Step 1: Establishing the Sample Space ($S$)

When two six-sided dice are tossed, the sample space $S$ consists of all ordered pairs $(x, y)$, where $x$ represents the result of the first die and $y$ represents the result of the second die. The total number of outcomes is $6 \times 6 = 36$.

$S = \{ (1,1), (1,2), \dots, (6,6) \}$

Step 2: Defining Event A and Event B

Event A: Getting an even number on the first die.

$A = \{ (x, y) \in S : x \in \{2, 4, 6\} \}$

$A = \{ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}$

Event B: Getting an odd number on the first die.

$B = \{ (x, y) \in S : x \in \{1, 3, 5\} \}$

$B = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \}$

Step 3: Calculating the Union of Events (A or B)

In set theory, "A or B" corresponds to the union of sets $A$ and $B$, denoted as $A \cup B$. This set contains all elements that are in $A$, in $B$, or in both [Per the Definition of Union in Probability Theory].

$A \cup B = \{ \omega : \omega \in A \text{ or } \omega \in B \}$

Since the first die must be either even $\{2, 4, 6\}$ or odd $\{1, 3, 5\}$, and every integer from 1 to 6 falls into one of these two categories, the union covers the entire sample space $S$.

Step 4: Logical Conclusion

Because $A \cap B = \emptyset$ (the first die cannot be both even and odd simultaneously) and $A \cup B = S$, the event "A or B" represents the Certain Event, as the first die will necessarily yield either an even or an odd number.

Final Solution: The event "A or B" is equal to the sample space S, containing all 36 possible outcomes of throwing two dice.

Step 1: Definition of the Sample Space

The experiment consists of tossing a fair, six-faced die. The sample space $S$ is defined as the set of all possible outcomes. Since a standard die has faces numbered 1 through 6, the sample space is:

$S = \{1, 2, 3, 4, 5, 6\}$

[Per the Classical Definition of Probability, the total number of outcomes $n(S) = 6$.]

Step 2: Identifying Event A

Event $A$ is defined as the occurrence of a "number less than 7." We must identify all elements $x \in S$ such that $x < 7$.

• Checking element 1: $1 < 7$ (True)
• Checking element 2: $2 < 7$ (True)
• Checking element 3: $3 < 7$ (True)
• Checking element 4: $4 < 7$ (True)
• Checking element 5: $5 < 7$ (True)
• Checking element 6: $6 < 7$ (True)

Step 3: Constructing the Subset

By set notation, event $A$ is a subset of the sample space $S$ [By the definition of an event as a subset of the sample space].

$A = \{1, 2, 3, 4, 5, 6\}$

Visual Representation of the Event

Analytical Conclusion

Since every outcome in the sample space satisfies the condition of being less than 7, event $A$ is equivalent to the sample space itself.

[In probability theory, such an event is classified as a Sure Event or a Certain Event because $A = S$.]

Final Solution: The event A is {1, 2, 3, 4, 5, 6}.

Step 1: Establishing the Sample Space ($S$)

When three unbiased coins are tossed simultaneously, each coin has $2$ possible outcomes: Heads ($H$) or Tails ($T$). The total number of outcomes is determined by the Fundamental Counting Principle: $2 \times 2 \times 2 = 8$.

The sample space $S$ is defined as:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Step 2: Defining Mutually Exclusive Events

Two events $A$ and $B$ are defined as mutually exclusive if they cannot occur simultaneously. Formally, $A \cap B = \emptyset$, where $\emptyset$ is the null (empty) set. [By the Axiomatic Definition of Probability].

Step 3: Selecting Specific Events

To satisfy the condition of mutual exclusivity, we must identify two subsets of $S$ that share no common elements.

• Event $A$ (Getting exactly three heads): $A = \{HHH\}$
• Event $B$ (Getting no heads / Getting three tails): $B = \{TTT\}$

Step 4: Verification of Mutual Exclusivity

To verify that $A$ and $B$ are mutually exclusive, we calculate their intersection:

$A \cap B = \{HHH\} \cap \{TTT\} = \emptyset$

[By the Principle of Disjoint Sets, since no outcome exists in both sets, the probability $P(A \cap B) = 0$.]

Visual Representation of Sample Space

Conclusion

Since the set of outcomes where three heads appear and the set of outcomes where three tails appear have no intersection, they are mutually exclusive.

