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So how come we don't even use the C(80,11)? It just seems like that number is important--[[User:Krwade|Krwade]] 16:41, 17 February 2009 (UTC) | So how come we don't even use the C(80,11)? It just seems like that number is important--[[User:Krwade|Krwade]] 16:41, 17 February 2009 (UTC) | ||
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| + | It does not with how many choices the computer has, what really matter is the number 11 I think, because we only pick 7 numbers out of 80. | ||
Revision as of 03:08, 18 February 2009
I am confused with what numbers I should put on top of my probability fraction. I have three numbers: # of ways for someone to choose 7 #s from 80 = C(80,7), # of ways for the computer to choose 11 numbers from 80 = C(80,11), and # of ways someone could have chosen the 7 from the 11 = C(11, 7). I think I understand that the number C(11, 7) is the total number of lottery numbers that are winners. And that the total number of lottery number is C(80, 11), but what do I do with C(80,7)?
~James Gilmore
I'm not sure if this is right, but my answer was C(11,7)/C(80,7). -Zoe
So how come we don't even use the C(80,11)? It just seems like that number is important--Krwade 16:41, 17 February 2009 (UTC)
It does not with how many choices the computer has, what really matter is the number 11 I think, because we only pick 7 numbers out of 80.
