What is the probability that if two cards are drawn from a standard deck without replacement that the first is red and the second is a heart? Two cards are drawn at random (without replacement) from a regular deck of 52 cards. What is the probability that the first card is a red and the second card is heart?
Let $A$ be the event that a red card is drawn and $B$ be the event the second card drawn is a heart.
We have;
$n(S) = C(52,2)$
$P(A) = \frac{26}{52}$
How is $P(B)$ calculated? Is a  solution even possible?
 A: In a deck of cards, there are four suits: clubs, diamonds, hearts, and spades. Diamonds and hearts are red; clubs and spades are black.  There are $13$ cards of each suit.  
We want to find the probability that the first card is red and the second card is a heart when two cards are drawn without replacement from a standard deck.
There are two possibilities:


*

*The first card is a diamond and the second card is a heart.

*Both cards are hearts.


Let $H$ denote the event that a heart is drawn; let $D$ denote the event that a diamond is drawn.
The first card is a diamond and the second card is a heart:  The probability of drawing a diamond on the first draw is $\Pr(D) = 13/52$.  Of the $51$ cards that remain, $13$ are hearts.  Hence, the probability of drawing a heart given that a diamond was selected on the first draw is $\Pr(H \mid D) = 13/51$.  Hence, the probability that the first card is a diamond and the second card is a heart is 
$$\Pr(H \mid D)\Pr(D) = \left(\frac{13}{51}\right)\left(\frac{13}{52}\right)$$
Both cards are hearts:  The probability of drawing a heart on the first draw is $\Pr(H) = 13/52$.  Of the $51$ cards that remain, $12$ are hearts.  Hence, the probability of drawing a heart given that a heart was drawn on the first draw is $\Pr(H \mid H) = 12/51$.  Thus, the probability that both cards are hearts is 
$$\Pr(H \mid H)\Pr(H) = \left(\frac{12}{51}\right)\left(\frac{13}{52}\right)$$
Since these cases are mutually exclusive and exhaustive, the desired probability can be found by adding the probabilities for the two cases.
A: Hint:
Try to break it down by cases:
1) Case $1$: The first card is a heart and the second card is a heart
2) Case $2$: The first card is not a heart and the second card is a heart
Edit:
$$P(B)=P(B|H)P(H)+P(B|H^c)P(H^c)$$
A: Edit: The following answer proposes a solution to the probability of the second card being drawn being hearts, interpreting that the question is not linked with the statement that the first card is red. If you take the question to actually need the first card to be red, please refer to N. F. Taussig's answer.
Let's define the events
$F = \{ \text{first card is hearts}\}$,
$S = \{  \text{second card is hearts}\}$
We want to calculate the probability that the second card is hearts, that is, we are looking for $P(S)$. 
Now, we can break this experiment up into two distinct scenarios: either the first card you picked was hearts, or it wasn't ($F$ and $F^C$, respectively).
So, by conditional probability we have that
$$P(S) = P(S |F)P(F)+P(S |F^C)P(F^C)$$
hence
$$P(S) = \left(\dfrac{12}{51}\right)\left( \dfrac{13}{52} \right)+\left(\dfrac{13}{51}\right)\left( \dfrac{39}{52} \right)= \frac{1}{4}$$
since we have $13$ cards of each suit. Notice that this is the same as the probability of just picking a random card and getting hearts (why?).
