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a number cube has faces numbered 1 to 6. what is true about rolling the…

Question

a number cube has faces numbered 1 to 6.
what is true about rolling the number cube one time? choose three correct answers.
if a subset a
represents the
complement of
rolling an even
number, then
a = {1, 3}.
if a is a subset
of s, a could be
{0, 1, 2}.
if a subset a
represents the
complement of
rolling a 5, then
a = {1, 2, 3, 4, 6}
s = {1, 2, 3, 4, 5,
if a is a subset
of s, a could be
{5, 6}.

Explanation:

Step1: Analyze the sample space \( S \)

The sample space \( S \) for rolling a number cube (1 - 6) is \( S=\{1,2,3,4,5,6\} \).

Step2: Evaluate each option

  • First option (complement of even numbers):

Even numbers in \( S \) are \( \{2,4,6\} \). The complement (odd numbers) is \( \{1,3,5\} \), not \( \{1,3\} \). So this is incorrect.

  • Second option (subset \( \{0,1,2\} \)):

\( 0
otin S \), so \( \{0,1,2\} \) is not a subset of \( S \). Incorrect.

  • Third option (complement of rolling a 5):

The event "rolling a 5" is \( \{5\} \). Its complement is \( S - \{5\}=\{1,2,3,4,6\} \). Correct.

  • Fourth option (subset \( \{5,6\} \)):

\( 5,6 \in S \), so \( \{5,6\} \) is a subset of \( S \). Correct.

  • Fifth option (the set \( S \)):

\( S = \{1,2,3,4,5,6\} \) is the sample space, so it is a valid subset (trivially, as every set is a subset of itself). Wait, wait—wait, the fourth option (bottom left) is \( \{5,6\} \), the third is complement of 5, and the last (top right) is \( S \). Wait, let's re - check:
Wait, the problem says "choose three correct answers". Let's re - evaluate:

  • Top right: \( S=\{1,2,3,4,5,6\} \) is the sample space, so it is a subset (of itself). Correct.
  • Third option (top middle - right): complement of 5 is \( \{1,2,3,4,6\} \). Correct.
  • Bottom left: \( \{5,6\} \) is a subset of \( S \). Correct.

Wait, my initial step - by - step had a miscalculation. Let's redo:

  1. Top left (complement of even numbers):

Even numbers in \( S \) are \( \{2,4,6\} \). Complement is \( \{1,3,5\} \), not \( \{1,3\} \). Incorrect.

  1. Top middle (subset \( \{0,1,2\} \)):

\( 0
otin S \), so not a subset. Incorrect.

  1. Top middle - right (complement of rolling a 5):

Event "rolling a 5" is \( \{5\} \). Complement: \( S-\{5\}=\{1,2,3,4,6\} \). Correct.

  1. Top right (\( S \)):

\( S \) is the sample space, so it is a subset of itself. Correct.

  1. Bottom left (subset \( \{5,6\} \)):

\( 5,6 \in S \), so \( \{5,6\} \) is a subset. Correct.

So the three correct ones are:

  • The top - middle - right (If a subset A represents the complement of rolling a 5, then \( A = \{1,2,3,4,6\} \))
  • The top - right (\( S=\{1,2,3,4,5,6\} \))
  • The bottom - left (If A is a subset of \( S \), A could be \( \{5,6\} \))

Wait, but let's confirm the sample space. The number cube has faces 1 - 6, so \( S=\{1,2,3,4,5,6\} \).

  • The set \( S \) is a subset of itself (since for all \( x\in S \), \( x\in S \)), so the top - right is correct.
  • The complement of rolling a 5: since the event of rolling a 5 is \( \{5\} \), the complement is all outcomes not equal to 5, which is \( \{1,2,3,4,6\} \), so the top - middle - right is correct.
  • The bottom - left: \( \{5,6\} \) is a subset because both 5 and 6 are in \( S \), so the bottom - left is correct.

Answer:

The three correct answers are:

  • The box with "If a subset A represents the complement of rolling a 5, then \( A=\{1,2,3,4,6\} \)"
  • The box with " \( S = \{1,2,3,4,5,6\} \)"
  • The box with "If A is a subset of \( S \), A could be \( \{5,6\} \)"

(In terms of their positions: the third box from the left (top row, middle - right), the fourth box (top row, right), and the bottom - left box)