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October 1, 2014

October 1, 2014

Posted by **meep** on Saturday, January 12, 2013 at 3:51pm.

- math -
**Reiny**, Saturday, January 12, 2013 at 4:44pmnumber of committes without restriction

= C(12,3) =220

number of committess with three boys

= C(5,3) = 10

prob( 3boys) = 10/220 = 1/22

number of committees with 2 boys, 1 girs

= C(5,2) x C(7,1)

= 70

prob (2boys, 1 girl ) = 70/220 = 7/22

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