# Applied Probability with Four Disjoint Subsets.

I have a rather longwinded question, so please bear with me!

A number of employees conduct duties at company X. Each type of employee meets by department 4 times yearly. Occasionally, a departmental chief will have an employee present a case to the rest of the employees about a problem that has occurred at company X.

Within a 2 year period there were 9 scheduled meetings for a particular department at company X. On 4 occasions employees were asked to present cases.

This particular department is comprised of employees from 4 different subcategories: A,B,C,D, where:

Group A contains 13 members

Group B contains 2 members

Group C contains 1 member

Group D contains 1 member

A given employee belongs to only one group.

The 4 presentations over 2 years have been from groups B, C and D.

Surprisingly, no members of Group A have been called upon to present despite having the largest number of member employees (who consequently take care of more cases).

What is the probability of this occurring by chance alone?

What is the probability of another fifth presentation being from either B,C, or D, and not A by chance alone?

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First, a warning; it is dangerous to refer to events that have actually occurred as having a probability. They have happened (or not) and therefore have a posterior probability of 1 (or 0).

The question I think you want is the anterior probability - what is the chance that the next sequence of 4 speakers in the future will have no speakers from Group A.

One thing to understand from a random process is that given enough time every sequence will occur - that is what random means. So in an infinitely long string of speakers there will be runs of 4, 40, 4,000, 3x10$^8$, 4.72x10$^{378}$ etc. Some of these are incredibly unlikely but infinity is more than incredibly big.

To the questions (in reverse order):

The probability that a person selected at random is not from Group A is $\frac{4}{17}$.

The probability that this happens in the next 4 times is $\left(\frac{4}{17}\right)^4\approx 0.003$.

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I see, thank you very much. So there are a lot of red herrings in this question then. For example, it doesn't matter at all how many of the meetings had cases presented. Let's say instead that it did and we wanted to find out, given that 4/9 meetings had cases presented, what the probability was out of the total number of meetings. Would we simply multiply our answer by 4/9? –  114 Jul 3 '13 at 19:13
No - but this is a new question, mark this one as answered and ask a new one. –  Dale M Jul 3 '13 at 22:21
That's reasonable, thanks for your help. –  114 Jul 4 '13 at 0:42