Final Solution: Event $A = \{HHH\}$ (getting three heads) and Event $B = \{TTT\}$ (getting three tails) are mutually exclusive.

Step 1: Defining the Sample Space ($S$)

When a standard six-faced die is rolled, the set of all possible outcomes, denoted as the sample space $S$, is defined as:

$S = \{1, 2, 3, 4, 5, 6\}$

[By the definition of a discrete sample space for a fair die]

Step 2: Identifying Event Outcomes

Event $E$ is defined as the event where the die shows a 4. Thus:

$E = \{4\}$

Event $F$ is defined as the event where the die shows an even number. The even numbers in the sample space $S$ are 2, 4, and 6. Thus:

$F = \{2, 4, 6\}$

Step 3: Calculating the Intersection of Events ($E \cap F$)

To determine if two events are mutually exclusive, we must calculate the intersection of the two sets. Two events are mutually exclusive if and only if $E \cap F = \emptyset$, meaning the events share no common outcomes.

$E \cap F = \{4\} \cap \{2, 4, 6\}$

$E \cap F = \{4\}$

[Per the set-theoretic definition of intersection]

Step 4: Evaluating Mutually Exclusive Criteria

Since $E \cap F = \{4\}$, and $\{4\} \neq \emptyset$ (the intersection is not the empty set), there is an outcome common to both events.

Analytical Conclusion

Because the die showing "4" satisfies both the condition of being a "4" and the condition of being an "even number," the events are not mutually exclusive.

Final Solution: No, E and F are not mutually exclusive because E $\cap$ F = {4} $\neq$ $\emptyset$.

Initial Setup & Sample Space Determination

When tossing three unbiased coins simultaneously, each coin has two possible outcomes: Heads ($H$) or Tails ($T$). The total number of elementary outcomes in the sample space $S$ is calculated as $2^n$, where $n=3$.

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The total number of outcomes $|S| = 8$.

Theoretical Definitions

• Mutually Exclusive Events: Two or more events are mutually exclusive if the occurrence of one prevents the occurrence of all others. Formally, $A \cap B = \emptyset$ for all distinct pairs.
• Exhaustive Events: A set of events is exhaustive if their union is equal to the entire sample space: $\bigcup_{i=1}^{n} E_i = S$.

Step 1: Defining the Three Events

We define three events based on the number of heads appearing in the toss:

• Event $A$ (Zero Heads): The event that no heads appear.
  $A = \{TTT\}$
• Event $B$ (Exactly one Head): The event that exactly one head appears.
  $B = \{HTT, THT, TTH\}$
• Event $C$ (Two or more Heads): The event that two or more heads appear.
  $C = \{HHH, HHT, HTH, THH\}$

Step 2: Verification of Mutual Exclusivity

To confirm mutual exclusivity, we check the intersection of all pairs:

• $A \cap B = \emptyset$ [Since $A$ has 0 heads and $B$ has 1 head]
• $A \cap C = \emptyset$ [Since $A$ has 0 heads and $C$ has $\ge 2$ heads]
• $B \cap C = \emptyset$ [Since $B$ has exactly 1 head and $C$ has $\ge 2$ heads]

[By the definition of intersection of disjoint sets, the events are mutually exclusive].

Step 3: Verification of Exhaustive Nature

To confirm the events are exhaustive, we take the union of $A, B, \text{ and } C$:

$A \cup B \cup C = \{TTT\} \cup \{HTT, THT, TTH\} \cup \{HHH, HHT, HTH, THH\}$

$A \cup B \cup C = \{TTT, HTT, THT, TTH, HHH, HHT, HTH, THH\} = S$

[Since the union of the events spans all possible outcomes in $S$, the collection is exhaustive].

Final Solution: The three events $A = \{TTT\}$, $B = \{HTT, THT, TTH\}$, and $C = \{HHH, HHT, HTH, THH\}$ are mutually exclusive and exhaustive.

Step 1: Establishing the Sample Space

When three coins are tossed, each coin has two possible outcomes: Head ($H$) or Tail ($T$). Since there are three coins, the total number of elementary outcomes is given by $2^n$ where $n=3$.

The sample space $S$ is defined as:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Total number of outcomes, $n(S) = 8$.

Step 2: Defining Events for Non-Mutually Exclusive Status

Two events $A$ and $B$ are defined as not mutually exclusive if they have at least one common outcome, i.e., $A \cap B \neq \emptyset$ [Per the definition of Mutually Exclusive Events in Probability Theory].

