Basic die question (probability) Suppose we roll a fair die twice. Let $X$ be the number of sixes obtained and $Y$ be the number of threes recorded. Then, we have:
$P(X = 0 \cap Y = 0) = \frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$
Could someone explain how they got this step by step please?
 A: No 6s and no 3s means 4 possibilities left for each die.  So you have 16 configurations with no 6s and no 3s.  There are 36 altogether, giving
$16/36 = 4/9$ chance of no 6s and no 3s.
A: Presumably different rolls of a die are independent, and if you get $X=0$ and $Y=0$, that means that both rolls you got neither a three nor a six. But the probability of this for one roll is $4/6$. But since the rolls are independent, we square this to get $4/9$.
A: Since $P(A \cap B) = P(A)+P(B)-P(A \cup B)$,
$P(X=0 \cap Y=0) = P(X=0)+P(Y=0)-P(X=0 \cup Y=0) = \frac{25}{36}+\frac{25}{36}-\frac{34}{36}$.
This also happens to equal $\left(\frac{4}{6}\right)^2$, but it is easier to understand in the context of this problem. $\frac{4}{6}\times\frac{4}{6}$ is not helpful because it implies several things that are incorrect; note that $P(X=0) \ne \frac{4}{6}$, $P(Y=0) \ne \frac{4}{6}$ and $P(X=0 \cap Y=0) \ne P(X=0)P(Y=0)$ ie they are not independent events. $\frac{4}{6}\times\frac{4}{6}$ is an incorrect answer to this question even though it gives the same numeric result. To demonstrate that they are equal would require simple set theory, which is likely outside the scope of this problem. 
