Question

New employees at a fast food restaurant have to learn to perform their tasks while serving customers, so they are likely to serve orders with an error. For a new employee, the probability of making an error while serving the order is 15%, while for an experienced employee the probability of making an error while serving is 3%. Suppose the restaurant is in an employee recruitment campaign, which is why 30% of employees at any time of day are new. Also presume that the orders to be dispatched are randomly distributed among the employees.

 What is the probability that a random order will be dispatched in error?
 If you receive your order and it contains an error, what is the probability that it was dispatched by a new employee?
 What is the probability that the order will be filled by a new employee and contain no errors?

Answer 1

Contingency table based on the information:

Data	Distribution	Error	No error	Orders with error	No error
New emp	0.30	0.15	0.85	0.0450	0.2550
Exp emp	0.70	0.03	0.97	0.0210	0.6790
total	1.00	-	-	0.0660	0.9340

(1) P(with error) = 0.30 * 0.15 + 0.70 * 0.03 = 0.0450 + 0.0210

P(without error) = 0.0660

(2) P(new employee / order with error) = P(new employee and order with error)/ P(order with error)

P(new employee / order with error) = 0.0450 / 0.0660

P(new employee / order with error) = 0.6818

(3) P(new employee and no error) = 0.30 * 0.85 = 0.2550

P(new employee and no error) = 0.2550