Question

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

Answer 1

Answer: =0.2465

Explanation:

Firstly the probability that a random requisition is entered by data entry specialist 1 equals 0.3, by data

entry specialist 2 equals 0.45, by data entry specialist 3 equals 0.25.

probability(p1) =

(0.3 × 0.03

)/(0.3 × 0.03 + 0.45 × 0.05 + 0.25 × 0.02 )= 0.2465