SOLUTION: 1) A warehouse employs 26 workers on first​ shift, 15 workers on second​ shift, and 12 workers on third shift. Eight workers are chosen at random to be interviewed abou
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Question 1125784: 1) A warehouse employs 26 workers on first shift, 15 workers on second shift, and 12 workers on third shift. Eight workers are chosen at random to be interviewed about the work environment.
a) Find the probability of choosing exactly two second shift workers and two third shift workers.
I was able to understand everything up to this point. I think I am just having issues because classwork is piling up and my brain is a little foggy.
Thank you so much to anyone that can help!
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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.
http://www.probabilityformula.org/combination-probability-formula.html
note that nCx is the same as c(n,x)
they are both equal to n! / (x! * (n-x)!)
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