Question:
Three dice (six faces: each face -> number 1 to 6) are rolled. What is the number of possible outcomes such that at least one die shows number 2?
My attempt:
One die has to show two. The possible ways in which the other two can show their numbers is $6 \cdot 6 = 36$. But, these dices can be interchanged in $3$ ways. Thus, answer $=36 \cdot 3 = 108$.
But it is wrong. I need help identifying where I went wrong.
Best Answer
You are double counting the outcomes. You have to assume there is exactly one 2 instead of at least one 2, then exactly two 2's, and finally exactly three 2's:
Exactly one 2: $1\times 5\times 5\times \binom{3}{1}=75$
Exactly two 2's: $1\times 1\times 5\times \binom{3}2=15$
Exactly three 2's: $1\times 1\times1 =1$
This gives $75+15+1=91$ desirable outcomes.
An easier method would be to just find the number of outcomes with no 2's, then subtract that from the total number of outcomes:
$$6\cdot 6\cdot 6 - 5\cdot 5\cdot 5=216-125=91$$