Question

then Var(X)= ( na2 Instruction: The ollowing questions need you show all your work in details. Question 3. (5 points) A machine produces defective parts with three different probabilities depending on its state of repair. If the machine is in good working order, it produces defective parts with probability 0.02. If it is wearing down, it produces defective parts with roduces defective parts with probability 0.3. The probability 0.1.Hit needs maintenance, it p probability that the machine is in good working order is 0.8, the probability that it is wearing down is 0 1, and the probability that it needs maintenance is 0.1. Compute the probability that a randomly selected part will be defective.

Answer 1

Let D denote the event that a part is defective.
Let G denote the machine is in good working order.
Let W denote the machine is wearing down.
Let M denote the machine is in need of maintenance.

By law of Tota probability we have

P(D) = P(D|G)*P(G)+P(D|W)*P(W)+P(D|M)*P(M)

We are given

P(D|G)= 0.02
P(G)=0.8
P(D|W)=0.1
P(W) = 0.1
P(D|M)=0.3
P(M) = 0.1

Using these value we get

P(D) = 0.02*0.8 + 0.1*0.1+0.3*0.1= 0.056

The probability of picking a defective part is 5.6%