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Problem 6 [10 marks]. There is a 25% probability that a customer will arrive at

ID: 3305795 • Letter: P

Question

Problem 6 [10 marks]. There is a 25% probability that a customer will arrive at an ATM at any given minute. Independently, there is a 35% probability that the person at the ATM will complete their business within the same minute interval. Assuming that the line runs in discrete time (0 minutes, 1 minute, 2 minutes, 3 minutes, etc.), what is the probability that there will be 1 person in ine (in the queue) at t 1 minute and at t 2 minutes if a customer arrives at the ATM at 0 minutes? Note: The customer using the ATM is counted as someone waiting in line/in the queue, and at most 1 customer will arrive in any minute interval Hint: Use a tree diaqram to map out all the possible customer changes and their probabilities at t 0 minutes, and extent to t = 1 and t = 2 minutes

Explanation / Answer

Solution

Probability that there is one person in line at t = 1

One person in line at t = 1is possible under two cases:

a) The person who is given to have arrived at 0 minutes has not left the ATM for which the probability is 0.65 [given 35% probability of a customer completing the ATM work within 1 minute ] or

b) The person who is given to have arrived at 0 minutes has left the ATM for which the probability is 0.35, one person arrives in time 1 for which the probability is 0.35, and this person has not left the ATM within that minute for which the probability is 0.65 [given 35% probability of a customer completing the ATM work within 1 minute and given 25% probability of a customer arriving at the ATM within 1 minute].

Thus, probability of case (b) = 0.35 x 0.25 x 0.65 = 0.056875.

Combining both cases, probability that there is one person in line at t = 1 is:

0.65 + 0.056875 = 0.706875 ANSWER 1

Probability that there is one person in line at t = 2

One person in line at t = 2 is possible under three cases:

a) There is one person in line at t = 1, no person arrives in t = 2 and the person already in line has still not completed the ATM work. For this combination, the probability is:

0.706875[from above] x 0.75 x 0.65 = 0.34460

b) There is one person in line at t = 1, one person arrives in t = 2 and the person already in line has completed the ATM work. For this combination, the probability is:

0.706875[from above] x 0.25 x 0.35 = 0.06712

c) There is no person in line at t = 1, one person arrives in t = 2 and this person has not left the ATM within that minute. For this combination, the probability is:

(1 - 0.706875)[from above] x 0.25 x 0.65 = 0.0476.

Combining all three cases, probability that there is one person in line at t = 2 is:

0.34460 + 0.06712 + 0.04763 = 0.45935 ANSWER 2