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?
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}