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: $ \frac{{6 \choose 5}(34)}{{40 \choose 6}} $ --Krwade 16:55, 17 February 2009 (UTC)
