My problem is this:
A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
Here is my work so far:
probability that there is at least one ace in the 13: 1-(C(52,13)-C(48,13))/C(52,13) = .3038
probability that there is an ace left at the end of dealing 12 cards: 12!/13! = .076923
probability of both of those happening: .3038*.076923 = .02337 = 2.34%
does this look correct? Thank you so much in advance. Nolan