What is the probability that a group of 3 dentists will have an available appointment slot at 2 PM?

Assuming that a dentist has 8, 1 hour appointment time frames available in a day. Now assume that 3 independent dentists each having a utilization rate of 75% (that is, on average 6 of 8 slots a day are filled) decide to work togethor as a group.

What is the probability on any given day that the 2 PM time slot will be available?

How many dentists would have to work togethor at this utilization rate so that the probability would exceed 90%?

Please help me get my mind around this concept.
Thanks for the answers, great help.

Would you mind explaining how to handle the case where instead of only 1 appointment time slot, 2 PM in this example, the patient also choose another second choice time slot, so now it could be something 2 PM first choice and 11 AM second choice. How would you do the availability calculations for this sample case?

Here is how I would work this problem.

Prob(2pm available) = 1 – Prob(2pm slot full for all three dentists) = 1 – Prob(2pm slot for a given dentist is full)^3 = 1 – .75^3 = .578125 or about 58%.

Now for the second part, we want to find the minimum number of dentists that would have to work together so that the chance 2pm is available is 90%, or in other words, the chance it is full is at most 10%. So you want to find the smallest integer d such that:.75^d < .1

To solve for c, take the log (any kind of log will do) of both sides to get: d*log(.75) < log(.1)

divide both sides by log(.75) (which is negative so it flips the < to a >) to get d > log(.1)/log(.75) = 8.0039.

So technically you need 9 or more dentists to have the probability that 2pm is free to be >90%, but 8 will get you very close to 90%.

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4 Responses to “What is the probability that a group of 3 dentists will have an available appointment slot at 2 PM?”

Supermom on April 23rd, 2010 at 5:03 pm

Here’s the problem with statistics.

No matter what number you get on paper, it’s wrong because it doesn’t take into account the popularity/personality of a given dentist, or the fact that the 3pm appointment time is much more popular with people than the 10am appointment slot. Or if it’s the time of year when people have paid their deductibles and are getting all their appointments taken care of, or the weather, or the rest of the stuff that we call real life.

Sorry you’re stuck with Stats, and sorry I can’t help you with the math. Just distracting you during your homework with the reminder that this class isn’t really going to be very useful to you in your future life in the real world.

Good luck!
References :
hackmaster_sk on April 23rd, 2010 at 5:13 pm

So what we need to do is find out what the probability of 1 dentist being free at 2 is. Since we have a 75% slots occupied, there is a 25% probability that any given dentist is free at 2.

Now that the 3 dentists are working to gether, we have (3×6) occupied out of (3×8), which is still 75%, so it will always be 25% probability that a slot is free.
References :
The Face of Reason on April 23rd, 2010 at 5:35 pm

37/64
For any dentist the probability that the 2 slot is NOT open is 3/4. Therefore, the probability that for all three the 2 slot is NOT open is (3/4)^3 = 27/64. Therefore, the probability that at least one of the three has the slot open is 1 – 27/64 = 37/64

Proceeding in this manner, for 9 dentists, the probability that the slot is NOT open is (3/4)^9 = .075…, and thus the probability that it is open is 1 – .075.. = .9249…; whereas for 8 it is only 1 – (3/4)^8 = .89988…
9 would exceed 90%
References :
Ben on April 23rd, 2010 at 6:22 pm

Here is how I would work this problem.

Prob(2pm available) = 1 – Prob(2pm slot full for all three dentists) = 1 – Prob(2pm slot for a given dentist is full)^3 = 1 – .75^3 = .578125 or about 58%.

Now for the second part, we want to find the minimum number of dentists that would have to work together so that the chance 2pm is available is 90%, or in other words, the chance it is full is at most 10%. So you want to find the smallest integer d such that:.75^d < .1

To solve for c, take the log (any kind of log will do) of both sides to get: d*log(.75) < log(.1)

divide both sides by log(.75) (which is negative so it flips the < to a >) to get d > log(.1)/log(.75) = 8.0039.

So technically you need 9 or more dentists to have the probability that 2pm is free to be >90%, but 8 will get you very close to 90%.
References :

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