Three data entry specialists enter requisitions into a compu

Three data entry specialists enter requisitions into a computer. Specialist 1 processes 30 percent of the requisitions, specialist 2 processes 45 percent, and specialist 3 processes 25 percent. The proportions of incorrectly entered requisitions by data entry specialists 1, 2, and 3 are 0.03, 0.05, and 0.02, respectively. Suppose that a random requisition is found to have been incorrectly entered. What is the probability that it was processed by data entry specialist 1? By data entry specialist 2? By data entry specialist 3?

I figured out the first part is .247. I am having trouble with the second and third specalist

Solution

P(A₁) = 30   P(A₂) = 45,    P(A₃) = 25

P(B|A₁) = 0.03   P(B|A₂) = 0.05 P(B|A₃) = 0.02

P(B) = ?_iP(B|A_i)P(A_i) = (0.03)(30) + (0.05)(45) + (0.02)(25) = 3.75

By Bayes\' theorem,

P(A₁/B) = (P(B/A₁) P(A₁)) / P(B) = ((0.03)30) / 3.75 = 0.247

P(A₂/B) = (P(B/A₂) P(A₂)) / P(B) = ((0.05)45)/3.75 = 0.6

P(A₃/B) = (P(B/A₃) P(A₃)) / P(B) = ((0.02)25)/3.75 = 0.133