What is the conditional probability that the second card is a Spade given that the second-to-last card is a Spade? What is the conditional probability that the second card is a Spade given that the second-to-last card is a Spade? Cards are dealt without replacement.
I know conditional probability is $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ My question is how do we find $P(B)$ and $P(A \cap B)$. Finding the P(B) means what is the 51st card place down is spades. Would $$P(B) = \frac{\binom{51}{13}}{\binom{52}{13}} \cdot \frac{13}{51}$$ and $$P(A \cap B) = \left(\frac{13}{52} + \frac{12}{51} + \frac{1}{2}\right)?$$
Any hint or advice will help. Thanks.

Let $A=$ {2nd card is a spade} and $B=$ {penultimate card is a spade}.
All the condition tells you is that for each of the other 51 positions, you have one fewer spade that it could be. So $P(A)=P(B)=\frac{13}{52}=\frac{1}{4}$, but $P(A|B)=P(B|A)=\frac{12}{51}=\frac{4}{17}$.
From there, it's easy to get $P(A \cap B)$ using the formula $P(A \cap B) = P(A|B) \cdot P(B)$.
