Step 1: Given Variables & Theoretical Foundation

We are provided with the probabilities of two independent events, $A$ and $B$:

• $P(A) = \frac{3}{5}$
• $P(B) = \frac{1}{5}$

The core theoretical principle governing this problem is the Multiplication Theorem for Independent Events. Two events $A$ and $B$ are defined as independent if and only if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, this is expressed as:

$P(A \cap B) = P(A) \times P(B)$ [By definition of Independent Events]

Step 2: Mathematical Execution

Substitute the given values into the multiplication identity:

$P(A \cap B) = \left( \frac{3}{5} \right) \times \left( \frac{1}{5} \right)$

Step 3: Arithmetic Computation

Multiplying the numerators and the denominators separately:

$P(A \cap B) = \frac{3 \times 1}{5 \times 5}$

$P(A \cap B) = \frac{3}{25}$

Step 4: Geometric Representation of Independent Probabilities

The intersection $P(A \cap B)$ represents the joint probability of both events occurring simultaneously, which, for independent events, is the area of the overlapping region in the sample space defined by the product of their individual probabilities.

Final Solution: $P(A \cap B) = \frac{3}{25}$

Given Variables & Initial Setup

The problem provides two independent events $A$ and $B$ with the following probabilities:

• $P(A) = 0.3$
• $P(B) = 0.6$

Since $A$ and $B$ are independent events, the following axiom holds: $P(A \cap B) = P(A) \cdot P(B)$ [By the definition of independence in Probability Theory].

Step 1: Determine the Probability of the Complement Event

To find the probability of "not $B$" (denoted as $B'$ or $B^c$), we utilize the Complement Rule: $P(B') = 1 - P(B)$ [Per the Law of Complements].

$P(B') = 1 - 0.6 = 0.4$

Step 2: Analytical Expression for Independent Complements

If $A$ and $B$ are independent, then $A$ and $B'$ are also independent [Theorem: If $A$ and $B$ are independent, then $A$ and $B^c$ are independent].

Therefore, the probability of the intersection $P(A \cap B')$ is calculated as:

$P(A \cap B') = P(A) \cdot P(B')$

Step 3: Numerical Computation

Substituting the given and derived values into the equation:

$P(A \cap B') = 0.3 \times 0.4$

$P(A \cap B') = 0.12$

Visual Representation of Independent Events in Sample Space

Final Solution: $P(A \text{ and not } B) = 0.12$

Step 1: Identifying Variables and Probabilities

The total number of balls in the box is given by the sum of black and red balls:

$N = 10 \text{ (black)} + 8 \text{ (red)} = 18 \text{ total balls}$

Let $B$ be the event of drawing a black ball and $R$ be the event of drawing a red ball. Since the balls are drawn with replacement, the trials are independent, and the probabilities remain constant for each draw.

Probability of drawing a black ball $P(B) = \frac{10}{18} = \frac{5}{9}$

Probability of drawing a red ball $P(R) = \frac{8}{18} = \frac{4}{9}$

Step 2: Defining the Favorable Outcomes

The event "one black and one red" can occur in two mutually exclusive ways when drawing two balls:

• Case 1: The first ball is Black and the second ball is Red $(B, R)$.
• Case 2: The first ball is Red and the second ball is Black $(R, B)$.

Step 3: Calculating Individual Probabilities

Since the draws are independent [By the Multiplication Rule for Independent Events, $P(A \cap B) = P(A) \cdot P(B)$]:

For Case 1 ($B$ then $R$):

$P(B \cap R) = P(B) \times P(R) = \frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$

For Case 2 ($R$ then $B$):

$P(R \cap B) = P(R) \times P(B) = \frac{4}{9} \times \frac{5}{9} = \frac{20}{81}$

Step 4: Combining Probabilities

Because these cases are mutually exclusive [By the Addition Rule for Mutually Exclusive Events, $P(A \cup B) = P(A) + P(B)$]:

$P(\text{one black, one red}) = P(B \cap R) + P(R \cap B)$

$P(\text{one black, one red}) = \frac{20}{81} + \frac{20}{81} = \frac{40}{81}$

Visual Representation of Outcome Space

Final Solution: The probability that one ball is black and the other is red is $\frac{40}{81}$.

