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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?

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    $\begingroup$ $Pr(A\cap B)\neq Pr(A)Pr(B)$ in general. They are only equal if $A$ and $B$ are independent events. You seem to have tried using the formula here $Pr(E\mid F)=\frac{Pr(E\cap F)}{Pr(F)}$ and tried to split up the top as though it were independent. The correct observation should have been $Pr(E\cap F)=\frac{3}{8}$ giving you a final answer of $\frac{3}{8}/\frac{5}{8}=\frac{3}{5}$ $\endgroup$ – JMoravitz Jul 24 '17 at 16:21
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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.

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  • $\begingroup$ wait then is choosing a vowel dependent on choosing a letter in the first half of the alphabet? $\endgroup$ – space Jul 25 '17 at 4:22
  • $\begingroup$ In this case, yes, and in general, yes. The probability of choosing a vowel, given that your choice is in the first half of the alphabet, is not the same as the probability of choosing a vowel out of the second half. $\endgroup$ – G Tony Jacobs Jul 25 '17 at 4:24
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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.

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  • $\begingroup$ I really like that idea! Thank you. $\endgroup$ – space Jul 25 '17 at 4:23
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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.

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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.

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