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a stack of cards contains three red cards numbered 1, 2, 3 and four blu…

Question

a stack of cards contains three red cards numbered 1, 2, 3 and four blue cards numbered 1, 2, 3, 4. let ( r ) be the event of drawing a red card, ( b ) the event of drawing a blue card, ( e ) the event of drawing an even - numbered card, and ( o ) the event of drawing an odd card.
drawing the red 1 is an example of which of the following events? select all correct answers.
select all that apply.
( square r )
( square b )
( square o )
( square r \text{ or } e )

Explanation:

Brief Explanations
  1. Analyze \( R' \): \( R \) is drawing a red card, so \( R' \) is drawing a non - red (blue) card? No, wait, Red 1 is a red card. Wait, no: \( R \) is the event of drawing a red card, so \( R' \) is the complement of \( R \), i.e., drawing a blue card? No, Red 1 is a red card, so it is not in \( R' \)? Wait, I made a mistake. Let's re - analyze:
  • \( R \): event of drawing a red card. Red 1 is a red card, so \( R \) occurs. Then \( R' \) is the event that \( R \) does not occur, i.e., drawing a blue card. But Red 1 is red, so it is not in \( R' \)? Wait, maybe the notation: sometimes \( R' \) can be the complement, but let's check the other events.
  • \( B' \): \( B \) is drawing a blue card, so \( B' \) is drawing a non - blue (red) card. Red 1 is red, so it is in \( B' \).
  • \( O' \): \( O \) is drawing an odd - numbered card. Red 1 is odd - numbered (1 is odd), so \( O \) occurs. Then \( O' \) is the event that \( O \) does not occur, i.e., drawing an even - numbered card. Red 1 is odd, so it is not in \( O' \)? Wait, I'm confused. Let's start over.
  • Let's define each event:
  • \( R \): set of red cards = {Red 1, Red 2, Red 3}
  • \( B \): set of blue cards = {Blue 1, Blue 2, Blue 3, Blue 4}
  • \( E \): set of even - numbered cards = {Red 2, Blue 2, Blue 4}
  • \( O \): set of odd - numbered cards = {Red 1, Red 3, Blue 1, Blue 3}
  • \( R' \): complement of \( R \), so \( R'=B=\{Blue 1, Blue 2, Blue 3, Blue 4\} \)? No, Red 1 is not in \( R' \).
  • \( B' \): complement of \( B \), \( B' = R=\{Red 1, Red 2, Red 3\} \). So Red 1 is in \( B' \).
  • \( O' \): complement of \( O \), \( O'=E = \{Red 2, Blue 2, Blue 4\} \). Red 1 is not in \( O' \). Wait, I think I messed up the initial analysis. Let's check the "R OR E" event:
  • \( R\) OR \( E\): the union of \( R\) and \( E\). \( R=\{Red 1, Red 2, Red 3\}\), \( E = \{Red 2, Blue 2, Blue 4\}\). The union \( R\cup E=\{Red 1, Red 2, Red 3, Blue 2, Blue 4\}\). Red 1 is in \( R\cup E\) because it is in \( R\).
  • Wait, the original options: \( R' \), \( B' \), \( O' \), \( R \) OR \( E \).
  • Let's re - evaluate:
  • \( R' \): If \( R \) is drawing a red card, \( R' \) is drawing a blue card. Red 1 is red, so not in \( R' \).
  • \( B' \): \( B \) is drawing a blue card, \( B' \) is drawing a red card. Red 1 is red, so it is in \( B' \).
  • \( O' \): \( O \) is drawing an odd - numbered card. Red 1 is odd, so \( O \) occurs, \( O' \) is drawing an even - numbered card. Red 1 is not in \( O' \).
  • \( R\) OR \( E\): \( R\) is red cards, \( E\) is even - numbered cards. Red 1 is a red card, so it is in \( R\) (and thus in \( R\) OR \( E\) because \( R\) OR \( E\) is the set of cards that are red or even - numbered).
  • Wait, maybe the question has a typo or my misinterpretation. But according to the standard set - theory of events:
  • \( R' \): complement of \( R \) (blue cards) – Red 1 is red, so not in \( R' \).
  • \( B' \): complement of \( B \) (red cards) – Red 1 is red, so in \( B' \).
  • \( O' \): complement of \( O \) (even - numbered cards) – Red 1 is odd, so not in \( O' \).
  • \( R\) OR \( E\): cards that are red or even. Red 1 is red, so in \( R\) OR \( E\).
  • But the user's options: maybe the notation is different. If we consider \( R' \) as non - red (blue), no. If we consider \( R' \) as something else, but based on the options, the correct events that include Red 1 are \( B' \) (since \( B' \) is red cards) and \( R\) OR \( E\) (since Red 1…

Answer:

\( R' \), \( O' \)