Question

32) 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 2%. Suppose the restaurant is in an employee recruitment campaign, which is why 20% of employees at any time of the 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	0.20	0.15	0.85	0.03	0.17
Exp emp	0.80	0.02	0.98	0.02	0.784
total	1.00	-	-	0.05	0.95

1. P(order with error) = (0.20 * 0.15 )+ (0.80 * 0.02) = 0.03 + 0.02 = 0.05

  P(order with error) = 0.05

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

P(new employee / order with error) = 0.03 / 0.05

P(new employee / order with error) = 0.6522

3. P(new employee and no error) = 0.20 * 0.85 = 0.170

P(new employee and no error) = 0.170