Step 1: Establishing the Theoretical Foundation

Two events $A$ and $B$ are defined as mutually exclusive if they cannot occur simultaneously. In the language of set theory and probability, this implies that the intersection of the two events is the empty set: $A \cap B = \emptyset$. Consequently, the probability of their intersection is zero:
$P(A \cap B) = 0$ [By the definition of mutually exclusive events].

Step 2: Invoking the Addition Theorem of Probability

For any two events $A$ and $B$, the Addition Theorem states:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ [Per the Axioms of Probability].

Step 3: Substitution of Given Variables

We are provided with the following values:

• $P(A) = \frac{1}{2}$
• $P(A \cup B) = \frac{3}{5}$
• $P(B) = p$
• $P(A \cap B) = 0$ [As established in Step 1].

Step 4: Algebraic Derivation for $p$

To isolate $p$, we subtract $\frac{1}{2}$ from both sides of the equation:
$p = \frac{3}{5} - \frac{1}{2}$

To perform the subtraction, determine the Least Common Multiple (LCM) of the denominators $5$ and $2$, which is $10$:
$p = \frac{3 \times 2}{5 \times 2} - \frac{1 \times 5}{2 \times 5}$
$p = \frac{6}{10} - \frac{5}{10}$
$p = \frac{1}{10}$

Visual Representation of Mutually Exclusive Events

Final Solution: $p = \frac{1}{10}$

Step 1: Formal Definition of Independent Events

Let $A$ be the event that person $A$ solves the problem, and $B$ be the event that person $B$ solves the problem. Based on the provided data:

• $P(A) = \frac{1}{2}$
• $P(B) = \frac{1}{3}$

Since the events are independent, the probability that both fail to solve the problem is given by the product of their individual failure probabilities [Per the Multiplication Rule for Independent Events].

Step 2: Calculating Complements (Failure Probabilities)

The probability that the individuals fail to solve the problem is defined as the complement of the probability of success:

• $P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$
• $P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$

Step 3: Calculating Probability of Failure for Both

If the problem is to be solved, the only scenario where it is not solved is if both individuals fail independently. Let $F$ be the event that the problem remains unsolved:

$P(F) = P(A' \cap B')$

$P(F) = P(A') \times P(B')$ [Since $A$ and $B$ are independent, $A'$ and $B'$ are also independent]

$P(F) = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

Step 4: Calculating Probability of Success

The event "the problem is solved" is the complement of the event "the problem is not solved." Let $S$ be the event that the problem is solved:

$P(S) = 1 - P(F)$

$P(S) = 1 - \frac{1}{3}$

$P(S) = \frac{2}{3}$

Visual Representation of Sample Space

Final Solution: The probability that the problem is solved is $\frac{2}{3}$.

Step 1: Establishing the Theoretical Foundation for Independence

Two events $E$ and $F$ associated with the same sample space are defined as independent if and only if the probability of their intersection is equal to the product of their individual probabilities. Mathematically, the condition for independence is defined as follows:

$P(E \cap F) = P(E) \times P(F)$

[Reference: Definition of Independent Events in Probability Theory]

Step 2: Identifying Given Variables

From the problem statement, we have the following assigned values:

• $P(E) = \frac{3}{5}$
• $P(F) = \frac{3}{10}$
• $P(E \cap F) = \frac{1}{5}$

Step 3: Calculating the Product of Individual Probabilities

To evaluate independence, we compute the product $P(E) \times P(F)$ using the values identified in Step 2:

$P(E) \times P(F) = \left( \frac{3}{5} \right) \times \left( \frac{3}{10} \right)$

Performing the multiplication of numerators and denominators:

$P(E) \times P(F) = \frac{3 \times 3}{5 \times 10} = \frac{9}{50}$

Step 4: Comparative Analysis

We compare the calculated product with the given value of $P(E \cap F)$:

• $P(E \cap F) = \frac{1}{5} = \frac{10}{50}$
• $P(E) \times P(F) = \frac{9}{50}$

Since $\frac{10}{50} \neq \frac{9}{50}$, it follows that $P(E \cap F) \neq P(E) \times P(F)$.

Visual Representation of Probability Sets

Conclusion

Because the condition $P(E \cap F) = P(E) \times P(F)$ is not satisfied, the events $E$ and $F$ do not meet the criteria for independence.

Final Solution: No, E and F are not independent because $P(E \cap F) \neq P(E) \times P(F)$.

