Really struggling with this question and could use some guidance.
If a poker hand (five cards) is known to contain at least three aces, what is the probability that it contains all four aces?
I have been taking an approach using conditional probability to solve this and am coming up short. Thanks in advance
If there are at least three aces in the hand, either it contains exactly three aces or exactly four aces. Thus $$P(4A|\ge3A)=\frac{P(4A)}{P(3A)+P(4A)}=\frac{\#(4A)}{\#(3A)+\#(4A)}=\frac{48}{\binom43\binom{48}2+48}=\frac{48}{4560}=\frac1{95}$$ where $\binom43\binom{48}2$ consists of choosing three aces from four and two cards from the 48 non-aces.