I just cooked up this problem and I get different answers when I try w/
and w/o combinations. Where am I going wrong?
Form a 4 member team w/ and 4 girls and 3 boys. What is the prob there are atleast 2 girls in the team ?
With combinations
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P(atleast 2 girls) = (4c2 + 4c3 + 4c4)/7c4 [ (2 girls + 3 girls + 4 girls) / total no. of ways ]
= 11/35
W/o combinations
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P(atleast 2 girls) = 1 - ( P (0 girls) + P ( 1 girl) )
P (0 girls) = 0
P (1 girl) = 4 * (3/7)(2/6)(1/5)(4/4) [ BBBG,BBGB,BGBB,GBBB ]
= 4/35
P(atleast 2 girls) = 1 - 4/35 = 31/35
Any good source for higher-level probability/combination questions?