Venn Diagram and Operation on Sets Questions in English

(B) Let $A = \{1, 2, 3, 4\}$.
Let $B = \{x : x \text{ is a natural number and } 4 \le x \le 6\} = \{4, 5, 6\}$.
To check if the sets are disjoint,we find their intersection: $A \cap B = \{1, 2, 3, 4\} \cap \{4, 5, 6\} = \{4\}$.
Since $A \cap B \neq \emptyset$,the sets are not disjoint.

(B) Two sets are disjoint if their intersection is the empty set,i.e.,$A \cap B = \emptyset$.
Given sets: $A = \{a, e, i, o, u\}$ and $B = \{c, d, e, f\}$.
The intersection is $A \cap B = \{a, e, i, o, u\} \cap \{c, d, e, f\} = \{e\}$.
Since $A \cap B \neq \emptyset$,the sets are not disjoint.

(A) The difference of two sets $A$ and $B$,denoted by $A - B$,is the set of elements which belong to $A$ but do not belong to $B$.
Given $A = \{3, 6, 9, 12, 15, 18, 21\}$ and $B = \{4, 8, 12, 16, 20\}$.
The common element between $A$ and $B$ is $\{12\}$.
Removing the element $12$ from set $A$,we get $A - B = \{3, 6, 9, 15, 18, 21\}$.

(A) The set $A-C$ consists of elements that are in $A$ but not in $C$.
Given $A = \{3, 6, 9, 12, 15, 18, 21\}$ and $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$.
Comparing the two sets,the common elements are $\{6, 12\}$.
Removing these common elements from set $A$,we get $A-C = \{3, 9, 15, 18, 21\}$.

(A) The set difference $A-D$ consists of elements that are in $A$ but not in $D$.
Given $A = \{3, 6, 9, 12, 15, 18, 21\}$ and $D = \{5, 10, 15, 20\}$.
We remove the elements of $D$ that are present in $A$.
The only common element is $15$.
Therefore,$A-D = \{3, 6, 9, 12, 18, 21\}$.

(A) The difference of two sets $B - A$ is defined as the set of elements that are in $B$ but not in $A$.
Given $A = \{3, 6, 9, 12, 15, 18, 21\}$ and $B = \{4, 8, 12, 16, 20\}$.
We identify the elements present in $B$ that are also in $A$. The only common element is $12$.
Removing $12$ from set $B$,we get $B - A = \{4, 8, 16, 20\}$.

(A) The difference of two sets $C - A$ is defined as the set of elements that are in $C$ but not in $A$.
Given $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$ and $A = \{3, 6, 9, 12, 15, 18, 21\}$.
The common elements between $C$ and $A$ are $\{6, 12\}$.
Removing these elements from $C$,we get $C - A = \{2, 4, 8, 10, 14, 16\}$.

(A) The difference of two sets $D - A$ is defined as the set of elements that are in $D$ but not in $A$.
Given $D = \{5, 10, 15, 20\}$ and $A = \{3, 6, 9, 12, 15, 18, 21\}$.
We identify the elements of $D$ that are also present in $A$. The element $15$ is common to both sets.
Removing $15$ from $D$,we get $D - A = \{5, 10, 20\}$.

(B) The difference of two sets $B$ and $C$,denoted by $B - C$,is the set of elements which are in $B$ but not in $C$.
Given $B = \{4, 8, 12, 16, 20\}$ and $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$.
Elements of $B$ that are also in $C$ are $\{4, 8, 12, 16\}$.
Removing these elements from $B$,we get $B - C = \{20\}$.

(A) The difference of two sets $B$ and $D$,denoted by $B - D$,is the set of all elements that are in $B$ but not in $D$.
Given $B = \{4, 8, 12, 16, 20\}$ and $D = \{5, 10, 15, 20\}$.
We remove the elements of $D$ from $B$.
The element $20$ is present in both $B$ and $D$.
Therefore,$B - D = \{4, 8, 12, 16\}$.

