Conditional probability question, joint probability for discrete random variables I have a homework question that I want to make sure I'm getting it right.
This is a joint probability table for the proportions of survey respondents who smoke and who have had heart attacks.
                          Smoker Non-Smoker
Heart Attack      0.03      0.03              
No Heart Attack0.44      0.50              
If a person is a smoker, are they more likely to have had a heart attack than someone who is not a smoker?
So I know that $P(A|B) = \dfrac{P(A \cap B)}{P(B)}$
So I could think of the question like: Find the conditional probability that a randomly selected person is a victim of a heart-attack, given that they’re a smoker/ non-smoker.
$ P(X = Smoker \cap Y = Heart Attack) = 0.03  $
    $     P(X = NonSmoker \cap Y = Heart Attack) = 0.03  $
$P(Y) = 0.06$
That would mean that a smoker and a non-smoker would both have a $\dfrac{0.03}{0.06}$ chance of having a heart attack?
This defies my intuition because $\dfrac{3}{47}$ smokers get heart attacks, while only $\dfrac{3}{53}$ non-smokers get heart attacks.
 A: You are given that they are a smoker/non-smoker. So your math is correct for what you have expressed, however you've done the conditional probabilities backwards. You are interested in $P(\text{ heart attack }| \text{ smoker })$ and $P(\text{ heart attack }| \text{ non-smoker })$. Calculating these will give you a more intuitive result for your question. 
In the general case, consider $P(A|B)$ reading as "the probability of $A$ given $B$". Then your formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ should hopefully make some intuitive sense looking at your denominator. 
A: 
$ P(X = Smoker$ | $Y = Heart Attack) = 0.03  $
    $     P(X = NonSmoker$ | $Y = Heart Attack) = 0.03  $
    $P(Y) = 0.06$

No, from the table you have joint probabilities:
$ P(X = \textsf{Smoker} \cap Y = \textsf{Heart Attack}) = 0.03  $
$ P(X = \textsf{Non-Smoker}\cap Y = \textsf{Heart Attack}) = 0.03 $
$ \therefore P(Y=\textsf{Heart Attack}) = 0.06$
Then you calculate ${\large P}\big(X=\textsf{Smoker}\mid Y=\textsf{Heart Attack}\big) = 0.50$
This is the probability that someone is a smoker given that they have had a heart attack.
Now, according to the table it is slightly more likely for a smoker to have a heart attack than for a non-smoker. 
$$P(Y=\textsf{Heart Attack}\mid X=\textsf{Smoker}) = \frac{0.03}{0.03+0.44} = \frac 3{47} \\ P(Y=\textsf{Heart Attack}\mid X=\textsf{Non-Smoker}) = \frac{0.03}{0.03+0.50} = \frac 3{53}$$
However there are only slightly more non-smokers than smokers over all.   This happens to balance out so that there are an equal number of smokers and non-smokers who have had heart-attacks.
