Question

A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P(computer turns out to bedefective, given that it is produced in plant T1)=10P (computer turns out to be defective, given that it is produced in plant T2), where P(E) denotes the probability of an event E.A computer produced in the factory is randomly selected and it does not turn out to be defective. Then, the probability that it is produced in plant T2, is A. 36 73 B. 47 79 C. 78 93 D. 75 83 Select the correct answer from above options

Answer 1

Correct Answer - C Consider the following events : E1 =Computer is produced by plant T1 E2 =Computer is produced by plant T2 A=Computer produced is defective It is given that P(E1)= 20 100 = 2 10 ,P(E2)= 80 100 = 8 10 and P(A)= 7 100 . It is also given that P(A/E1)=10P(A/E2) . ∴P(A)=P(A/E1)P(E1)+P(A/E2)P(E2) ⇒ 7 100 =10P(A/E2)× 2 10 +P(A/E2)× 8 10 ⇒ 7 100 = 28 10 P(A/E2)⇒P(A/E2)= 1 40 So, P(A/E1)=10P(A/E2)= 10 40 = 1 4 ∴P(E2/ ˉ A )= P(E2)P( ˉ A /E2) P( ˉ A ) = 8 10 ×(1- 1 40 ) (1- 7 100 ) = 8 10 × 39 40 93 100 = 78 93