Question

The employees have been divided into 3 groups: manufacturing line employees(mfg), hourly office employees(oh), and salaried office employees(os)

80% of the employees are mfg, 15% oh, and 5% os

The probability of an mfg employee being late for work is 1%, for OH it is 50% and for OS it is 75%

A) If an employee is selected at random, what is the probability( between 0 and 1) that they were late for work today?

B)if we know an employee was late today, what is the probability( between 0 and 1 ) that they are an OS employee?

Answer 1

A)

Given,

P(MFG) = 0.8 ; P(OH) = 0.15 ; P(OS) = 0.05

P(Late | MFG) = 0.01 ; P(Late | OH) = 0.5 ; P(Late | OS) = 0.75

By Law of total probability,

P(Late) = P(Late | MFG) P(MFG) + P(Late | OH) P(OH) + P(Late | OS) P(OS)

= 0.01 * 0.8 + 0.5 * 0.15 + 0.75 * 0.05

= 0.1205

B)

P(OS | Late) = P(Late | OS) P(OS) / P(Late) (Bayes Theorem)

= 0.75 * 0.05 / 0.1205

= 0.3112033