<s>[INST] When $2$ $6$-sided die are rolled, what is the probability that the first is odd and the difference is $0$? The probability the the first die is odd is $\frac{3}{6}=\frac{1}{2}$.
The probability that first is odd and diff is $0$ is $\frac{3}{6}\cdot \frac{1}{6} = \frac{3}{36} = \frac{1}{12}$.
Is $\frac{1}{12}$ correct? 
 [/INST] Set $X_1$: outcome of the first die. There are three successful outcomes: $X_1 = 1, X_1 = 3, X_1 = 5$; since the probabilities are uniform, $P(X_1= \{1,3,5\}) = \frac{1}{6}$. 'Difference equal to $0$' means the second outcome is the same; hence, we need 
$P(X_2 \cap X_1) = P(X_2 = 1|X_1=1)P(X_1=1) + P(X_2 = 3|X_1=3)P(X_1=3) + P(X_2 = 5|X_1=5)P(X_1=5) = \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} + \frac{1}{6} \times \frac{1}{6} = \frac{3}{36}$
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