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.
JustMe22
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11 years ago
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probability.docx
