Answer :
We are given the universal set
[tex]$$
S = \{2, 4, 6, 9, 10, 14, 15, 16, 18\},
$$[/tex]
and two subsets defined as follows:
- Subset [tex]$A$[/tex] contains all the even numbers in [tex]$S$[/tex].
- Subset [tex]$B$[/tex] contains all the square numbers in [tex]$S$[/tex].
Let’s work through the problem step by step.
─────────────────────────────
Step 1. Identify the Elements in Each Subset
1.1. Even numbers in [tex]$S$[/tex] (subset [tex]$A$[/tex]):
An even number is divisible by [tex]$2$[/tex]. In [tex]$S$[/tex], these numbers are
[tex]$$
A = \{2, 4, 6, 10, 14, 16, 18\}.
$$[/tex]
1.2. Square numbers in [tex]$S$[/tex] (subset [tex]$B$[/tex]):
A square number is an integer that is the square of an integer. In [tex]$S$[/tex], the square numbers are:
- [tex]$4$[/tex] because [tex]$2^2=4$[/tex],
- [tex]$9$[/tex] because [tex]$3^2=9$[/tex],
- [tex]$16$[/tex] because [tex]$4^2=16$[/tex].
Thus,
[tex]$$
B = \{4, 9, 16\}.
$$[/tex]
─────────────────────────────
Step 2. Construct the Venn Diagram
In the Venn diagram, we represent [tex]$A$[/tex] and [tex]$B$[/tex] with overlapping circles.
2.1. Intersection [tex]$A \cap B$[/tex]:
Elements that belong to both [tex]$A$[/tex] and [tex]$B$[/tex] are the even numbers that are also perfect squares. From the lists, these are:
[tex]$$
A \cap B = \{4, 16\}.
$$[/tex]
2.2. Elements only in [tex]$A$[/tex] (in [tex]$A$[/tex] but not in [tex]$B$[/tex]):
Remove the intersection from [tex]$A$[/tex]:
[tex]$$
A \setminus B = \{2, 6, 10, 14, 18\}.
$$[/tex]
2.3. Elements only in [tex]$B$[/tex] (in [tex]$B$[/tex] but not in [tex]$A$[/tex]):
Remove the intersection from [tex]$B$[/tex]:
[tex]$$
B \setminus A = \{9\}.
$$[/tex]
2.4. Elements outside both [tex]$A$[/tex] and [tex]$B$[/tex] (i.e., in [tex]$S \setminus (A \cup B)$[/tex]):
These are the elements in [tex]$S$[/tex] that are neither even nor square numbers. Looking at [tex]$S$[/tex], we have:
[tex]$$
S \setminus (A \cup B) = \{15\}.
$$[/tex]
A diagram representing the sets would look like this:
[tex]\[
\begin{array}{c}
\textbf{Venn Diagram:} \\[6pt]
\text{Universal set } S = \{2, 4, 6, 9, 10, 14, 15, 16, 18\} \\[6pt]
\framebox[1.0\textwidth]{
\begin{minipage}{0.9\textwidth}
\quad Circle for $A$: \quad $\{2, 6, 10, 14, 18\}$ in $A$ only; Intersection: $\{4,16\}$ \\[6pt]
\quad Circle for $B$: \quad $\{9\}$ in $B$ only; Intersection: $\{4,16\}$ \\[6pt]
\quad Outside both circles: $\{15\}$
\end{minipage}
}
\end{array}
\][/tex]
─────────────────────────────
Step 3. Calculate the Probabilities
The total number of elements in [tex]$S$[/tex] is
[tex]$$
|S| = 9.
$$[/tex]
3.1. Probability [tex]$P(A \cup B)$[/tex]:
The union [tex]$A \cup B$[/tex] consists of all the elements that are in [tex]$A$[/tex] or in [tex]$B$[/tex] (or both). Combining all the numbers:
[tex]$$
A \cup B = \{2, 4, 6, 9, 10, 14, 16, 18\}.
$$[/tex]
There are [tex]$8$[/tex] elements in [tex]$A \cup B$[/tex], so
[tex]$$
P(A \cup B) = \frac{8}{9} \approx 0.8889.
