Three data entry specialists enter requisitions into a compu

Three data entry specialists enter requisitions into a computer. Specialist 1 processes 38 percent of the requisitions, specialist 2 processes 20 percent, and specialist 3 processes 42 percent. The proportions of incorrectly entered requisitions by data entry specialists 1, 2, and 3 are 0.03, 0.02, and 0.05, 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? (Round your answers to 3 decimal places.)

     
P(specialist 1 | incorrect)  
P(specialist 2 | incorrect)  
P(specialist 3 | incorrect)

Solution

Let I = incorrect

Hence,

P(I) = P(1) P(I|1) + P(2) P(I|2) + P(3) P(I|3)

= 0.38*0.03 + 0.20*0.02 + 0.42*0.05

P(I) = 0.0364

Hence,

P(1|I) = P(1) P(I|1)/P(I) = 0.38*0.03/0.0364 = 0.313 [ANSWER]

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P(2|I) = P(2) P(I|2)/P(I) = 0.20*0.02/0.0364 = 0.110 [ANSWER]

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p(3|I) = P(3)P(I|3)/P(I) = 0.42*0.05/0.0364 = 0.577 [ANSWER]