Question

drag the tiles to the correct boxes to complete the pairs. not all tiles will be used.
a six - sided number cube is rolled. event a consists of rolling an even number. event b consists of rolling a number greater than four. match the correct sample space to each event.
{2, 4, 5, 6} {4, 5, 6} {2, 4, 6} {5, 6} {6}
{1, 3, 5}
event sample space
intersection of a and b
b
union of a and b
complement of a

Explanation:

First, let's define the events:

• Event A: rolling an even number. The possible outcomes are 2, 4, 6. So \( A = \{2, 4, 6\} \).
• Event B: rolling a number greater than four. The possible outcomes are 5, 6. So \( B = \{5, 6\} \).

Now let's find each required set:

Step 1: Intersection of A and B (\( A \cap B \))

The intersection of two sets is the set of elements that are in both sets. So we find the common elements in \( A = \{2, 4, 6\} \) and \( B = \{5, 6\} \). The common element is 6? Wait, no: 6 is in both, and also check again. Wait, \( A = \{2,4,6\} \), \( B = \{5,6\} \). So the intersection is \( \{6\} \)? Wait, no, wait: 6 is in both, but wait, 4 is not in B, 2 is not in B, 5 is not in A, 6 is in both. Wait, no, wait, maybe I made a mistake. Wait, Event B is numbers greater than four: 5, 6. Event A is even numbers: 2,4,6. So the intersection is the numbers that are even and greater than four. So even numbers greater than four: 6. Wait, no, 6 is greater than four and even. Wait, but also, is 4 greater than four? No, 4 is equal to four, not greater. So 5 is odd, 6 is even. So intersection of A and B is \( \{6\} \)? Wait, no, wait the tiles have {6}, but also let's check again. Wait, maybe I messed up Event B. Wait, a six - sided cube has numbers 1 - 6. Number greater than four: 5, 6 (since 5 > 4 and 6 > 4). Event A: even numbers: 2,4,6. So \( A\cap B \): elements in both A and B. So 6 is in both, 2 is not in B, 4 is not in B, 5 is not in A. So \( A\cap B=\{6\} \).

Step 2: Set for Event B

Event B is rolling a number greater than four. The numbers on a six - sided cube greater than four are 5 and 6. So \( B = \{5, 6\} \).

Step 3: Union of A and B (\( A\cup B \))

The union of two sets is the set of all elements that are in either set. So we combine the elements of \( A = \{2, 4, 6\} \) and \( B=\{5, 6\} \). So \( A\cup B=\{2, 4, 5, 6\} \) (we take all elements from A and B, without duplication: 2,4,6 from A and 5,6 from B; so combined we have 2,4,5,6).

Step 4: Complement of A (\( A' \))

The complement of a set A (in the sample space \( S=\{1,2,3,4,5,6\} \)) is the set of elements in S that are not in A. Since \( A = \{2,4,6\} \), the complement is \( \{1,3,5\} \) (the odd numbers, since A is the even numbers).

Now let's match:

• Intersection of A and B: The elements common to A and B. As we saw, \( A = \{2,4,6\} \), \( B=\{5,6\} \), so \( A\cap B = \{6\} \).
• Event B: Numbers greater than four, so \( B=\{5, 6\} \).
• Union of A and B: Elements in A or B, so \( A\cup B=\{2,4,5,6\} \).
• Complement of A: Elements not in A (i.e., odd numbers), so \( A'=\{1,3,5\} \).

Wait, but let's re - evaluate the intersection. Wait, maybe I made a mistake. Let's list all elements:

• Event A (even numbers): \( A=\{2,4,6\} \)
• Event B (numbers > 4): \( B = \{5,6\} \)

Intersection (\( A\cap B \)): elements in both A and B. So 6 is in both, 2 is not in B, 4 is not in B, 5 is not in A. So \( A\cap B=\{6\} \)

Event B: \( B = \{5,6\} \)

Union (\( A\cup B \)): elements in A or B. So 2,4,6 (from A) and 5,6 (from B). Combining them, we get 2,4,5,6 (since 6 is already in A, we don't repeat it). So \( A\cup B=\{2,4,5,6\} \)

Complement of A: elements in the sample space (1 - 6) not in A. So 1,3,5. So \( A'=\{1,3,5\} \)

Matching:

• Intersection of A and B: {6}
• Event B: {5, 6}
• Union of A and B: {2, 4, 5, 6}
• Complement of A: {1, 3, 5}

Answer:

• Intersection of A and B: \(\{6\}\)
• Event B: \(\{5, 6\}\)
• Union of A and B: \(\{2, 4, 5, 6\}\)
• Complement of A: \(\{1, 3, 5\}\)