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| − | Is this one set up by <math>frac{6 \choose 5}{40 \choose 6}</math> | + | Is this one set up by <math>frac{{6 \choose 5}}{{40 \choose 6}}</math> |
-Brian | -Brian | ||
You have it mostly right. The problem is saying that there are 6 winning numbers out of 40. and they want to know the probability that person will choose 5 of those winning numbers and 1 other arbitrary number so your almost right but you have to take into account the arbitrary number so there would be 34 choices for that so it would be: (6 choose 5)(34)/(40 choose 6) --[[User:Krwade|Krwade]] 16:55, 17 February 2009 (UTC) | You have it mostly right. The problem is saying that there are 6 winning numbers out of 40. and they want to know the probability that person will choose 5 of those winning numbers and 1 other arbitrary number so your almost right but you have to take into account the arbitrary number so there would be 34 choices for that so it would be: (6 choose 5)(34)/(40 choose 6) --[[User:Krwade|Krwade]] 16:55, 17 February 2009 (UTC) | ||
Revision as of 20:03, 17 February 2009
I feel this is similar to the previous problem (28) but the phrase 'choosing 5 (but not six) numbers' throws me. How many numbers is the play picking? At first I thought it was 5, but wouldn't that make choosing 6 impossible? - Jnikowit
Is this one set up by $ frac{{6 \choose 5}}{{40 \choose 6}} $
-Brian
You have it mostly right. The problem is saying that there are 6 winning numbers out of 40. and they want to know the probability that person will choose 5 of those winning numbers and 1 other arbitrary number so your almost right but you have to take into account the arbitrary number so there would be 34 choices for that so it would be: (6 choose 5)(34)/(40 choose 6) --Krwade 16:55, 17 February 2009 (UTC)