Let us define the following two events:

• Event A: Getting at least two heads.
• Event B: Getting at least two tails.

Step 3: Calculating Event Sets

Set A (At least two heads):

$A = \{HHH, HHT, HTH, THH\}$

Set B (At least two tails):

$B = \{HTT, THT, TTH, TTT\}$

Step 4: Analyzing the Intersection

To verify if these events are not mutually exclusive, we evaluate the intersection $A \cap B$.

By observing the sets $A$ and $B$:

$A \cap B = \emptyset$ (These specific sets are mutually exclusive).

Revision: To satisfy the requirement of "not mutually exclusive," we must redefine the events to ensure a non-empty intersection.

Step 5: Redefining Events for Non-Mutual Exclusivity

Let us define new events:

• Event C: Getting exactly two heads.
• Event D: Getting a head on the first coin.

Set C: $\{HHT, HTH, THH\}$

Set D: $\{HHH, HHT, HTH, HTT\}$

Step 6: Final Verification

Checking the intersection of $C$ and $D$:

$C \cap D = \{HHT, HTH\}$

Since $C \cap D \neq \emptyset$, the events share common outcomes ($HHT$ and $HTH$).

Final Solution: Two events that are not mutually exclusive are: Event C (getting exactly two heads) and Event D (getting a head on the first coin), because $C \cap D = \{HHT, HTH\}$, which is not an empty set.

Step 1: Defining the Sample Space ($S$)

For a standard six-faced fair die, the possible outcomes of a single throw are represented by the set of integers from 1 to 6. This represents the total sample space ($S$).

$S = \{1, 2, 3, 4, 5, 6\}$

[Definition: The sample space is the set of all possible outcomes of a random experiment.]

Step 2: Defining the Event Space ($B$)

The problem defines event $B$ as the occurrence of "a number greater than 7". We represent this as a subset of the sample space $S$ such that every element $x \in B$ satisfies $x > 7$.

Mathematically, we look for $B = \{x \in S : x > 7\}$.

Step 3: Logical Verification of Elements

Testing each element of $S$ against the condition $x > 7$:

• Is $1 > 7$? False.
• Is $2 > 7$? False.
• Is $3 > 7$? False.
• Is $4 > 7$? False.
• Is $5 > 7$? False.
• Is $6 > 7$? False.

[Logical Deduction: Since no element in the defined sample space $S$ satisfies the condition, the set $B$ contains no elements.]

Step 4: Classification of the Event

An event that contains no outcomes from the sample space is termed an Impossible Event or the Null Event, denoted by the symbol $\phi$ or $\{\}$.

Final Solution: The event B is an impossible event, represented as $B = \phi$.

Step 1: Establishing the Sample Space ($S$)

When three unbiased coins are tossed simultaneously, each coin has $2$ possible outcomes ($H$ for Heads, $T$ for Tails). By the Fundamental Counting Principle, the total number of outcomes is $2^3 = 8$.

The sample space $S$ is defined as: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Step 2: Defining Mutually Exclusive and Exhaustive Criteria

• Mutually Exclusive: Two events $A$ and $B$ are mutually exclusive if $A \cap B = \emptyset$ (i.e., they cannot occur simultaneously).
• Exhaustive: Two events $A$ and $B$ are exhaustive if $A \cup B = S$ (i.e., at least one of the events must occur).
• Requirement: We seek two events such that $A \cap B = \emptyset$ (mutually exclusive) but $A \cup B \neq S$ (not exhaustive).

Step 3: Constructing the Events

Let us define Event $A$ and Event $B$ as follows:

• Event $A$: Getting exactly three heads. $A = \{HHH\}$
• Event $B$: Getting exactly three tails. $B = \{TTT\}$

Step 4: Analytical Verification

Visual Representation of Sample Space Distribution

Final Solution: Event $A = \{HHH\}$ (getting exactly three heads) and Event $B = \{TTT\}$ (getting exactly three tails) are mutually exclusive because $A \cap B = \emptyset$, and they are not exhaustive because $A \cup B \neq S$.

Step 1: Establishing the Sample Space ($S$)

When rolling a pair of fair six-sided dice, each die has $6$ possible outcomes. The total number of outcomes in the sample space is $6 \times 6 = 36$. We define the sample space $S$ as the set of all ordered pairs $(x, y)$, where $x$ and $y \in \{1, 2, 3, 4, 5, 6\}$.

