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?
Best Answer
$P(1,5)= P(2,4)=P(3,3)=P(4,2)=P(5,1) = 1/36$
Thus, you need $P(3,3)/P(6) = 1/5$. ($P(sum= 6|(3,3))=1$, and sum can be 6 in those 5 ways above with total probability 5/36)