(A) The difference of two sets $C$ and $B$,denoted by $C - B$,is the set of all elements that are in $C$ but not in $B$.
Given $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$ and $B = \{4, 8, 12, 16, 20\}$.
Removing the elements of $B$ from $C$,we get:
$C - B = \{2, 6, 10, 14\}$.

(A) The difference of two sets $D$ and $B$,denoted by $D - B$,is the set of all elements that are in $D$ but not in $B$.
Given $D = \{5, 10, 15, 20\}$ and $B = \{4, 8, 12, 16, 20\}$.
We remove the elements of $B$ from $D$.
The element $20$ is common to both sets.
Therefore,$D - B = \{5, 10, 15, 20\} - \{4, 8, 12, 16, 20\} = \{5, 10, 15\}$.

(A) The difference of two sets $C$ and $D$,denoted by $C - D$,is the set of all elements that are in $C$ but not in $D$.
Given $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$ and $D = \{5, 10, 15, 20\}$.
We remove the elements of $D$ that are present in $C$.
The common element between $C$ and $D$ is $\{10\}$.
Therefore,$C - D = \{2, 4, 6, 8, 10, 12, 14, 16\} - \{5, 10, 15, 20\} = \{2, 4, 6, 8, 12, 14, 16\}$.

(B) The difference of two sets $D$ and $C$,denoted by $D - C$,is the set of elements which belong to $D$ but do not belong to $C$.
Given $D = \{5, 10, 15, 20\}$ and $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$.
We identify elements in $D$ that are also in $C$: the element $10$ is present in both sets.
Removing $10$ from set $D$,we get $D - C = \{5, 15, 20\}$.

(A) The difference of two sets $Y - X$ is defined as the set of all elements that are in $Y$ but not in $X$.
Given $X = \{a, b, c, d\}$ and $Y = \{f, b, d, g\}$.
The elements present in $Y$ are $f, b, d, g$.
The elements of $Y$ that are also in $X$ are $b$ and $d$.
Therefore,$Y - X = \{f, g\}$.

(B) The intersection of two sets $X$ and $Y$,denoted by $X \cap Y$,consists of all elements that are common to both $X$ and $Y$.
Given $X = \{a, b, c, d\}$ and $Y = \{f, b, d, g\}$.
The common elements in both sets are $b$ and $d$.
Therefore,$X \cap Y = \{b, d\}$.

(B) False.
Two sets are disjoint if their intersection is the empty set,i.e.,$A \cap B = \emptyset$.
Given sets are $A = \{2, 3, 4, 5\}$ and $B = \{3, 6\}$.
Since $3 \in A$ and $3 \in B$,the intersection is $A \cap B = \{3\}$.
Since $A \cap B \neq \emptyset$,the sets are not disjoint.

(B) False.
Two sets are disjoint if their intersection is the empty set,i.e.,$A \cap B = \emptyset$.
Here,let $A = \{a, e, i, o, u\}$ and $B = \{a, b, c, d\}$.
Since $a \in A$ and $a \in B$,we have $A \cap B = \{a\}$.
Since $A \cap B \neq \emptyset$,the sets are not disjoint.

(A) The statement is $True$.
Two sets are said to be disjoint if their intersection is the empty set,denoted by $\varnothing$.
Let $A = \{2, 6, 10, 14\}$ and $B = \{3, 7, 11, 15\}$.
The intersection of $A$ and $B$ is $A \cap B = \{2, 6, 10, 14\} \cap \{3, 7, 11, 15\} = \varnothing$.
Since the intersection is empty,the sets are disjoint.

Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,$A = \{2, 4, 6, 8\}$,and $B = \{2, 3, 5, 7\}$.
First,find $A \cup B = \{2, 3, 4, 5, 6, 7, 8\}$.
Then,$(A \cup B)^{\prime} = U \setminus (A \cup B) = \{1, 9\}$.
Next,find $A^{\prime} = U \setminus A = \{1, 3, 5, 7, 9\}$ and $B^{\prime} = U \setminus B = \{1, 4, 6, 8, 9\}$.
Then,$A^{\prime} \cap B^{\prime} = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\} = \{1, 9\}$.
Since $(A \cup B)^{\prime} = \{1, 9\}$ and $A^{\prime} \cap B^{\prime} = \{1, 9\}$,it is verified that $(A \cup B)^{\prime} = A^{\prime} \cap B^{\prime}$.