Step 2: Defining the Events A, B, and C

Event A: The sum of the numbers is greater than 8.
The possible sums are $\{9, 10, 11, 12\}$.

• Sum = 9: $(3,6), (4,5), (5,4), (6,3)$
• Sum = 10: $(4,6), (5,5), (6,4)$
• Sum = 11: $(5,6), (6,5)$
• Sum = 12: $(6,6)$

Event B: 2 occurs on either die.
This includes cases where 2 is on the first die, the second die, or both. $B = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2)\}$

Event C: The sum is at least 7 and a multiple of 3.
Possible sums $\ge 7$ are $\{7, 8, 9, 10, 11, 12\}$. Multiples of 3 in this set are $\{9, 12\}$.

• Sum = 9: $(3,6), (4,5), (5,4), (6,3)$
• Sum = 12: $(6,6)$

Step 3: Determining Mutually Exclusive Pairs

Two events $X$ and $Y$ are mutually exclusive if $X \cap Y = \emptyset$ [Definition of Mutually Exclusive Events].

Step 4: Visualizing the Sample Space Subset Relationships

Final Solution: The pairs of events that are mutually exclusive are (A, B) and (B, C), because their intersections contain no common elements, whereas (A, C) is not mutually exclusive as $A \cap C = C$.

Step 1: Defining the Sample Space ($S$)

When two dice are thrown simultaneously, each die has 6 faces $\{1, 2, 3, 4, 5, 6\}$. The sample space $S$ consists of all ordered pairs $(x, y)$, where $x$ is the outcome of the first die and $y$ is the outcome of the second die. The total number of outcomes is $n(S) = 6 \times 6 = 36$.

Step 2: Defining Event A and Event B

Event $A$ is defined as "getting an even number on the first die." The possible outcomes for $x$ are $\{2, 4, 6\}$. The outcomes for $y$ can be any integer from $1$ to $6$. $A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$. [Cardinality: $n(A) = 3 \times 6 = 18$].

Event $B$ is defined as "getting an odd number on the first die." The possible outcomes for $x$ are $\{1, 3, 5\}$. The outcomes for $y$ can be any integer from $1$ to $6$. $B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$. [Cardinality: $n(B) = 3 \times 6 = 18$].

Step 3: Determining the Complement of Event B ($B'$)

The complement event $B'$ consists of all outcomes in the sample space $S$ that are not in event $B$. By definition, $B' = S - B$. Since $S$ is the set of all pairs $(x, y)$ where $x \in \{1, 2, 3, 4, 5, 6\}$, and $B$ contains all pairs where $x \in \{1, 3, 5\}$, then $B'$ must contain all pairs where $x \in \{2, 4, 6\}$.

Step 4: Logical Comparison

As derived in Step 3, the outcomes belonging to $B'$ are exactly the pairs where the first die shows an even number ($x \in \{2, 4, 6\}$). Comparing the set $A$ (defined in Step 2) and the set $B'$ (derived in Step 3): $A = \{(x, y) \mid x \in \{2, 4, 6\}, y \in \{1, \dots, 6\}\}$ $B' = \{(x, y) \mid x \in \{2, 4, 6\}, y \in \{1, \dots, 6\}\}$

Since $A = B'$, the statement is logically valid. [By the Law of Complementation in Set Theory, since the sample space is partitioned into $A$ (even first) and $B$ (odd first), $B^c$ is the set of all outcomes not in $B$, which is $A$.]

Final Solution: True. The event $A$ (getting an even number on the first die) is identical to the complement of event $B$ (getting an odd number on the first die) because the first die can only result in either an even or an odd number, and these events are mutually exclusive and collectively exhaustive.

Step 1: Defining the Sample Space ($S$)

When two six-sided dice are thrown, each die has 6 outcomes ($1, 2, 3, 4, 5, 6$). The total number of outcomes in the sample space $S$ is $6 \times 6 = 36$.

$S = \{ (x, y) : x, y \in \{1, 2, 3, 4, 5, 6\} \}$

Step 2: Defining Event A and Event B

• Event A: Getting an even number on the first die. $A = \{ (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$
• Event B: Getting an odd number on the first die. $B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) \}$

Step 3: Analyzing "A and B" (Intersection $A \cap B$)

The event "A and B" denotes the intersection of set $A$ and set $B$ ($A \cap B$). This event consists of outcomes that belong to both Event A and Event B simultaneously [Per the Definition of Intersection in Set Theory].

