probability of different events for a hand of cards hand $H$ of 3 random cards are dealt from an ordinary deck of 52. Let
$E_1$ denote the event that $H$ has at least one Ace, $E_2$ denote the event that $H$ has at least two Aces, and let $E_{AS}$ denote the event that $H$ includes the Ace of
Spades.


*

*What are $Pr(E_1), Pr(E_2)$ and $Pr(E_{AS})$?

*What is the conditional probability $Pr(E_2 | E_1)$?

*What is the conditional probability $Pr(E_2 | E_{AS})$?


Solution:


*

*$$\begin{align}
Pr(E_1)    & = 1 - \frac{48 \choose 3}{52 \choose 3}\\
Pr(E_2)    & = \frac{\left[{4 \choose 3}+{4 \choose 2}{48 \choose 1}\right]} {52 \choose 3}\\ 
Pr(E_{AS}) & = \frac{51 \choose 2}{52 \choose 3}
\end{align}$$

*$$\begin{align}
Pr(E_2|E_1) & = \frac{Pr(E_2 ∩ E_1)}{Pr(E_1)} \\
&  = \frac{Pr(E_2)}{Pr(E_1)} \\
&  = \frac{\left[{4 \choose 3}+{4 \choose 2}{48 \choose 1}\right]}{ \left[{52 \choose 3}-{48 \choose 3}\right]}
\end{align}$$

*$$ Pr(E_2|E_{AS}) = 1 - \left[\frac{48 \choose 2}{51 \choose 2}\right]$$

i'm having trouble understanding the solution of this problem. so far the only $Pr(E)$ i understand is 1: (taking the probability that the hand does not contain an ace ${48 \choose 3}$ dividing it by the sample space ${52 \choose 3}$, this gives us the probability that the hand does not contain an ace, taking its complement we end up with the probability that the hand does contain an ace) . however for all the other $Pr$ i'm lost, can someone help and explain how the other probabilities are computed.
 A: How many ways are there to get at least $2$ aces? You could get $3$ aces or you could get $2$ aces and a non-ace. There are $\binom43$ different sets of $3$ aces. There are $\binom42$ sets of $2$ aces and $\binom{48}1$ ways to choose a single non-ace, so there are $\binom42\binom{48}1$ ways to choose a hand with exactly $2$ aces. Altogether, then, there are $\binom43+\binom42\binom{48}1$ hands with at least $2$ aces, and since there are $\binom{52}3$ hands, 
$$\operatorname{Pr}(E_2)=\frac{\binom43+\binom42\binom{48}1}{\binom{52}3}\;.$$
To get a hand that includes the ace of spades you must get the ace of spades and any $2$ other cards. The only choice here is in the other cards: ther are $51$ candidates, and you can have any $2$ of them, so there are $\binom{51}2$ ways to pick the other $2$ cards. Thus, there are $\binom{51}2$ hands that include the ace of spades, and
$$\operatorname{Pr}(E_{AS})=\frac{\binom{51}2}{\binom{52}3}\;.$$
For the next one you use the formula that you have in the question:
$$\operatorname{Pr}(E_2\mid E_1) =\frac{\operatorname{Pr}(E_2\cap E_1)}{\operatorname{Pr}(E_1)}\;.$$
You know the denominator, so you have only to compute the numerator. The event $E_1\cap E_2$ is the event that the hand has at least $1$ ace and at least $2$ aces. Clearly that’s the same as having at least $2$ aces: $E_1\cap E_2=E_2$. Thus,
$$\operatorname{Pr}(E_2\mid E_1) =\frac{\operatorname{Pr}(E_2)}{\operatorname{Pr}(E_1)}\;,$$ 
and you need only substitute the known values of $\operatorname{Pr}(E_1)$ and $\operatorname{Pr}(E_2)$.
You could use the same method to calculate $\operatorname{Pr}(E_2\mid E_{AS})$, but it’s simpler in this case to take a different approach. Suppose that you know that $E_{AS}$ has occurred, i.e., that you have the ace of spades. What is the probability that $E_2$ has not occurred? That’s the probability that you have no other ace. Among the $\binom{51}2$ hands that include the ace of spades, how many have no other ace? There must be $\binom{48}2$, the number of ways of picking two non-aces. Thus, given that you have the ace of spades, the probability that it’s your only ace is
$$\frac{\binom{48}2}{\binom{51}2}\;,$$
and the probability that you have at least $2$ aces is therefore the complementary probability,
$$1-\frac{\binom{48}2}{\binom{51}2}\;.$$
