How do you construct a probability tree for overlapping events? I'm new to probability so apologies if this question is very basic.
I was preparing a tree for an experiment involving drawing two cards from a deck of 52, one after the other.
The events of interest were the probability of drawing an ace followed by a black card.
The first part of the tree was simple - 4/52 for an ace, 48/52 for a non-ace.
The second part though, got me puzzled. The probability of drawing a black card on the second draw depends on whether the initial card was black. But the way the tree was formulated, the color of the 1st card is not revealed.
What exactly is going on here? Is the tree impossible to complete because it was not properly formulated?
Thanks for any tips / insights!
 A: We work through this using conditional probabilities.
The probability of drawing an ace as the first card is, as you've calculated, $1/13$.  Given that the first card is an ace, the probability that it is black, denoted by $p(\text{card 1 is black} | \text{card 1 is ace}) = 1/2$, and similarly for the probability that card 1 is not black; evidently, the probability that the first card is both an ace and black is equal to $2/52$, and similarly for the probability that the first card is both an ace and red.
Given that card 1 is black, the probability that card 2 is black is $12/51$.  If it isn't black, the probability that card 2 is black is $13/51$.
Combining all this using the rules of conditional probability gives us:
$\begin{eqnarray}
p(\text{ace followed by black}) &=& p(\text{2 is black | 1 is ace}) p(\text{1 is ace}) \\&=& p(\text{2 is black | 1 is ace and black}) p(\text{1 is ace and black)}) \\&&+ p(\text{2 is black | 1 is ace and red}) p(\text{1 is ace and red)}) 
\\ &=& (12/51)(2/52) + (13/51)(2/52) 
\\ &=& {50 \over 51*52} = 0.01886
\end{eqnarray}$
A: You need a slightly more sophisticated tree

