You are manager of a ticket agency that sells concert tickets. You assume that people will call 4 times in an attempt to buy tickets and then give up. Each telephone ticket agent is available to receive a call with probability 0.1. If all agents are busy when someone calls, the caller hears a busy signal. Find , the minimum number of agents that you have to hire to meet your goal of serving 98% of the customers calling to buy tickets.

10 Operators

Step-by-step explanation:

Given:

- The probability that a call is received p = 0.1

- Total number of tries till no call is received  = 4

- Total number of operators required = n

Find:

The minimum number of agents that you have to hire to meet your goal of serving 98% of the customers calling to buy tickets.

Solution:

- We know that each caller is willing to make 4 attempts to get through. An attempt is  a failure if all n operators are busy, which occurs with probability:

                                      (1 - p)^n = q ( failure probability)

- Assuming  call attempts are independent, a caller will suffer four failed attempts with probability:

                                     ( 1 - p )^4n = q^4

- Now, we are given that we want to serve 98% of the customers. Hence, we have the tolerance of only 2% to fail per call. Hence, we can set an inequality as follows:

                                      ( 1 - p )^4n = q^4 < 0.02

- Plug in the values and solve:

                                      ( 1 - 0.1 )^(4n) < 0.02

Taking natural logs:

                                      4n*Ln(0.9) < Ln(0.02)

                                         n > 37.1298 / 4

                                          n > 9.28 ≈ 10

- Hence, the minimum number of operators should n = 10 to meet the quality standards.

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