D|T2)=1P(D|T2)=10.07=0.93.
Next, we calculate P(D), the probability that a randomly selected computer is not defective. This can be calculated using the law of total probability:
P(D)=P(D|T1)·P(T1)+P(D|T2)·P(T2).
Since the complement of being defective is not being defective, we have:
P(D|T1)=1P(D|T1)=10.10P(D|T2).
Substituting the given values, we have:
P(D|T1)=10.10(0.07)=0.993.
Now, we can calculate P(D):
P(D)=P(D|T1)·P(T1)+P(D|T2)·P(T2).
P(D)=0.993(0.20)+0.93(0.80)=0.1986+0.744=0.9426

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