How often will everyone be there?

[almostdone]

Hello, I need to learn how to calculate the probability that all the workers will be in the office at the same time. Assume there is a small business with 8 employees, they each work 5 days a week but only go into the office some days. Say they all go into the office 3 days a week but which days of the week they go in is unpredictable. What is the probability that they will all be in the office on the same day of the week? Please help me with the formula for this question. I also would like to calculate the probability of everyone being in the office on a given day if some of them go in 2 days and others 3 days, etc.

 

[Cubist]

Hi, you should be able to learn how to do this quite quickly. Please tell us if you're a student studying probability, or is this a real-world problem? (So that we know what kind of help you need). Can you attempt to calculate the probability that ONE worker would be in the office (on any one day) assuming that they work 3 out of 5 days? This is like rolling a five sided die, and working out the probability of getting a 1,2 or 3.

 

[Dr.Peterson]

One problem I see is that "unpredictable" does not imply independence, or a uniform distribution. It's conceivable that everyone would choose (as I currently do in my part-time job) to take off Mondays and Fridays. There may also be some trouble caused by the fact that one isn't working 3 days out of every 5 on average, but exactly 3 days out of each 5, so that probability isn't quite the right model (but I'm being a little nitpicky there).

But doing what Cubist suggests may give a reasonable approximation. Please do give it a start, so we can see where you will need help.

 

[almostdone]

Hi, you should be able to learn how to do this quite quickly. Please tell us if you're a student studying probability, or is this a real-world problem? (So that we know what kind of help you need). Can you attempt to calculate the probability that ONE worker would be in the office (on any one day) assuming that they work 3 out of 5 days? This is like rolling a five sided die, and working out the probability of getting a 1,2 or 3.

Hello, This is a real-world problem. I'm working on a project for a coworking space. They have 8 seats to rent out and I want to be able to calculate the probability that all seats will be full if the members only come in an average of 2 days a week, 3 days a week, etc. Basically, I want to predict how often all seats will be full.

As for your question.. I understand that the probability that one worker would be in the office on any day would be 60%. assuming there was only one worker. However, the calling population of 8 is stumping me.

 

[almostdone]

One problem I see is that "unpredictable" does not imply independence, or a uniform distribution. It's conceivable that everyone would choose (as I currently do in my part-time job) to take off Mondays and Fridays. There may also be some trouble caused by the fact that one isn't working 3 days out of every 5 on average, but exactly 3 days out of each 5, so that probability isn't quite the right model (but I'm being a little nitpicky there).

But doing what Cubist suggests may give a reasonable approximation. Please do give it a start, so we can see where you will need help.

Hello, you're right. Once, I figure this out I should take into consideration the likelihood of Tuesday - Thursday being busier days. However, this is a hypothetical calling population based on the average flex space member of a coworking shop going in 2 or 3 days per week. I was using a traditional office example for simplicity but now I'm seeing one must be very specific with probability problems!

I'm going back and forth on whether this is a probability problem or a spin-off of waiting-line analysis

 

[Cubist]

As for your question.. I understand that the probability that one worker would be in the office on any day would be 60%. assuming there was only one worker. However, the calling population of 8 is stumping me.

Good, 60% is correct! Let p=0.6 be the probability that a person will be in the office. The probability that exactly k people, from a possible n people, will be in the office (assuming every person turns up with the same probability) would be...

[math] \binom{n}{k} \times p^k \times (1-p)^{(n-k)}[/math]
[math]=\frac{n!}{k! \times (n-k)!} \times p^k \times (1-p)^{(n-k)}[/math]
Example, 10 possible workers. What's the probability that exactly 8 turn up?

[math]=\frac{10!}{8! \times (10-8)!} \times 0.6^8 \times (1-0.6)^{(10-8)}[/math]
[math]≈ 0.121 \text{ or } 12.1\%[/math]
NOTE1: You might also want to check the sum of all the probabilities that more than 8 turn up (the not enough seats scenario). Probability that 9 OR 10 turn up is approx 0.04031 + 0.00605 = 0.04636

NOTE2: Post#3 is important. Combinatorics/ permutations should give you a better answer.

NOTE3: Also you probably need to consider national holiday weeks (with only 4 possible working days)

 

[SEstudent22]

Cubist showed you the formula for this kind of problem. If you want to know the name it's called a Bernoulli distribution, which is a special case of the Binomial distribution for number of trials = 1.

 

[Cubist]

Cubist showed you the formula for this kind of problem. If you want to know the name it's called a Bernoulli distribution, which is a special case of the Binomial distribution for number of trials = 1.

The formula in post#6 is the Binomial distribution's PMF, not Bernoulli But I should've provided a link myself, well done for thinking of this!

 

[almostdone]

Good, 60% is correct! Let p=0.6 be the probability that a person will be in the office. The probability that exactly k people, from a possible n people, will be in the office (assuming every person turns up with the same probability) would be...

[math] \binom{n}{k} \times p^k \times (1-p)^{(n-k)}[/math]
[math]=\frac{n!}{k! \times (n-k)!} \times p^k \times (1-p)^{(n-k)}[/math]
Example, 10 possible workers. What's the probability that exactly 8 turn up?

[math]=\frac{10!}{8! \times (10-8)!} \times 0.6^8 \times (1-0.6)^{(10-8)}[/math]
[math]≈ 0.121 \text{ or } 12.1\%[/math]
NOTE1: You might also want to check the sum of all the probabilities that more than 8 turn up (the not enough seats scenario). Probability that 9 OR 10 turn up is approx 0.04031 + 0.00605 = 0.04636

NOTE2: Post#3 is important. Combinatorics/ permutations should give you a better answer.

NOTE3: Also you probably need to consider national holiday weeks (with only 4 possible working days)

Thank you this is so helpful!
What if n & k are the same number? I want to calculate the probability of 8 people from a possible of 8 people being in the office but that throws a zero into the equation?

 

[Deleted member 4993]

Thank you this is so helpful!
What if n & k are the same number? I want to calculate the probability of 8 people from a possible of 8 people being in the office but that throws a zero into the equation?

Use:
0! = 1 and

a^0 = 1