# Recent content by sandman

1. ### Chance without replacement

I'm ready to try this Going back to this scenario, 2 good balls I want out of 4 so the chance to not draw either on first turn = (3-n)/(5-n) = 2/4 = 50%. So On average I only need to draw 1 ball to get either or? This is part I'm stuck on due to only needing 2 rs and 3 bs how should I handle...
2. ### Chance without replacement

Is this supposed to be (98-n)/(100-n)? EDIT: I"m sorry, just matters that first term would be 98/100 depending on how you start n I apologize i don't know how to insert symbols and hope math lingo is right Does this work? For average for drawing the one remaining ball would be summation of (1...
3. ### Chance without replacement

My main issue is how do you know when its better to switch to new bag or when to keep going until you get next ball. For example I assume if I drew either red or blue on first marble draw then I should switch right? rather then continue digging through bag for a 1 in 98? After you successfully...