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23 February, 14:24

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.

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  1. 23 February, 17:27
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    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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