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...
 A: When you roll two dice there are $6\times6=36$ possible results, each equally likely to occur. That is the denominator of the probability you are looking for (the possible outcomes). In order to determine the numerator of the probability we should count the favourable results, that is the results where the minimum of the two dice is equal to 3. 
In order to count them correctly (and not get confused) we set firstly the first dice to be equal to 3 and then give to the second all the "admissible" values so that the minimum of the two remains 3. Thus, dice two can take the values 3,4,5 and 6. So there are 4 results.
Now we repeat by setting the second dice equal to 3. But we have to be cautious not to include the results $(3,3)$ again. So the admissible values for dice two, are 4,5 and 6. This results in a total of 7 favourable results out of 36 possible.
$$P(\min\{D_1,D_2\}=3)=\frac{\left|(3,3),(3,4),(3,5),(3,6),(4,3)(5,3)(6,3)\right|}{36}=\frac{7}{36}$$ 
A: The probability is 7/36 in case when we differ the dices.
A: 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.
