Problem 7 · Three dice and a sum
Conditional probability by counting triples: $24/45 = 8/15$.
Integer answer, at most 4 digitsWe roll 3 dice, and it turns out that one of them shows the sum of the other two. What is the probability that at least one of the dice shows a 2? Express the answer as an irreducible fraction and give the product of numerator and denominator.
Reasoned solution
Key idea: this is a conditional probability: count the triples satisfying the condition and, among them, those containing a 2.
The valid triples are the permutations of multisets $\{a, b, a+b\}$ with $a + b \le 6$. By pairs $a \le b$: $(1,1), (2,2), (3,3)$ give multisets with repetition ($3$ orderings each) and $(1,2), (1,3), (1,4), (1,5), (2,3), (2,4)$ give three distinct values ($6$ orderings each):
Which contain a $2$? Those from $\{1,1,2\}$ ($3$), $\{1,2,3\}$ ($6$), $\{2,2,4\}$ ($3$), $\{2,3,5\}$ ($6$) and $\{2,4,6\}$ ($6$): $24$ in total.