<s>[INST] Two dice are rolled and the sum of the face values is six. What is the probability that at least one of the dice came up a three? 
Two dice are rolled and the sum of the face values is six.  What is the probability that at least one of the dice came up a three?

Attempt: I feel I'm having trouble formalizing the expression for what I believe to be happening. This is a question concerning conditional probabilities so as such I will need first the probability of obtaining a sum of $6$:   
$$P(6) = P(6|(4,2))P(4,2) + P(6|(5,1))P(5,1) + P(6|(3,3))P(3,3)$$ 
So in terms of notation, the ordered pairs represent the way in which a sum of $6$ could be attained (ex: $(4,2)$, where you roll a $4$ on the first die and $2$ on the second).  With that being taken into account:
$$\left(\frac{2}{5}\right)\left(\frac{2}{36}\right) + \left(\frac{2}{5}\right)\left(\frac{2}{36}\right) + \left(\frac{1}{5}\right)\left(\frac{1}{36}\right)  = 0.0495$$
Now what was requested was the probability of a $3$ coming up given the sum of 6:  
\begin{align}P((3,3)|6) &= \frac{P(6|(3,3))P(3,3)}{P(6) = P(6|(4,2))P(4,2) + P(6|(5,1))P(5,1) + P(6|(3,3))P(3,3)}\\\\
&= \frac{0.0056}{0.0495}\\\\
&= 0.113\end{align}
But I still feel I'm not treating the condition on the sum of $6$ properly.. Is this correct?
 [/INST] I would use a table to first see this intuiutively
$$\begin{array}{|cc|cccccc|}\hline
&&&&&D_1\\
&&1&2&3&4&5&6\\
\hline
&1&2&3&4&5&\color{red}6&7\\
&2&3&4&5&\color{red}6&7&8\\
D_2&3&4&5&\color{blue}6&7&8&9\\
&4&5&\color{red}6&7&8&9&10\\
&5&\color{red}6&7&8&9&10&11\\
&6&7&8&9&10&11&12\\
\hline\end{array}$$
The red numbers show all combinations where the dice sum to $6$ and the blue shows the one event we are interested in, where at least one dice shows a $3$
So we have $\dfrac 15$ elements which satisfy the given constraints
Now consider it mathematically
We know that $P(\text{sum}=6)=\dfrac 5{36}$
We also know that $P((3,3))=\dfrac 1{36}$
We can therefore say that \begin{align}P(\text{sum}=6\text{ AND one dice shows }3)&=\frac{P((3,3))}{P(\text{sum}=6)} \\\\
&=\frac{\frac1{36}}{\frac5{36}}\\\\
&=\frac1{36}\cdot\frac{36}5\\\\
&=\frac 15\end{align}
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