| Line 20: | Line 20: | ||
computer picks 6 numbers and the rest of 34 we are to pick.. in order to hit 5 numbers correctly. | computer picks 6 numbers and the rest of 34 we are to pick.. in order to hit 5 numbers correctly. | ||
| − | so it will be <math>\frac{{6 \choose 5}{34 \choose 1}}{{40 \choose 6}}</math> | + | so it will be <math>\frac{{6 \choose 5}{34 \choose 1}}{{40 \choose 6}}</math>--[[User:Kangw|Kangw]] 20:41, 18 February 2009 (UTC) |
Revision as of 16:41, 18 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: $ \frac{{6 \choose 5}(34)}{{40 \choose 6}} $ --Krwade 16:55, 17 February 2009 (UTC)
I don't see why multiply by 34..?
Oh, I get it--"5 but not 6" means they actually aren't allowed to get the 6th number correct. Now, I'm still not entirely convinced about the 34, but I do get that it has something to do with picking an incorrect 6th number. Should it not be * 34/35 instead of * 34 (since we're looking for probability)? -Zoe
Aha. it is 34 because we should not get the number in a range with in which computers pick..
computer picks 6 numbers and the rest of 34 we are to pick.. in order to hit 5 numbers correctly.
so it will be $ \frac{{6 \choose 5}{34 \choose 1}}{{40 \choose 6}} $--Kangw 20:41, 18 February 2009 (UTC)
