A worker is instructed to place an order for a shipment of p

A worker is instructed to place an order for a shipment of parts before a deadline. Even if the order is placed before the deadline, there is still only a probability of 0.80 that it will get processed. If the order is sent out after the deadline, the probability that the order will not get processed is 0.95. Suppose the probability that the worker places the deadline is 0.40. If the order does not get processed, what is the probability that the employee placed it before the deadline?

Solution

Let

D = makes the deadline
D\' = does not make the deadline
P = processed
P\' = not processed

Thus,

P(D|P\') = P(D n P\') / P(P\')

As

P(P\') = P(D) P(P\'|D) + P(D\') P(P\'|D\') = 0.40*(1-0.80) + (1-0.40)*(0.95) = 0.65

and

P(D n P\') = P(D) P(P\'|D) = 0.40*(1-0.80) = 0.08

Then

P(D|P\') = P(D n P\') / P(P\') = 0.123076923 [answer]