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.

- 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.

andromache andromache · Answer 2 · 2022-02-10T17:36:50+01:00

The minimum number of agents is 10

Given that:

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

Calculation:

The probability is[tex](1 - p)^n = q[/tex]( failure probability)
A caller will suffer four failed attempts with probability should be [tex]1 - p )^{4n} = q^4[/tex]

Now

[tex]( 1 - p )^4n = q^4 < 0.02\\\\( 1 - 0.1 )^(4n) < 0.02\\\\4n\times Ln(0.9) < Ln(0.02)\\\\n > 37.1298 \div 4\\\\[/tex]

n > 9.28 ≈ 10

Learn more about the probability here: https://brainly.com/question/795909?referrer=searchResults