Question

Question 3: In your company, there is a manufacturing process that produces a part that inspectors grade as as either 1) Excellent, 2) Fair, or 3) Poor. The probabilities that a part is graded as excellent, fair, and poor are 1/8, 3/8, and 1/2; respectively. A new inspector, Steve, has just started working for your company. After performing three inspections, what is the probability that Steve grades an excellent part before a poor part?

Answer 1

P(the part is graded as excellent) = P(E) = 1/8

P(the part is graded as fair) = P(F) = 3/8

P(the part is graded as poor) = P(G) =1/2

A new inspector perform 3 inspections , He grades an excellent part before the poor part

Case 1 - He grades an excellent part in the first inspection and poor part in the second or 3rd inspection

Lets say he grades the poor part in the 2nd inspection .So, he could grade excellent , fair or poor in the 3rd inspection

P(G/E) = P(E)*P(G)*P(E) + P(E)*P(G)*P(F) + P(E)*P(G)*P(G)

= 1/8*1/2*1/8 + 1/8*1/2*3/8 + 1/8*1/2*1/2

= 1/16

Now, lets say he grades the poor part in the 3rd inspection . So , he grades the excellent in the 2nd inspection also

P(G/E) = P(E)*P(E)*P(G) = 1/8*1/8*1/2 = 1/128

Case 2 - He grades the excellent part in the second inspection

So , he grades either the excellent , the fair or the poor in the 1st inspection. As we have already included the excellent part in the first case so , we will exclude that case

P(G/E) = P(F)*P(E)*P(G) + P(G)*P(E)*P(G) = 3/8*1/8*1/2 + 1/2*1/8*1/2

= 3/128 + 1/32 = 13/128

Adding both the cases we get = 1/16 +1/128 + 13/128 = 1/16 +14/128 = 11/64