Question 1125784
26 on first shift.
15 on second shift.
12 on third shift.


8 workers are chosen at random.


you want the probability of getting exactly 2 from the second shift and 2 from the third shift.


in order for this to happen, you would need 2 from the second shift, 2 from the third shift, and 4 from the first shift.


the number of ways of getting 8 from the total of 26 + 15 + 12 = 53 is c(53,8).


the number of ways of getting 2 from the 15 on the second shift is c(15,2).


the number of ways of getting 2 from the 12 on the third shift is c(12,2).


the number of ways of getting 4 from the first shift is c(26,4).


the probability should therefore be (c(15,2) * c(12,3) * c(26,4)) / c(53,8).


that would be equal to (105 * 66 * 14950) / 886322710 = .1168913973.


i think that's the way to do it, but can't be 100% sure because there's no way to go back and tally up the total ways.


one method i use to is try a simpler example where you can tally up the ways.


that's difficult in this problem, but i did try a much simpler problem with only 2 possibilities (first shift and second shift only) with much less workers, and the formula seems to work with that much simpler problem.


that, however, doesn't mean it will work with the more complex problem, although i have used the method before in other problems where it was satisfactory.


here's a referenced on the topic.


<a href = "http://www.probabilityformula.org/combination-probability-formula.html" target = "_blank">http://www.probabilityformula.org/combination-probability-formula.html</a>


note that nCx is the same as c(n,x)


they are both equal to n! / (x! * (n-x)!)