Step 4: Logical Evaluation of the Intersection

An outcome $(x, y)$ belongs to $A \cap B$ if $x$ is simultaneously even and odd.
Since the set of even numbers $\{2, 4, 6\}$ and the set of odd numbers $\{1, 3, 5\}$ are mutually exclusive, there exists no element $x$ such that $x \in \{2, 4, 6\} \cap \{1, 3, 5\}$.

Therefore, $A \cap B = \emptyset$ [By the Property of Mutually Exclusive Events].

Step 5: Conclusion on Event (A and B)

Since the intersection results in an empty set, the event "A and B" is an impossible event. There are no outcomes in the sample space where the first die is both even and odd.

Final Solution: $A \cap B = \phi$ (The event is impossible).

Step 1: Defining the Sample Space

When two six-sided dice are thrown, the sample space $S$ consists of all ordered pairs $(x, y)$, where $x, y \in \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $n(S) = 6 \times 6 = 36$.

Step 2: Defining Events A and B

Event A (Even number on the first die): The first die can be $2, 4,$ or $6$. $A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$ $n(A) = 3 \times 6 = 18$.

Event B (Odd number on the first die): The first die can be $1, 3,$ or $5$. $B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$ $n(B) = 3 \times 6 = 18$.

Step 3: Checking Mutually Exclusive Property

Two events are mutually exclusive if $A \cap B = \phi$ [Definition of Mutually Exclusive Events]. Evaluating the intersection: $A \cap B = \{(x, y) \in S : x \text{ is even and } x \text{ is odd}\}$ Since no integer can be simultaneously even and odd, $A \cap B = \phi$. Conclusion: A and B are mutually exclusive.

Step 4: Checking Exhaustive Property

Two events are exhaustive if $A \cup B = S$ [Definition of Exhaustive Events]. $A \cup B$ contains all outcomes where the first die is either even or odd. Since every possible result on a die must be either even or odd, $A \cup B$ covers all $36$ outcomes in $S$. $n(A \cup B) = n(A) + n(B) = 18 + 18 = 36$. Since $n(A \cup B) = n(S)$, the union forms the complete sample space.

Step 5: Visual Representation

Reasoning Summary

Final Solution: The statement is True. Events A and B are mutually exclusive because their intersection is an empty set (it is impossible for a number to be both even and odd simultaneously), and they are exhaustive because their union constitutes the entire sample space S.

1. Sample Space Determination

When three unbiased coins are tossed simultaneously, each coin has 2 possible outcomes ($H$ or $T$). By the Fundamental Counting Principle, the total number of outcomes is $2^3 = 8$. The sample space $S$ is defined as:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

2. Event Set Formalization

We define the sets based on the criteria provided:

• Event A (Three heads show): $A = \{HHH\}$
• Event B (Two heads and one tail show): $B = \{HHT, HTH, THH\}$
• Event C (Three tails show): $C = \{TTT\}$
• Event D (Head shows on the first coin): $D = \{HHH, HHT, HTH, HTT\}$

3. Theoretical Definition of a Simple Event

[Per the axioms of Probability Theory], an event is classified as a simple event (or elementary event) if and only if it consists of exactly one outcome of the sample space $S$. Mathematically, an event $E$ is simple if $|E| = 1$.

4. Evaluation of Given Events

5. Geometric Representation of Sample Outcomes

6. Final Logical Conclusion

Since events A and C contain exactly one elementary outcome from the sample space, they satisfy the definition of a simple event.

Final Solution: The simple events are A and C.

Step 1: Establishing the Sample Space ($S$)

When three coins are tossed, each coin has 2 possible outcomes: Head ($H$) or Tail ($T$). The total number of outcomes is $2^3 = 8$. The sample space $S$ is defined as:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Step 2: Defining the Events

Based on the conditions provided, we define the subsets of $S$ for events $A, B, C,$ and $D$:

• Event $A$ (Three heads): $A = \{HHH\}$
• Event $B$ (Two heads, one tail): $B = \{HHT, HTH, THH\}$
• Event $C$ (Three tails): $C = \{TTT\}$
• Event $D$ (Head on the first coin): $D = \{HHH, HHT, HTH, HTT\}$

Step 3: Analytical Criterion for Mutually Exclusive Events

Two events $E_1$ and $E_2$ are mutually exclusive if and only if $E_1 \cap E_2 = \emptyset$ [Per the definition of disjoint sets in Probability Theory].

