Precal Probability Homework 
The letters in the word AARDVARK are printed on square pieces of cardboard with one letter per card.
The eight letters are placed in a hat and one letter is chosen at random. Find the following probabilities:
a) P(the letter chosen is a vowel given that the letter falls in the first half of the alphabet)

So this is a homework problem I'm having trouble with... the teacher has provided an answer (different from mine!) but not an explanation, and he'll be gone a while, so I'd like to know what I did wrong.
The probability that the letter falls in the first half of the alphabet is $5/8$, as there are 3 As, 1 D and 1 K.
The probability that the letter chosen is a vowel from AARDVARK is $3/8$.
Then using conditional probability, I get $(5/8)*(3/8) / (5/8) = (3/8)$. However, the answer is apparently $3/5$. What did I do wrong?
 A: The conditional probability $P(A|B)$ is given by the formula $\frac{P(A\cap B)}{P(B)}$. That numerator is the probability of choosing a vowel that is also in the first half of the alphabet $\left(\frac38\right)$. The denominator is simply the probability of choosing a letter in the first half of the alphabet $\left(\frac58\right)$.
It looks like you tried to use the formula $P(A\cap B)=P(A)P(B)$. That only works for independent events, and in this case, choosing a vowel and choosing a letter in the first half of the alphabet are not independent.
A: When you condition, in this case, it is like having a new hat with (only) the cards with A, A, A, D, K  (letters in the first half of the alphabet). You can easily see that probability of a vowel is then 3/5, under this condition. The formula $P(A|B)=P(A\cap B)/P(B)$, with events $A:=$ letter sampled is a vowel and $B:=$letter sampled is from the first half of the alphabet, just reflects this understanding.
A: You've miscalculated $P(\mathrm{vowel}\cap\mathrm{first\,half\,of\,alphabet})$ - it's simply $3/8$, not $5/8 * 3/8$. The reason you can't just multiply $P(\mathrm{vowel})$ and $P(\mathrm{first\,half\,of\,alphabet})$ is because the letter being a vowel and the letter being in the first half of the alphabet aren't independent events - in this case, the 3 vowels are all A, and hence all in the first half of the alphabet.
A: You are given that the letter drawn is $A,D,$ or $K$.  If you drew $R$ or $V$ you would draw again and ignore the first draw.  Three of the $A,D,K$ letters are vowels.
