Probability related to an Appointment Scheduling Simulation

AI Thread Summary
The discussion focuses on simulating a web booking system for a dentist with 8 daily appointments, where three dentists average 6 bookings each. The key problem is calculating the probability that at least one of three preferred time slots is available for prospective clients using a self-booking system. The calculations show that the probability of at least one slot being available with one dentist is approximately 0.9943, while the probability with two dentists is about 0.9157. The approach involves combinatorial probability to determine the likelihood of appointment slots being filled. This simulation highlights the complexities of appointment scheduling and the effectiveness of web-based booking systems.
wolfego
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I am trying to simulate the performance of a web booking/scheduling system and although there are many features of this problem that intrigue me the one currently giving me fits is demonstrated below.

Assume that a dentist has 8 appointments per day and that each appointment time is equally desireable from the perspective of the patient.

Assume that 3 dentists working together as a group each have appointments via the traditional call and book method averaging 6 of the 8 daily appointment slots. In other words, some days they have more or less than 6 bookings but in the aggregrate they each have 6 bookings per day.

Then using a web based (last-minute/day-before) self-booking system where a prospective client picks 3 preferred time slots what is the probability that one of those 3 time slots would be available with at least 1 dentist? With 2 dentists? How can one solve this and similar problems?

Thanks,
Bernie
 
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Since the distribution is not given,a fair assumption is to take exactly 6 bookings that day
Let us say he books 1,2,3 as his time slot(of 1,2,3,4,5,6,7,8)
We first compute the probability that all the 3 time-slots are filled with all the dentists
Pr(of the 6 appointments,3 appointments fall in 1,2,3 for 1 dentist)=5C3/8C3=P(slots are filled for dentist 1)=5/28
So the probability that the slots are filled for all the dentists=(5/28)^3=
So the probability that atleast 1 slot is empty = 1-(5/28)^3
0.9943

For it to be available with 2 dentists=1-(5/28)^3-3C1*(5/28)^2*(23/28)
=0.9157
 
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