Probability that the second and third card extracted are both aces?

Question

I would like to know if the following reasoning is correct: the situation is the extraction without replacement of 3 cards out of 40, where there are 4 aces. I would say the cardinality of the sample space is $\frac {40!}{(40-3)!}$ because order matters here, and P(C2 and C3 are aces) = P(C2 is an ace) × P(C3 is an ace | C2 is an ace) = $\frac{4\cdot \binom {39}{1}}{40\cdot39\cdot38} \frac{3\cdot \binom {38}{1}}{40\cdot39\cdot38} $

Answer 1

Let $E$ denote the event that the second card is an ace and let $F$ denote the event that the third card is an ace.

Then:$$P (E\cap F)=P (E)P(F|E)=\frac4{40}\frac3{39}$$

Answer 2

If you need exactly two aces (second and third card only) then first pick the non-ace ($36$) then pick the two aces in order $(4\cdot 3)$.

If you need at least two aces (second and third cards definitely, but the first can be an ace also) then choose the second and third cards as aces $(4 \cdot 3)$ then choose the first card as anything else $(38)$.

Answer 3

You have got is just slightly wrong. The probability that the second card is an ace will be $\frac3{40}$ because it is $\frac3{39}$ times the probability that the first card was not an ace, plus $\frac2{39}$ times the probability that the first card was an ace: $$ \frac3{39}\frac{37}{40}+\frac2{39}\frac3{40} = \frac3{39}\frac{37}{40}+\frac2{39}\frac3{40} = \frac3{39}\frac{37}{40}+\frac3{39}\frac2{40} = \frac3{39}\left(\frac{37}{40}+\frac2{40} \right)=\frac3{40} $$ Then the probability that the third card is also an ace will be $\frac2{39}$ to the answer probability is $\frac1{260}$.

If you want to do it by counting cases, the number of success cases is $3$ (which A is in slot 2) times $2$ (which other A is in slot 3) times $388$ (which card is in slot $1$).
$$ \frac{3\cdot 2\cdot 38}{40\cdot 39 \cdot 38} = \frac1{260}$$