I have the following probability problem I am trying to figure out how to go about. It seems to be a conditional probability but it seems more intuitive than that.

Suppose that a hand of 5 cards is selected at random, without replacement, from a standard deck of 52 playing cards.

(a) If it is known that the hands contains the ace of spades, what is the probability that the hand contains exactly two aces?

(b)If it is known that the hand contains at least one ace, what is the probability the hand contains exactly two aces?

I am on part (a)

Well If I just look at the first probability, which is, the 5-card hand contains an ace of spade I see it as this

1 ace of spades out of 4 total aces

4 non-aces out of 48 remaining non-ace cards

then

Pr(the hand contains the ace of spades)=

$\displaystyle (^{4}_1)*(^{48}_4)$

correct?

Then the hand containing two aces given you know the hand contains the ace of spades would be.

$\displaystyle (^{4}_2)*(^{48}_3)$

If it seems that the condition of knowing the hand contains an ace of spades does not really affect the probability the hand could contain two aces.

Intutively for me this makes some sense but I feel like I leaving something out. Also, it seems like part (b) would be exact the same as part (a)

All help is appreciated