draw a spade card given the first card is black Suppose two cards are dealt from a deck of $52$. What is the probability that the second card is a spade given that the first card is black?
I am thinking it may be $26/52 \times 4/25 = 104/1300$.
 A: I think it's best to break it into two events.  The probability that the second card is a spade given that the first card is a spade is 12/51.  The probability that the second card is a spade given the first card is a club is 13/51.  By the law of total probability the answer should be 1/2 (12/51) + 1/2 (13/51).
A: Conditional probability states that $$P(A | B) = \frac{P(A\cap B)}{P(B)}$$
It reads "The probability that $A$ happens given $B$ has happened is the probability of $A$ and $B$ both happening divided by the probability that $B$ happened, as $B$ is the event we are conditioning on.
For this question, let $A$ be the event that a spade is drawn on the second draw, and $B$ is a black is drawn on the first draw.
The probability of a black on the first draw, $P(B)$, is easy enough, it is $1/2$, as we have an equal chance of either red or black.
The numerator , $P(A \cap B)$, is a little more complicated to find, as we have two cases we need to consider:


*

*the first card was a black spade

*the first card was black but was something other than a spade (a club)
The first case has a probability of $\frac{13}{52} * \frac{12}{51}$ because we started with $13$ spades originally, but then a black spade was picked on the first draw, so we only have $12$ left out of the remaining $51$ cards.
See if you can continue from here with the second case, and then once you have that value, you can then add it to the answer for the first case and then you will have $P(A \cap B)$ which then you will be able to find the final answer for the original problem.
A: Given that the second card is a spade, there are $51$ possibilities for the first card, of which $25$ are black.
So define the event $B$ that the first card is black and $S$ that the second card is a spade.  Then
$$P(B\cap S)=P(S)P(B\mid S)=\frac14\times\frac{25}{51}$$
and
$$P(S\mid B)=\frac{P(S\cap B)}{P(B)}=\frac{\frac14\times\frac{25}{51}}{\frac12}=\frac{25}{102}\ .$$
