Answer
$C\cup B$
$n(C\cup B)=$12
Work Step by Step
$E\cap F$ represents the event that both $E$ and $F$ occur.
$E\cup F$ represents the event that $E$ occurs or $F$ occurs (or both)
The complement of $E,\ E^{\prime}$ represents the event that $E$ does not occur.
A: the red die shows l
B: the numbers add to 4
C: at least one of the numbers is 1
D: the numbers do not add to 11
$ S=\{(1,1),(1,2),...,(1,6),\\$
$(2,1),(2,2),...,(2,6),\\$
$...\\$
$(6,1),(6,2),...,(6,6)\}$,
where (x,y) represents "x on red die, y on green die"
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The event here is: (C occurs) OR (B occurs): $C\cap B$
$C=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\$
$(2,1),(3,1),(4,1),(5,1),6,1)\}\\\\$
$B=\{(1,3),(2,2),(3,1)\}\\\\$
$C\cap B=\{(1,3),(3,1)\}\\\\$
$n(C\cup B)=n(C)+n(B)-n\{C\cap B\}=11+3-2=12$