$$[/tex]
3.2. Probability [tex]$P(A \cap B)$[/tex]:
The intersection [tex]$A \cap B$[/tex] is
[tex]$$
A \cap B = \{4, 16\}.
$$[/tex]
There are [tex]$2$[/tex] elements in [tex]$A \cap B$[/tex], so
[tex]$$
P(A \cap B) = \frac{2}{9} \approx 0.2222.
$$[/tex]
3.3. Probability [tex]$P(A' \cap B')$[/tex]:
The set [tex]$A' \cap B'$[/tex] consists of the elements that are in neither [tex]$A$[/tex] nor [tex]$B$[/tex]. As found above,
[tex]$$
A' \cap B' = \{15\}.
$$[/tex]
There is [tex]$1$[/tex] element in this set, so
[tex]$$
P(A' \cap B') = \frac{1}{9} \approx 0.1111.
$$[/tex]
─────────────────────────────
Final Answers
1. Venn Diagram Representation:
- In the circle of [tex]$A$[/tex]: [tex]$A \setminus B = \{2, 6, 10, 14, 18\}$[/tex]
- In the circle of [tex]$B$[/tex]: [tex]$B \setminus A = \{9\}$[/tex]
- In the intersection: [tex]$A \cap B = \{4, 16\}$[/tex]
- Outside both circles: [tex]$A' \cap B' = \{15\}$[/tex]
2. Probability that an element is in [tex]$A$[/tex] or [tex]$B$[/tex]:
[tex]$$
P(A \cup B) = \frac{8}{9} \approx 0.8889.
$$[/tex]
3. Probability that an element is in both [tex]$A$[/tex] and [tex]$B$[/tex]:
[tex]$$
P(A \cap B) = \frac{2}{9} \approx 0.2222.
$$[/tex]
4. Probability that an element is in neither [tex]$A$[/tex] nor [tex]$B$[/tex]:
[tex]$$
P(A' \cap B') = \frac{1}{9} \approx 0.1111.
$$[/tex]
This completes our step-by-step solution.
[tex]$$
S = \{2, 4, 6, 9, 10, 14, 15, 16, 18\},
$$[/tex]
and two subsets defined as follows:
- Subset [tex]$A$[/tex] contains all the even numbers in [tex]$S$[/tex].
- Subset [tex]$B$[/tex] contains all the square numbers in [tex]$S$[/tex].
Let’s work through the problem step by step.
─────────────────────────────
Step 1. Identify the Elements in Each Subset
1.1. Even numbers in [tex]$S$[/tex] (subset [tex]$A$[/tex]):
An even number is divisible by [tex]$2$[/tex]. In [tex]$S$[/tex], these numbers are
[tex]$$
A = \{2, 4, 6, 10, 14, 16, 18\}.
$$[/tex]
1.2. Square numbers in [tex]$S$[/tex] (subset [tex]$B$[/tex]):
A square number is an integer that is the square of an integer. In [tex]$S$[/tex], the square numbers are:
- [tex]$4$[/tex] because [tex]$2^2=4$[/tex],
- [tex]$9$[/tex] because [tex]$3^2=9$[/tex],
- [tex]$16$[/tex] because [tex]$4^2=16$[/tex].
Thus,
[tex]$$
B = \{4, 9, 16\}.
$$[/tex]
─────────────────────────────
Step 2. Construct the Venn Diagram
In the Venn diagram, we represent [tex]$A$[/tex] and [tex]$B$[/tex] with overlapping circles.