(N/A) By De Morgan's Law,$A^{\prime} \cap B^{\prime} = (A \cup B)^{\prime}$.
This represents the region in the universal set $U$ that is outside both sets $A$ and $B$.
In the Venn diagram,this is the shaded region covering the entire rectangle excluding the circles $A$ and $B$.

(N/A) By De Morgan's Law,we know that $A^{\prime} \cup B^{\prime} = (A \cap B)^{\prime}$.
This represents the region in the universal set $U$ that is outside the intersection of sets $A$ and $B$.
Therefore,the Venn diagram consists of the entire universal set $U$ except for the region where sets $A$ and $B$ overlap.

(A) It is given that:
$n(X \cup Y) = 18, n(X) = 8, n(Y) = 15$
We know the formula for the number of elements in the union of two sets:
$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$
Substituting the given values into the formula:
$18 = 8 + 15 - n(X \cap Y)$
$18 = 23 - n(X \cap Y)$
Rearranging to solve for $n(X \cap Y)$:
$n(X \cap Y) = 23 - 18$
$n(X \cap Y) = 5$
Therefore,the number of elements in $X \cap Y$ is $5$.

(C) Let the three sets of students playing the three games be $A$,$B$,and $C$.
According to the problem,some students play two games,which means the intersection of any two sets is non-empty (i.e.,$A \cap B \neq \emptyset$,$B \cap C \neq \emptyset$,$A \cap C \neq \emptyset$).
However,no student plays all three games,which means the intersection of all three sets must be empty (i.e.,$A \cap B \cap C = \emptyset$).
In diagram $P$,there are only two sets,so it does not represent three games.
In diagram $Q$,the three circles are arranged such that there is a central region where all three overlap,meaning $A \cap B \cap C \neq \emptyset$.
In diagram $R$,the three circles are arranged such that there is no common region shared by all three circles,meaning $A \cap B \cap C = \emptyset$,while pairs of circles still overlap.
Therefore,only diagram $R$ justifies the statement. Since $R$ is not present in any of the options $A$,$B$,or $D$,the correct option is $C$.

(N/A) To show: $A = (A \cap B) \cup (A - B)$
Let $x \in A.$
We know that $x$ must either be in $B$ or not in $B$.
Case $I$: If $x \in B$,then $x \in A \cap B$,so $x \in (A \cap B) \cup (A - B).$
Case $II$: If $x \notin B$,then $x \in A - B$,so $x \in (A \cap B) \cup (A - B).$
Thus,$A \subseteq (A \cap B) \cup (A - B).$ ..........$(1)$
Since $(A \cap B) \subseteq A$ and $(A - B) \subseteq A$,their union must also be a subset of $A$.
Thus,$(A \cap B) \cup (A - B) \subseteq A.$ ..........$(2)$
From $(1)$ and $(2)$,$A = (A \cap B) \cup (A - B).$
To show: $A \cup (B - A) = A \cup B$
Let $x \in A \cup (B - A).$
This implies $x \in A$ or $(x \in B \text{ and } x \notin A).$
By distributive law,this is $(x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \notin A).$
Since $(x \in A \text{ or } x \notin A)$ is always true,we get $x \in A \cup B.$
Thus,$A \cup (B - A) \subseteq A \cup B.$ ..........$(3)$
Conversely,let $y \in A \cup B.$
This implies $y \in A$ or $y \in B.$
If $y \in A$,then $y \in A \cup (B - A).$
If $y \notin A$ and $y \in B$,then $y \in B - A$,so $y \in A \cup (B - A).$
Thus,$A \cup B \subseteq A \cup (B - A).$ ..........$(4)$
From $(3)$ and $(4)$,$A \cup (B - A) = A \cup B.$

(A) To prove: $A \cup (A \cap B) = A$
We know that for any sets $A$ and $B$,the intersection $A \cap B$ is a subset of $A$,i.e.,$(A \cap B) \subset A$.
Also,$A \subset A$.
Therefore,the union of these two sets is $A \cup (A \cap B) \subset A$ ........... $(1)$
Conversely,any element $x \in A$ implies $x \in A \cup (A \cap B)$ by the definition of union,so $A \subset A \cup (A \cap B)$ ........... $(2)$
From $(1)$ and $(2)$,by the definition of set equality,we conclude that $A \cup (A \cap B) = A$.

