<s>[INST] Let $S = \{1,2,3,4,5,6\}$ be a sample space of equally likely outcomes Let $S = \{1,2,3,4,5,6\}$ be a sample space of equally likely outcomes such that $P(\{s\}) = \frac{1}{6}$, $\forall s \in S$ let X = $I_{(0,9,3,1)}, Y = I_{(2,4,4,2)}$ and $W = X + Y$
Compute $P(W = 1)$
My attempt:
$$W(1) = X(1) + Y(1) = I_{(0,9,3,1)}(1) + I_{(2,4,4,2)}(1) = 1 + 0 = 1$$
Since $P(\{s\}) = 1/6$, then $P(1) = 1$. 
Would this be correct?  
 [/INST] Not entirely:
For one you are correct in saying:
$$
W(1) = 1
$$
But notice also that 
\begin{align}
W(2) &= X(2) +Y(2) = 0 + 1 = 1,  \\
W(3) &= X(3) +Y(3) = 1 + 0 = 1,  \\
W(4) &= X(4) +Y(4) = 0 + 1 = 1,  \\
W(5) &= X(5) +Y(5) = 0 + 0 = 0,  \\
W(6) &= X(6) +Y(6) = 0 + 0 = 0.
\end{align}
So we get that 
\begin{align}
P(W = 1) &= P \bigl(\{1, 2, 3, 4 \} \bigr) \\
&= P\bigl(\{1\}) +  P\bigl(\{2\}) +  P\bigl(\{3\}) +  P\bigl(\{4\}) \\
&= \frac 16 + \frac 16 +\frac 16 +\frac 16 = \frac 46.
\end{align}
</s>