Question

I guess I'm not following the below example Reiny answered previously. The reason for my confusion is if a coin has been removed, it stays removed, does it not? So how can you have a 1/16 chance on the second draw? If one coin has been removed, the second draw would have 15 coins left, right? Which would mean you'd have a 1/15 chance for drawing the penny, right? If this line of thought is wrong, please explain. My answer would be (10/16)(1/15) = 1/24

10 dimes
5 nickels
1 penny

Prob(dime, then penny)
= (10/16)(1/16)
= 5/128

Answer 1

(10/16)(1/15) = 1/24
Yes, that is correct if it stays removed.