The cashiers at a wholesale store are given a short training on how to work the

Question

The cashiers at a wholesale store are given a short training on how to work the register, and expected to learn it completely by serving clients. They would likely make an error scanning the items on their first few days. For a new employee the probability of making an error is of 10%, while an employee with experience will make with probability of 0.5%. Suppose the wholesale store currently hired new employees, for which 15% of the cashiers are new employees during each shift. Assume that the clients are distributed randomly between cashiers. a. What is the probability that a client is served by a new cashier and the order does not contain an error? b. If a client visits the store 8 consecutive days, what is the probability of having zero errors in your purchase on the 8 days? c. Suppose you are the new employee and you have a probability of scanning an item incorrectly of 10%. If you serve 7 clients, what is the probability of serving at least 6 clients without making an error?

Explanation / Answer

a) P(new cashier) = 0.15,P(experienced) = 0.85, P(error|new cashier) = 0.10, P(error|experienced) = 0.005

P(error|new cashier) = P(error and new cashier)/P(new cashier)

P(error and new cashier) = P(error|new cashier)*P(new cashier) = 0.10 * 0.15 = 0.015

b) P(error) = P(new cashier)*P(error|new cashier) + P(experienced)*P(error|experienced)

=0.15*0.10 + 0.85*0.005 = 0.01925

P(no error) = 1 - 0.01925 = 0.98075

P(errors = 0 for 8 visits) = 0.98075^8 = 0.85598 = 85.60%

c) This is a binomial distribution problem, each success being scanning without any error.

P(atleast 6 error free scanning) = 7C6 * 0.96 * 0.1 + 7C7 * 0.97 = 0.8503 = 85.03%

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