Conditional probability of picking particular second letters in Scrabble, given the first letter picked I was doing a question today and couldn't understand the answer. 
Here's the question, my attempt, and the answer:
Question
In a game of Scrabble, Dalene has the seven letters A, D, E, K, O, Q and S. She picks two of these letters at random.
Given that she picks the letter Q first, what is the probability that she picks the letter D or the letter K second?
My attempt
A = letter Q, B = letter D or K
$P(A) = \frac{1}{7}$
$P(B) = \frac{2}{6} = \frac{1}{3}$
So my answer is just $P(B) = \frac{1}{3}$
Official answer
However, the Correct answer gives $\frac{2}{7}$. 
Why is this? 
Thanks in advance!
 A: First, I'll tell you how I solve it, then I'll go into some extra detail on all of the variables you might try to use, just to make everything clear.
As noted by the OP, we're looking for $P(B|A)$. Since the example is given in terms of Scrabble, I am assuming that we're picking 2 tiles from the set, without replacement. In this case, since we're given that the Q is chosen first, there aren't any complex cases where we need to enumerate what happens if a D or a K is chosen first. The Q gets picked, and given that we have 6 remaining tiles to pick from, with 2 of those choices being D and K, the answer is simply $2\over 6$ or $1\over 3$.
Now, if you want to do this more formally, you might decide to use the definition of conditional probability, which states:
$P(B|A) = {P(A\cap B) \over {P(A)}}$.
One problem is that in some cases $P(A\cap B)$ can take a lot of careful counting to compute.  The good news is, this is a small example, and we can actually list all the possible outcomes and count out the results to compute probabilities pretty quickly. There are 7 choices for the first draw, and 6 for the second for a total of 42, and order of draw matters, so here are the possible draws:
AD, AE, AK, AO, AQ, AS, 
DA, DE, DK, DO, DQ, DS, 
EA, ED, EK, EO, EQ, ES, 
KA, KD, KE, KO, KQ, KS, 
OA, OD, OE, OK, OQ, OS, 
QA, QD, QE, QK, QO, QS, 
SA, SD, SE, SK, SO, SQ
So, let's see what our terms are.  
$P(A)$ is the # of times Q is drawn first / the total # of draws = 6/42.
We don't use $P(B)$, but for the sake of completeness (and to correct a mistake the OP made), here it is:
$P(B)$ is the number of times a D or a K is drawn second / the total number of draws = 12/42. (which is 2/7, not what the OP had.)
$P(A\cap B)$ is the number of times a Q is drawn first AND a D or a K is drawn second / the total number of draws = 2/42.
$P(B|A) = {P(A\cap B) \over {P(A)}} = {{2/42}\over{6/42}} = {1\over 3}$.
I won't speculate further (as I tried to do in comments) about why the "official" answer the OP was given is listed as 2/7. I assume they made an error, but guessing at which error they made just tends to confuse people (as demonstrated so eloquently in the comments), so I'll stop here.
