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.
 A: Your approach makes sense, i.e., count the number of ways to roll $10$ with three dice:
$(1,3,6), (1,4,5), (1,5,4), (1,6,3),\\
(2,2,6),(2,3,5),(2,4,4),(2,5,3),(2,6,2),\\
(3,1,6),(3,2,5),(3,3,4),(3,4,3),(3,5,2),(3,6,1),\\
(4,1,5),(4,2,4),(4,3,3),(4,4,2),(4,5,1),\\
(5,1,4),(5,2,3),(5,3,2),(5,4,1),\\
(6,1,3),(6,2,2),(6,3,1)
$
Those are $27$ possibilities, each with equal probability. Of them, $12$ contain a $2$, so that gives us $\frac{12}{27}=\frac49$.
Are you sure there isn't a mistake in the question? None of the given choices are even equivalent to any fraction with $27$ in the denominator. This is suspicous.
A: 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 $$\frac{12}{27}=\boxed{\frac{4}{9}\,}$$