Step 4: Intersection Analysis

Step 5: Visual Representation of Sample Space Distribution

Final Solution: The pairs of events that are mutually exclusive are (A, B), (A, C), (B, C), and (C, D).

Step 1: Establishing the Sample Space ($S$)

When two six-sided dice are thrown, each die has an outcome set of $\{1, 2, 3, 4, 5, 6\}$. The sample space $S$ consists of all ordered pairs $(x, y)$, where $x$ is the outcome of the first die and $y$ is the outcome of the second die. The total number of outcomes is $6 \times 6 = 36$ [By the Fundamental Counting Principle].

Step 2: Defining the Events $A$ and $C$

Event $A$ (Even number on the first die): $A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

Event $C$ (Sum of numbers $\le 5$): The possible sums are 2, 3, 4, and 5. $C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$

Step 3: Calculating $A$ but not $C$ ($A \cap C'$)

The event "A but not C" is represented in set theory as $A \setminus C$ or $A \cap C^c$. This requires identifying elements present in $A$ that are not present in $C$.

Elements of $A$ that are also in $C$: Intersection $A \cap C = \{(2,1), (2,2), (2,3), (4,1)\}$.

To find $A \setminus C$, we remove these elements from set $A$:

Step 4: Logical Visualization of Event Space

Step 5: Final Set Representation

The set $A \setminus C$ contains all outcomes where the first die is even and the sum of the dice is strictly greater than 5.

$A \setminus C = \{(2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

Final Solution: The event "A but not C" is the set of 14 outcomes: {(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)}

Step 1: Defining the Sample Space

When two six-faced dice are thrown, the sample space $S$ consists of all ordered pairs $(x, y)$, where $x$ is the result of the first die and $y$ is the result of the second die. The total number of outcomes is $6 \times 6 = 36$.

$S = \{ (1,1), (1,2), \dots, (6,6) \}$

Step 2: Defining Event A and Event B

Event A: Getting an even number on the first die.

• $A = \{ (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}$
• The cardinality of set A is $n(A) = 18$.

Event B: Getting an odd number on the first die.

• $B = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \}$
• The cardinality of set B is $n(B) = 18$.

Step 3: Determining Mutual Exclusivity

Two events are defined as mutually exclusive if and only if their intersection is the empty set, denoted as $A \cap B = \emptyset$. [Per the Set Theoretic Definition of Probability].

Analyzing $A \cap B$:

Set A contains pairs where the first element $x \in \{2, 4, 6\}$.

Set B contains pairs where the first element $x \in \{1, 3, 5\}$.

Since the set of even integers $\{2, 4, 6\}$ and odd integers $\{1, 3, 5\}$ have no common elements, $A \cap B = \emptyset$.

Visual Representation of Exclusivity

Conclusion

Because the first die cannot be both even and odd simultaneously, the events $A$ and $B$ cannot occur at the same time. Therefore, $A \cap B = \emptyset$.

Final Solution: True. The events A and B are mutually exclusive because it is impossible for the first die to show an even number and an odd number simultaneously, resulting in an empty intersection ($A \cap B = \emptyset$).

Step 1: Defining the Sample Space and Event Spaces

The sample space $S$ for two thrown dice consists of $6 \times 6 = 36$ ordered pairs $(x, y)$, where $x, y \in \{1, 2, 3, 4, 5, 6\}$.

We define the events as follows:

• Event A: Even number on the first die. $A = \{ (x, y) : x \in \{2, 4, 6\} \}$. There are $3 \times 6 = 18$ outcomes.
• Event B: Odd number on the first die. $B = \{ (x, y) : x \in \{1, 3, 5\} \}$. There are $3 \times 6 = 18$ outcomes.
• Event C: Sum of numbers on the dice $\le 5$. $C = \{ (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1) \}$.

Step 2: Identifying Complements

Per the Complement Rule in set theory, $E'$ contains all elements of $S$ not present in $E$ [By Set Complement Axiom].