(A) When two dice are thrown,the sample space $S$ contains $36$ outcomes.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Since $B$ is the set of all outcomes where the first die is odd,$B^{\prime} = A$.
$C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$
We need to find $A \cap B^{\prime} \cap C^{\prime} = A \cap A \cap C^{\prime} = A \cap C^{\prime}$.
$A \cap C^{\prime}$ represents the set of outcomes in $A$ that are not in $C$.
$A \cap C^{\prime} = \{(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

(B) The function $f(x, A)$ is the indicator function of the set $A$,denoted as $\chi_A(x)$.
By definition,$f(x, A \cup B) = 1$ if $x \in A \cup B$,and $0$ otherwise.
We know that $x \in A \cup B$ if and only if $x \in A$ or $x \in B$.
Using the properties of indicator functions:
$f(x, A \cup B) = \max(f(x, A), f(x, B))$.
Alternatively,using the inclusion-exclusion principle for indicator functions:
$f(x, A \cup B) = f(x, A) + f(x, B) - f(x, A \cap B)$.
Since $f(x, A \cap B) = f(x, A) \cdot f(x, B)$,we have $f(x, A \cup B) = f(x, A) + f(x, B) - f(x, A)f(x, B)$.
Comparing this with the given options,if we evaluate the truth table for all cases of $x \in A$ and $x \in B$,we find that the expression $f(x, A) + f(x, B) - f(x, A)f(x, B)$ correctly represents the union.

(A) Given: $P(E) = \frac{1}{8}, P(F) = \frac{1}{6}, P(G) = \frac{1}{4}, P(E \cap F \cap G) = \frac{1}{10}$.
$(A)$ We know $P(E) = P(E \cap F \cap G) + P(E \cap F \cap G^C) + P(E \cap F^C \cap G) + P(E \cap F^C \cap G^C)$.
Thus,$P(E \cap F \cap G^C) \leq P(E) - P(E \cap F \cap G) = \frac{1}{8} - \frac{1}{10} = \frac{5-4}{40} = \frac{1}{40}$. So,$(A)$ is $TRUE$.
$(B)$ Similarly,$P(E^C \cap F \cap G) \leq P(F) - P(E \cap F \cap G) = \frac{1}{6} - \frac{1}{10} = \frac{5-3}{30} = \frac{2}{30} = \frac{1}{15}$. So,$(B)$ is $TRUE$.
$(C)$ $P(E \cup F \cup G) = P(E) + P(F) + P(G) - [P(E \cap F) + P(F \cap G) + P(G \cap E)] + P(E \cap F \cap G)$.
Since $P(E \cap F), P(F \cap G), P(G \cap E) \geq P(E \cap F \cap G) = \frac{1}{10}$,we have $P(E \cup F \cup G) \leq \frac{1}{8} + \frac{1}{6} + \frac{1}{4} - 3(\frac{1}{10}) + \frac{1}{10} = \frac{3+4+6}{24} - \frac{2}{10} = \frac{13}{24} - \frac{1}{5} = \frac{65-24}{120} = \frac{41}{120} \leq \frac{13}{24}$. So,$(C)$ is $TRUE$.
$(D)$ $P(E^C \cap F^C \cap G^C) = 1 - P(E \cup F \cup G)$. Since $P(E \cup F \cup G) \geq P(E \cap F \cap G) = \frac{1}{10}$,$P(E^C \cap F^C \cap G^C) \leq 1 - \frac{1}{10} = \frac{9}{10} = 0.9$. The statement $P(E^C \cap F^C \cap G^C) \leq \frac{5}{12} \approx 0.416$ is not necessarily true. So,$(D)$ is $FALSE$.
Thus,the correct options are $A, B, C$.

