Help with probability of employee time off crossover

Morning all, I'm trying to figure out how to calculate the probability of x employees being on leave on the same day.

If I want to find the probability of x employees being on leave the same day, I currently have the equation down as:

$P(x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n} } $

N = available working days
n = total number of employees
K = average remaining days of leave

This creates a reducing P(x) as I increase x - is that correct for what I am trying to calculate? And then if so I assume there is a point in the relationship between N, n, and K where it would flip and generate an increasing probability that employees will be off on the same day?

Edit for detail

I am trying to create an equation where I can see what the likelihood is of x employees being away on the same day for any random day of the year. The above equation is what I currently have in an attempt to calculate that.

I understand that realistically there will be greater probabilities attached to specific days but am choosing to accept and ignore that.

Base numbers for a full year ahead is:
N = 245
n = 20
K = 23
x is then the only variable

Is the above equation correct (as I think it might be) for what I am trying to achieve?

If so, currently the probability decreases as x increases - which is what I would expect - am I correct in thinking there is a point in the relationship between N, K, and x where, as N decreases (and provided K hasn't decreased too much), the probability will act in the inverse and so increase as x increases?

  • How many days of the year is left? Your question is not well- formulated. One needs to spend time to clarify whatbyou are trying to do, and then answer it. I think you should definately offer a bounty for this.

    • Apologies for the question not being formulated well! So the intention was that N, n, K and x are all variable hence not initially putting figures to them, but the base figures for a full year ahead would be: N = 245 n = 20 K = 23 Which then leaves x as the primary variable I suppose. I've added an edit above so I can provide more explanation - I was, provided my equation is correct, looking for confirmation the probability is doing what is expected but can add a bounty if necessary/more to it.

  • Kav10 Kav10

    The equation you have is not correct! This needs time to clarify all details and provide a complete answer and deserves a reasonable bounty.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.