Question 974097
I'm going to use the <a href="http://www.mathwords.com/c/combination_formula.htm">Combination formula</a> n C r = (n!)/(r!*(n-r)!)


a)


First compute 15 C 3 to figure out how many ways there are to pick 3 people from a pool of 15


n C r = (n!)/(r!(n-r)!)


15 C 3 = (15!)/(3!*(15-3)!)


15 C 3 = (15!)/(3!*12!)


15 C 3 = (15*14*13*12!)/(3!*12!)


15 C 3 = (15*14*13)/(3!)


15 C 3 = (15*14*13)/(3*2*1)


15 C 3 = (2730)/(6)


15 C 3 = 455


So there are 455 ways to pick 3 people from a pool of 15


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Now compute 9 C 3 to figure out how many ways there are to pick 3 people from just the debate team.


n C r = (n!)/(r!(n-r)!)


9 C 3 = (9!)/(3!*(9-3)!)


9 C 3 = (9!)/(3!*6!)


9 C 3 = (9*8*7*6!)/(3!*6!)


9 C 3 = (9*8*7)/(3!)


9 C 3 = (9*8*7)/(3*2*1)


9 C 3 = (504)/(6)


9 C 3 = 84


There are 84 ways to pick 3 people from a pool of 9 (the debate team only)


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Divide the results:


(9 C 3)/(15 C 3) = 84/455 = <font color="red">0.184615</font>


The probability of having all 3 members selected from the debate team is approximately <font color="red">0.184615</font>


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b)


There are 15 - 9 = 6 people that aren't on the debate team


Compute 6 C 3 to get...


n C r = (n!)/(r!(n-r)!)


6 C 3 = (6!)/(3!*(6-3)!)


6 C 3 = (6!)/(3!*3!)


6 C 3 = (6*5*4*3!)/(3!*3!)


6 C 3 = (6*5*4)/(3!)


6 C 3 = (6*5*4)/(3*2*1)


6 C 3 = (120)/(6)


6 C 3 = 20


(6 C 3)/(15 C 3) = 20/455 = <font color="red">0.043956</font>


The probability of having no members of the debate team selected is roughly <font color="red">0.043956</font>



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