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I know that it takes some extra time to look over the Challenge Discussion Forum questions, but I hope you find them to be fun–well maybe at least interesting. The answer to the question posed this week demands that one have a good understanding of “combinations.” In this case, the answer can be found by considering how many ways we can select a 10 person group from the 45 boys and girls. This can be found using the Excel command “=combin(45,10)” Next, we need to determining how many ways we can select 3 boys from the 15 AND consequently 7 girls from the 30. This can be computed using “=combin(15,3)*combin(30,7)” in Excel. Consequently, the probability we seek can be computed by entering “=combin(15,3)*combin(30,7)/combin(45,10)” in a cell on an Excel spreadsheet. This value is about .2904.
A couple of points to summarize: First, we are dealing with a discrete situation here–there are no continuous variables. That is, only integer values of boys and girls can be selected. Next, yes the probability of selecting a boy or a girl changes as the process unfolds. For example, the probability of selecting a boy on the very first “draw” is 15/45, but if we select a boy on the first draw, then the probability of selecting a boy on the second “draw” is 14/44.
Now, we use combinations here, because selecting boys Smith and Kelly is exactly the same as selecting boys Kelly and Smith. Order is not important here–hence, combinations are to be used.
The Excel function =combin() is a great function to use to compute the number of combinations. Keep in mind factorials can be really big numbers–something Excel handles fairly easily.
Note that we can certainly find the probability of selecting 7 girls and 3 boys by finding
P(GGGGGGGBBB)=
(30/45)*(29/44)*(28/43)*(27/42)*(26/41)*(25/40)*(24/39)*(15/38)*(14/37)*(13/36).
But, this is only one way of getting 7 girls and 3 boys. There are lots of other ways of getting 7 girls and 3 boys; namely, 10C3 ways. So we have to multiply the above probability by this number of combinations. This yields the same .2904 that you arrive at using the formulas I provide in my solution post.
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