(B) Given set $A = \{x \mid x \in N, x \text{ is a prime number less than } 12\}$.
Since prime numbers less than $12$ are $2, 3, 5, 7, 11$,we have $A = \{2, 3, 5, 7, 11\}$.
Given set $B = \{x \mid x \in N, x \text{ is a factor of } 10\}$.
Since factors of $10$ are $1, 2, 5, 10$,we have $B = \{1, 2, 5, 10\}$.
The intersection $A \cap B$ consists of elements common to both sets $A$ and $B$.
$A \cap B = \{2, 3, 5, 7, 11\} \cap \{1, 2, 5, 10\} = \{2, 5\}$.

(A) Given $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n - 1) : n \in N\}$.
By binomial expansion,$4^n = (1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2!} (3^2) + \dots + 3^n$.
So,$4^n - 3n - 1 = 1 + 3n + \frac{9n(n-1)}{2} + \dots - 3n - 1 = \frac{9n(n-1)}{2} + \dots = 9 \left[ \frac{n(n-1)}{2} + \dots \right]$.
This shows that every element of $X$ is a multiple of $9$,and since $n(n-1)$ is always even,$\frac{n(n-1)}{2}$ is an integer.
Thus,$X \subseteq Y$.
Alternatively,testing values:
For $n=1, X = \{4^1 - 3(1) - 1\} = \{0\}$.
For $n=2, X = \{4^2 - 3(2) - 1\} = \{16 - 6 - 1\} = \{9\}$.
For $n=3, X = \{4^3 - 3(3) - 1\} = \{64 - 9 - 1\} = \{54\}$.
$X = \{0, 9, 54, \dots\}$.
$Y = \{0, 9, 18, 27, 36, 45, 54, \dots\}$.
Since all elements of $X$ are in $Y$,$X \cap Y = X$.

(A) First,we determine the elements of each set:
$A = \{ x : x \text{ is an integer and } x^2 - 9 = 0 \} = \{ -3, 3 \}$.
$B = \{ x : x \text{ is a natural number and } 2 \leq x < 5 \} = \{ 2, 3, 4 \}$.
$C = \{ x : x \text{ is a prime number } \leq 4 \} = \{ 2, 3 \}$.
Next,calculate $(B - C)$:
$B - C = \{ 2, 3, 4 \} - \{ 2, 3 \} = \{ 4 \}$.
Finally,calculate $(B - C) \cup A$:
$(B - C) \cup A = \{ 4 \} \cup \{ -3, 3 \} = \{ -3, 3, 4 \}$.

(C) Given that $A$ is a subset of $B$,denoted as $A \subset B$.
By the definition of subset,every element of $A$ is also an element of $B$.
Therefore,the union of $A$ and $B$ is simply $B$,i.e.,$A \cup B = B$.
Taking the number of elements on both sides,we get $n(A \cup B) = n(B)$.
Similarly,the intersection of $A$ and $B$ is $A$,i.e.,$A \cap B = A$,which implies $n(A \cap B) = n(A)$.

(C) When only event $A$ occurs,it is represented as $(A \cap B^c)$.
When only event $B$ occurs,it is represented as $(A^c \cap B)$.
Therefore,the event 'exactly one of the events $A$ or $B$ occurs' is the union of these two disjoint sets: $(A \cap B^c) \cup (A^c \cap B)$.

(C) We know that the difference of two sets $A-B = A \cap B^c$.
Substituting this into the expression,we get:
$A-(A-B) = A-(A \cap B^c)$
Using the property $X-Y = X \cap Y^c$:
$= A \cap (A \cap B^c)^c$
Applying De Morgan's Law $(A \cap B^c)^c = A^c \cup (B^c)^c = A^c \cup B$:
$= A \cap (A^c \cup B)$
Using the Distributive Law:
$= (A \cap A^c) \cup (A \cap B)$
Since $A \cap A^c = \emptyset$:
$= \emptyset \cup (A \cap B) = A \cap B$
Therefore,the correct option is $C$.