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In a Multiple Select Question, there are 4 options of which one or more can be correct. Let us define an event E that the option ‘A’ is correct. What is the cardinality of E?

MSQ Options are $= {A, B, C, D}

  1. $n(E1)= \{ A \}$ is $1$ Only $A$ is the correct answer
  2. $n(E2)= \{A, B\}$ is $2$ $A$ and $B$ are the correct answers
  3. $n(E3)= \{A, C\}$ is $2$ $A$ and $C$ are the correct answers
  4. $n(E4)= \{A, D\}$ is $2$ $A$ and $D$ are the correct answers
  5. $n(E5)= \{A, B, C\}$ is $3$ $A$, $B$ and $C$ are the correct answers
  6. $n(E6)= \{A, B, D\}$ is $3$ $A$, $B$ and $D$ are the correct answers
  7. $n(E7)= \{A, C, D\}$ is $3$ $A$, $C$ and $D$ are the correct answers
  8. $n(E8)= \{A, B, C, D\}$ is $4$ all options are the correct answers

Is this the answer? $=>$ The cardinality of $E$ is $=1+2+2+2+3+3+3+4 => 17$

I think the correct answer is $n(E)=\{\{ A \}, \{A, B\},\{A, C\},\{A, D\}, \{A, B, C\}, \{A, B, D\},\{A, B, C, D\}\} => 8$

Please correct me if I'm wrong about correct answer.

Abbas
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    More simply: each of the three other options, ${B,C,D}$ can (in principle) be either true or false independently of each other. There are $8$ ways to assign truth values to this triple. – lulu Jun 22 '22 at 12:06
  • To be clear, ${\color{red}{a},\color{blue}{b},{c,d,e,f}}$ is of cardinality $3$... you have the red thing, you have the blue thing, and you have the black thing. Just because that black "thing" happened to be a set with additional elements inside of it does not matter. We do not "flatten" the set first before asking about its cardinality. In the same way, each of your eight outcomes contribute only one to the overall size of $E$. – JMoravitz Jun 22 '22 at 12:07

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