Two cards are picked randomly from a deck. What is the probability that if the first card is Spades, the second card will be a Queen?
What I have so far:
Let $A$ be the event the first card is Spades.
Let $B$ be the event the second card is a Queen
$$P(B|A) = \frac{P(A\cap B)}{P(A)}$$ $$P(B|A) = \frac{P(A)P(B|A)}{P(A)}$$ $$P(B|A) = \frac{\frac{13}{52}P(B|A)}{\frac{13}{52}}$$
But now i get confused for $P(B|A)$ since its the problem im trying to solve in the first place... I know that $A$ and $B$ are dependent events since if a spade is drawn it could be the Queen spade.
So we we remove that possibility, which would make the probability of drawing the second queen be $3/51$. But, if that $P(B|A) = 3/51$ wouldnt that be the solution to the question in the first place? What happend to the $13/52$ parts?Also, if the first card is not a queen then shouldn't we consider that case as well?
Given the first card is a spade, there are two relevant cases: either the first card is the queen of spades (probability ${1 \over 13}$), or it is a different spade (probability ${12 \over 13}$).
In the first case, the probability the second card is a queen is ${3 \over 51}$. In the second case, the probability the second card is a queen is ${4 \over 51}$. Hence:
$$P({\rm queen}|{\rm spade}) = {1 \over 13}{3 \over 51}+ {12 \over 13}{4 \over 51} = {1 \over 13}$$