2.1. Intersection [tex]$A \cap B$[/tex]:
Elements that belong to both [tex]$A$[/tex] and [tex]$B$[/tex] are the even numbers that are also perfect squares. From the lists, these are:
[tex]$$
A \cap B = \{4, 16\}.
$$[/tex]
2.2. Elements only in [tex]$A$[/tex] (in [tex]$A$[/tex] but not in [tex]$B$[/tex]):
Remove the intersection from [tex]$A$[/tex]:
[tex]$$
A \setminus B = \{2, 6, 10, 14, 18\}.
$$[/tex]
2.3. Elements only in [tex]$B$[/tex] (in [tex]$B$[/tex] but not in [tex]$A$[/tex]):
Remove the intersection from [tex]$B$[/tex]:
[tex]$$
B \setminus A = \{9\}.
$$[/tex]
2.4. Elements outside both [tex]$A$[/tex] and [tex]$B$[/tex] (i.e., in [tex]$S \setminus (A \cup B)$[/tex]):
These are the elements in [tex]$S$[/tex] that are neither even nor square numbers. Looking at [tex]$S$[/tex], we have:
[tex]$$
S \setminus (A \cup B) = \{15\}.
$$[/tex]
A diagram representing the sets would look like this:
[tex]\[
\begin{array}{c}
\textbf{Venn Diagram:} \\[6pt]
\text{Universal set } S = \{2, 4, 6, 9, 10, 14, 15, 16, 18\} \\[6pt]
\framebox[1.0\textwidth]{
\begin{minipage}{0.9\textwidth}
\quad Circle for $A$: \quad $\{2, 6, 10, 14, 18\}$ in $A$ only; Intersection: $\{4,16\}$ \\[6pt]
\quad Circle for $B$: \quad $\{9\}$ in $B$ only; Intersection: $\{4,16\}$ \\[6pt]
\quad Outside both circles: $\{15\}$
\end{minipage}
}
\end{array}
\][/tex]
─────────────────────────────
Step 3. Calculate the Probabilities
The total number of elements in [tex]$S$[/tex] is
[tex]$$
|S| = 9.
$$[/tex]
3.1. Probability [tex]$P(A \cup B)$[/tex]:
The union [tex]$A \cup B$[/tex] consists of all the elements that are in [tex]$A$[/tex] or in [tex]$B$[/tex] (or both). Combining all the numbers:
[tex]$$
A \cup B = \{2, 4, 6, 9, 10, 14, 16, 18\}.
$$[/tex]
There are [tex]$8$[/tex] elements in [tex]$A \cup B$[/tex], so
[tex]$$
P(A \cup B) = \frac{8}{9} \approx 0.8889.
$$[/tex]
3.2. Probability [tex]$P(A \cap B)$[/tex]:
The intersection [tex]$A \cap B$[/tex] is
[tex]$$
A \cap B = \{4, 16\}.
$$[/tex]
There are [tex]$2$[/tex] elements in [tex]$A \cap B$[/tex], so
[tex]$$
P(A \cap B) = \frac{2}{9} \approx 0.2222.
$$[/tex]
3.3. Probability [tex]$P(A' \cap B')$[/tex]:
The set [tex]$A' \cap B'$[/tex] consists of the elements that are in neither [tex]$A$[/tex] nor [tex]$B$[/tex]. As found above,
[tex]$$
A' \cap B' = \{15\}.
$$[/tex]
There is [tex]$1$[/tex] element in this set, so
[tex]$$
P(A' \cap B') = \frac{1}{9} \approx 0.1111.
$$[/tex]
─────────────────────────────
Final Answers
1. Venn Diagram Representation:
- In the circle of [tex]$A$[/tex]: [tex]$A \setminus B = \{2, 6, 10, 14, 18\}$[/tex]
- In the circle of [tex]$B$[/tex]: [tex]$B \setminus A = \{9\}$[/tex]
- In the intersection: [tex]$A \cap B = \{4, 16\}$[/tex]
- Outside both circles: [tex]$A' \cap B' = \{15\}$[/tex]
2. Probability that an element is in [tex]$A$[/tex] or [tex]$B$[/tex]:
[tex]$$
P(A \cup B) = \frac{8}{9} \approx 0.8889.
$$[/tex]
3. Probability that an element is in both [tex]$A$[/tex] and [tex]$B$[/tex]:
[tex]$$
P(A \cap B) = \frac{2}{9} \approx 0.2222.
$$[/tex]
4. Probability that an element is in neither [tex]$A$[/tex] nor [tex]$B$[/tex]:
[tex]$$
P(A' \cap B') = \frac{1}{9} \approx 0.1111.
$$[/tex]
This completes our step-by-step solution.
