<s>[INST] Dice roll: Probability of getting a given sum + a particular set of numbers The problem goes as: "An unbiased usual six-sided die is thrown three times. The sum of the numbers coming up is $10$. What is the probability that $2$ has appeared at least once?"
Options:
$A. 1/36$
$B. 5/36$
$C. 91/216$
$D. 1/18$
My approach: I understood this as finding the probability of getting at least one $2$ (say event $A$) in each triplet subject to the condition that the sum of the numbers in the triplet $= 10$ (say event $B$). Using the rules of conditional probability,
$P(A|B)=P(A \cap B)/P(B)$
Manually listing all triplets that sum to $10$, I found $P(B)=27/216$ and counting the triplets that had at least one $2$ I found $P(A \cap B)=12/216$. Thus getting the answer $4/9$ which is definitely wrong looking at the options, I've no idea how to go ahead of this.
 [/INST] The ways to get a sum of $10$ are as follows:
$$1 + 3 + 6$$
$$1 + 4 + 5$$
$$1 + 5 + 4$$
$$1 + 6 + 3$$
$$\color{green}{2 + 2 + 6}$$
$$\color{green}{2 + 3 + 5}$$
$$\color{green}{2 + 4 + 4}$$
$$\color{green}{2 + 5 + 3}$$
$$\color{green}{2 + 6 + 2}$$
$$3 + 1 + 6$$
$$\color{green}{3 + 2 + 5}$$
$$3 + 3 + 4$$
$$3 + 4 + 3$$
$$\color{green}{3 + 5 + 2}$$
$$3 + 6 + 1$$
$$4 + 1 + 5$$
$$\color{green}{4 + 2 + 4}$$
$$4 + 3 + 3$$
$$\color{green}{4 + 4 + 2}$$
$$4 + 5 + 1$$
$$5 + 1 + 4$$
$$\color{green}{5 + 2 + 3}$$
$$\color{green}{5 + 3 + 2}$$
$$5 + 4 + 1$$
$$6 + 1 + 3$$
$$\color{green}{6 + 2 + 2}$$
$$6 + 3 + 1$$
There are $4 + 5 + 6 + 5 + 4 + 3 = 27$ total possibilities, of which $12$ involve a two. This gives $$\frac{12}{27}=\boxed{\frac{4}{9}\,}$$
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