I see another way to show that 1⁄5 is the correct solution:
P(2 Aces | Ace of Spades revealed)= P(2 Aces AND Ace of Spades revealed)/P(Ace of Spades revealed)
(note: for further calculations, I’m assuming that there are 5 possible hands and the probability for each hand is 1⁄5, since it already has been revealed that there is at least one Ace. The end result would be the same if you would also set aside a random card in case you have no Ace,but the probabilities in the steps before the end results would have to change accordingly)
P(2 Aces AND Ace of Spades reveled)=P(2 Aces)*1/2 = 1⁄5 * 1⁄2 =1/10
I see another way to show that 1⁄5 is the correct solution:
P(2 Aces | Ace of Spades revealed)= P(2 Aces AND Ace of Spades revealed)/P(Ace of Spades revealed)
(note: for further calculations, I’m assuming that there are 5 possible hands and the probability for each hand is 1⁄5, since it already has been revealed that there is at least one Ace. The end result would be the same if you would also set aside a random card in case you have no Ace,but the probabilities in the steps before the end results would have to change accordingly)
P(2 Aces AND Ace of Spades reveled)=P(2 Aces)*1/2 = 1⁄5 * 1⁄2 =1/10
P(Ace of Spades revealed)= 2⁄5 * 1 + 1⁄5 * 1⁄2 = 5⁄10
(1/10)/(5/10)=1/5