Conditional Probability 1)  A card is drawn from a standard deck of $52$ cards. The card is drawn from the pack and not replaced.  A second card is then drawn from the deck. Determine the probability:
$a)$    that the second card is a spade, given that the first card is a spade.
$b)$    That both cards are hearts.
$c)$    the first card is a diamond and the second card is a heart.
For part $a$, I put $P(\frac{S1}{S2})= \frac{4}{51}$
And part $c$, $\frac{4}{52}$ divides $\frac{4}{51}$ = $\frac{51}{52}$
I'm confused with part $b$ and will appreciate some help and can you guys kindly check over my part $a$ and $c$? 
Thank you very much. 
 A: $b)$ $P(\text{1st card is heart & second card is also heart})= \dfrac{13}{52}\cdot \dfrac{12}{51}= \dfrac{1}{17}$
For checking $a)$, the probability should be $\dfrac{12}{51}$ or $\dfrac{4}{17}$
For $c)$, the probability is $\dfrac{13}{52}\cdot \dfrac{13}{51} = \dfrac{13}{204}$
A: 
b) That both cards are hearts:



*

*The number of ways to draw $2$ out of $52$ cards is $\binom{52}{2}=1326$

*The number of ways to draw $2$ out of $13$ hearts is $\binom{13}{2}=78$

*Hence the probability of drawing $2$ hearts is $\frac{78}{1326}\approx5.88\%$

A: 
a) that the second card is a spade, given that the first card is a spade.

If you remove one card from the deck and it is a spade, then there are $12$ spades left in the deck of $51$ remaining cards.   So therefore the (conditional) probability that the next card is also a spade when given that the first card is a spade is:-

b) That both cards are hearts.

It is the probability that the first card is one of the $13$ hearts times the (conditional) probability that the second card is also a heart when given that the first was.

c) the first card is a diamond and the second card is a heart.

Similar, although this time removing one of the $13$ diamonds does not reduce the number of hearts left in the deck (just the size of the deck).
