Probability of two rolling dice As I am taking some distance course on probability theory, I have no other option to ask my question than here. If they sound too basic, please bear me.
Q: I roll two dice. What's the probability I get at least one six? A: P(first die is six) = $\frac{1}{6}$.
P(second die is six)= $\frac{1}{6}$.
P(both dice are six)= $\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}$.
P(at least one six)= $\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}=30.6\%$
This is what explained in the video. I think the question asked is "at least one six", so if I get six on both the dice, its ok for me, right? If that was the case, I need to add $\frac{1}{36}$ rather than subtracting, isn't it. Am I thinking right. Please clarify my doubt.
So, the solution will be 
P(at least one six)= $\frac{1}{6}+\frac{1}{6}+\frac{1}{36}= 36.2 \%$
A: The first $\frac 1 6$ counts the number of solutions where the first dice rolls a 6:
$$\frac{ |\{(6,1), (6,2), ... (6, 6)\}|} {36} = \frac 6 {36}$$ 
The second $\frac 1 6$ counts the number of solutions where the second dice rolls a 6:
$$\frac{ |\{(1,6), (2,6), ... (6, 6)\}|} {36} = \frac 6 {36}$$ 
You want to combine the solution sets:
$$\frac{ |\{(6,1), (6,2), ... (6, 6)\} \cup \{(1,6), (2,6), ... (6, 6)\}|} {36} = \frac {\text{???}} {36}$$ 
The first solution set has 6 elements, the second has 6 elements, but they share 1 element.
$$\frac{ |\{(6,1), (6,2), ... (6, 6)\} \cup \{(1,6), (2,6), ... (6, 6)\}|} {36} = \frac{ |\{(6,1), (6,2), ... (6, 5), (1,6), (2,6), ... (5, 6), (6,6)\}|} {36} = \frac {6 + 6 - 1} {36}$$ 
A: 
In this answer, dice is the plural form of die. I assume that you are using regular 6-sided dice.
For one die, the probability of rolling 3 or lower is 12. For three dice, you want at least two out of three with 3 or lower.
P(all dice 3 or lower)=12×12×12=18
P(two dice 3 or lower)=12×12×12×3=38
Hope this helps........