• $B'$ (Not B): Since $S = A \cup B$ and $A \cap B = \emptyset$, $B' = A$. Therefore, $B' = \{ (x, y) : x \in \{2, 4, 6\} \}$.
• $C'$ (Not C): These are outcomes where the sum $x + y > 5$.

Step 3: Logical Intersection Analysis

We seek the intersection $A \cap B' \cap C'$.

1. Since $B' = A$, the expression simplifies to $A \cap A \cap C'$, which is equivalent to $A \cap C'$.

2. We need to identify outcomes in $A$ (first die is even) that are not in $C$ (sum > 5).

Step 4: Compilation of Outcomes

The set $A \cap B' \cap C'$ is the set of all outcomes $(x, y)$ where $x \in \{2, 4, 6\}$ and $x + y > 5$.

The resulting set is:

$\{ (2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}$

Final Solution: $A \cap B' \cap C' = \{ (2,4), (2,5), (2,6), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) \}$

Step 1: Establishing the Sample Space ($S$)

When two six-sided dice are thrown, each die has an outcome set of $\{1, 2, 3, 4, 5, 6\}$. The total sample space $S$ consists of $6 \times 6 = 36$ ordered pairs $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

$S = \{(1,1), (1,2), \dots, (6,6)\}$ [By the Fundamental Counting Principle].

Step 2: Defining Event $B$ and Event $C$

Event $B$ is defined as "getting an odd number on the first die." Thus, $d_1 \in \{1, 3, 5\}$.

$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$

Event $C$ is defined as "getting the sum of the numbers on the dice $\le 5$."

$C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$

Step 3: Calculating the Union ($B \cup C$)

The event "B or C" is represented by the union $B \cup C$. This set includes all outcomes that are in $B$, or in $C$, or in both [Set Theory Definition of Union].

Step 4: Final Compilation of Outcomes

Combining the sets and removing duplicates:

$B \cup C = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$

Final Solution: The set $B \cup C = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$

Step 1: Defining the Sample Space ($S$)

When three coins are tossed simultaneously, each coin has two possible outcomes: Head ($H$) or Tail ($T$). The total number of outcomes is $2^n$, where $n=3$. Thus, $2^3 = 8$ outcomes.

The sample space $S$ is defined as:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

Step 2: Identifying Individual Events

Based on the problem statement, we define the sets for each event:

• Event A (Three heads): $A = \{HHH\}$
• Event B (Two heads, one tail): $B = \{HHT, HTH, THH\}$
• Event C (Three tails): $C = \{TTT\}$
• Event D (Head on first coin): $D = \{HHH, HHT, HTH, HTT\}$

Step 3: Theoretical Criteria for Compound Events

[Per the definition of Probability Theory]: An event is termed a Compound Event if it consists of more than one sample point from the sample space $S$. Conversely, an event consisting of exactly one sample point is a Simple (or Elementary) Event.

Step 4: Evaluating Each Event

Step 5: Visual Representation of Sample Space

Step 6: Conclusion

An event is compound if the cardinality of its set is greater than 1 ($|E| > 1$). Observing the analysis in Step 4, Events B and D contain multiple outcomes, satisfying the criteria for compound events.

Final Solution: The events B and D are Compound Events because they consist of more than one sample point from the sample space.

Step 1: Establishing the Sample Space ($S$)

When two six-faced dice are thrown, the sample space $S$ consists of all possible ordered pairs $(x, y)$, where $x$ is the result of the first die and $y$ is the result of the second die. $x, y \in \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes is $n(S) = 6 \times 6 = 36$.

Step 2: Defining the Events

Event $B$ is defined as "getting an odd number on the first die":

$B = \{ (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) \}$

Event $C$ is defined as "getting the sum of the numbers on the dice $\le 5$":

$C = \{ (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1) \}$

Step 3: Deriving the Event ($B \cap C$)

The notation $B \cap C$ refers to the intersection of events, which constitutes outcomes that belong to both set $B$ and set $C$ [Per the Set Theory Definition of Intersection].

• Identify pairs where the first digit is odd ($1, 3, 5$): $(1,1), (1,2), (1,3), (1,4), (3,1), (3,2), (5,1), (5,2), (5,3) \dots$
• Filter these pairs against the condition $x + y \le 5$:

Step 4: Visualizing the Intersection

Final Solution: The intersection of events B and C is $B \cap C = \{ (1,1), (1,2), (1,3), (1,4), (3,1), (3,2) \}$.