Probability of two aces when you get at least one ace "In drawing two cards from a deck (without returning the first card) what is the probability of two aces when you get at least one ace?"
Are my calculations correct?


*

*$4*51$ possible events with an ace on the first draw

*$52*4$ with an ace on the second draw

*$4*3$ with an ace on both draws


Thus we have $4(51+52+3) = 4(106)$ events with at least one ace.
$(4*3)/(4*106) = 3/106$ is the probability of two aces when you get at least one ace. 
 A: The idea is fine, but the count is not quite right. 
The probability of at least one Ace is $1-\frac{\binom{48}{2}}{\binom{52}{2}}$. The probability of $2$ Aces is $\frac{\binom{4}{2}}{\binom{52}{2}}$. For the conditional probability, divide.
Remark: A way to count that is closer to yours is to note that we can select, in order, $2$ non-Aces is $(48)(47)$ ways. So the number of ways to choose, in order, a hand with at least one Ace is $(52)(51)-(48)(47)$. Then a calculation like yours gives conditional probability $\frac{(4)(3)}{(52)(51)-(48)(47)}$.
A: Not quite.  If you have $4 \cdot 51 = 204$ events with an ace on the first draw, then you have only $48 \cdot 4 = 192$ with an ace on the second draw.  Furthermore, the $204$ already contains two-ace draws.  (Alternatively, you can count $4 \cdot 48 = 192$ events with only an ace on the first draw, and then add in the $4 \cdot 3 = 12$ events with aces on both draws.)  So the total count is $204+192 = 396$.  Then $(4 \cdot 3)/396 = 1/33$ is the desired probability.
A: You have that 
$$P( 2As | 1As ) = \frac{P( 2As \text{ and } 1As )}{P( 1As) }$$
But $2As \text{ and } 1As = 2As$, so you just need to find $ P(2As) $ and $P(1As)$
$$P(2As) = \frac{4 \choose 2}{ 52 \choose 2} $$
$$P(1As) = 1- P(0As) = 1- \frac{{2\choose 48} }{ 52 \choose 2} $$
So
$$P( 2As | 1As) = \frac{4\choose 2}{ { 52 \choose 2}  - {2\choose 48} } = \frac{6}{ \frac{52\times 51}{2} - \frac{48\times 47}{2}} = \frac{6}{198} = \frac{1}{33} $$
