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I'm having a bit of a tough time with one of my probability questions. Here's what was asked, and the answer:

Let $S$ = $\{1,2,3\}$. $X=I_{\{1\}}$, $Y=I_{\{2,3\}}$, $Z=I_{\{1,2\}}$, and $W=X-Y+Z$.
a. Compute $W(1)$.
b. Compute $W(2)$.
c. Compute $W(3)$.

The correct answer for a. is $1$, b. $0$, and c. $0$.

Here's my work:
$W(1) = \left\{ \begin{array}{rcl} 1 & \mbox{if} & s \ \epsilon \ \{1\} \\ 0 & \mbox{otherwise} \end{array}\right.$ + $\left\{ \begin{array}{rcl} 1 & \mbox{if} & s \ \epsilon \ \{2,3\} \\ 0 & \mbox{otherwise} \end{array}\right.$ + $\left\{ \begin{array}{rcl} 1 & \mbox{if} & s \ \epsilon \ \{1,2\} \\ 0 & \mbox{otherwise} \end{array}\right.$

$W(1) = 2$
$W(2) = 0$
$W(3) = -1$

What am I doing wrong?

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  • $\begingroup$ Your answers are correct. $\endgroup$
    – Kavi Rama Murthy
    Commented Sep 19, 2020 at 0:01
  • $\begingroup$ Are you sure? I got a=1, b=0, and c=0 from the solution manual. $\endgroup$
    – JerBear
    Commented Sep 19, 2020 at 0:06
  • $\begingroup$ A better way to answer this is to write $X-Y+Z=1_{\{1\}}-(1_{\{2\}}+1_{\{3\}})+(1_{\{1\}}+1_{\{2\}})=21_{\{1\}}-1_{\{3\}}$ $\endgroup$
    – Kavi Rama Murthy
    Commented Sep 19, 2020 at 0:11
  • $\begingroup$ @KaviRamaMurthy Thank you! $\endgroup$
    – JerBear
    Commented Sep 20, 2020 at 21:19

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