<s>[INST] Two dice are rolled, what is the probability that the minimum value of the two dice is $3$? Let $D_1$ and $D_2$ represent the values obtained from dice $1$ and dice $2$. We are looking for $P(\min(D_1, D_2) = 3)$
I.e.
$$
\begin{align}
& P([D_1 = 3, D_2 = 3, 4, 5, 6] \cup [D_2 = 3, D_1 = 3, 4, 5, 6]) 
= 2P([D_1 = 3, D_2 = 3, 4, 5, 6]) \\[8pt]
& = 2P([D_1 = 3 , D_2 = 3] \cup [D_1 = 3 , D_2 = 4] \cup [D_1 = 3 , D_2 = 5] \cup [D_1 = 3 , D_2 = 6]) \\[8pt]
& = 2P(D_1 = 3, D_2 = 3)^4 \\[8pt]
& = 2\left(\frac{1}{6}\right)^4 \\[8pt]
& = \frac{1}{648}
\end{align}
$$
Does that seem correct? It looks too extreme...
 [/INST] Be careful: the probability of the union is not equal to the product of the probabilities (that would be the intersection, and it's only true if the events are independent).  Nor is the probability of the union equal to the sum of the probabilities (unless the events are disjoint / "mutually exclusive").
\begin{align*}
P(\min(D_1, D_2 = 3))
&= P([D_1 = 3, D_2 = 3, 4, 5, 6] \cup [D_2 = 3, D_1 = 3, 4, 5, 6]) \\
&= P(D_1 = 3, D_2 = 3) \\
&\quad + P(D_1 = 3, D_2 = 4,5,6) \\
&\quad + P(D_1 = 4,5,6, D_2 = 3) \\
\end{align*}
This last step is valid because the different events are disjoint.  So then we have
\begin{align*}
&= P(D_1 = 3, D_2 = 3) + 2P(D_1 = 3, D_2 = 4,5,6) \\
&= P(D_1 = 3) P(D_2 = 3) + 2 P(D_1 = 3) P(D_2 = 4,5,6)
\textbf{ (because of independence)} \\
&= \frac16 \frac16 + 2 \frac16 \frac36 \\
&= \frac{7}{36}
\end{align*}
Of course, a direct count would have been easier: there are seven ways to achieve a minimum of 3 out of 36 possible rolls.
</s>