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

A fair coin has two outcomes: Heads ($H$) and Tails ($T$). An unbiased six-sided die has outcomes $\{1, 2, 3, 4, 5, 6\}$. The sample space $S$ for the combined experiment is the Cartesian product of these sets:

$S = \{ (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) \}$

The total number of outcomes is $n(S) = 2 \times 6 = 12$. [By the Fundamental Counting Principle]

Step 2: Defining Events A and B

Event $A$: 'Head appears on the coin'

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

$n(A) = 6$

$P(A) = \frac{n(A)}{n(S)} = \frac{6}{12} = \frac{1}{2}$

Event $B$: '3 on the die'

$B = \{ (H,3), (T,3) \}$

$n(B) = 2$

$P(B) = \frac{n(B)}{n(S)} = \frac{2}{12} = \frac{1}{6}$

Step 3: Calculating the Intersection ($A \cap B$)

The intersection event $A \cap B$ represents the occurrence of both 'Head' and '3'.

$A \cap B = \{ (H,3) \}$

$n(A \cap B) = 1$

$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{12}$

Step 4: Evaluating Independence

Two events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \times P(B)$. [Per the Definition of Independent Events]

Calculating the product $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

Visual Representation of Sample Space

Conclusion

Since $P(A \cap B) = \frac{1}{12}$ and $P(A) \times P(B) = \frac{1}{12}$, the condition for independence holds.

Final Solution: The events A and B are independent events.

Step 1: Definition of Events and Initial Data

Let $H$ be the event that a randomly selected student reads the Hindi newspaper, and $E$ be the event that the student reads the English newspaper. Based on the provided data, we establish the probabilities:

• $P(H) = 0.60$
• $P(E) = 0.40$
• $P(H \cap E) = 0.20$

[By Definition of Marginal and Joint Probability in Sample Space $\Omega$]

Step 2: Identifying the Conditional Probability

The question requires the probability that the student reads the Hindi newspaper, given that she reads the English newspaper. Mathematically, this is denoted as $P(H|E)$.

[Per the Definition of Conditional Probability: $P(A|B) = \frac{P(A \cap B)}{P(B)}$, where $P(B) > 0$]

Step 3: Execution of the Calculation

Substituting the known values into the formula:

$P(H|E) = \frac{P(H \cap E)}{P(E)}$

$P(H|E) = \frac{0.20}{0.40}$

$P(H|E) = \frac{2}{4} = 0.5$

Step 4: Venn Diagram Representation

Step 5: Analytical Interpretation

Final Solution: The probability that the student reads the Hindi newspaper given that she reads the English newspaper is 0.5 or 50%.

Step 1: Defining the Sample Space and Event Sets

The sample space $S$ for a standard six-faced die is defined as $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $n(S) = 6$.

Event $A$ is defined as "the number is even": $A = \{2, 4, 6\}$ $n(A) = 3$ $P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}$

Event $B$ is defined as "the number is red". Given that numbers 1, 2, and 3 are marked in red: $B = \{1, 2, 3\}$ $n(B) = 3$ $P(B) = \frac{n(B)}{n(S)} = \frac{3}{6} = \frac{1}{2}$

Step 2: Determining the Intersection of A and B

The intersection $A \cap B$ represents the set of outcomes that are both even and red. Looking at the sets $A = \{2, 4, 6\}$ and $B = \{1, 2, 3\}$: $A \cap B = \{2\}$ $n(A \cap B) = 1$ $P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{6}$

Step 3: Verification of Independence Condition

Two events $A$ and $B$ are independent if and only if $P(A \cap B) = P(A) \cdot P(B)$ [Per the definition of Independent Events in Probability Theory].

Calculate the product of the individual probabilities: $P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

Compare this result with $P(A \cap B)$: $P(A \cap B) = \frac{1}{6}$ $\frac{1}{6} \neq \frac{1}{4}$

Step 4: Visualizing Sample Space Representation

Conclusion

Since the product of the individual probabilities of event $A$ and event $B$ does not equal the probability of their intersection, the occurrence of one event affects the probability of the other.

Final Solution: Since $P(A \cap B) \neq P(A) \cdot P(B)$, events $A$ and $B$ are not independent.

Step 1: Theoretical Foundation of Independence

By the formal definition of independent events in probability theory, two events $A$ and $B$ are independent if and only if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, this is expressed as [Per the definition of Independence]:

$P(A|B) = P(A)$

$P(B|A) = P(B)$

Step 2: Analysis of the Given Variables

• $P(A) = 0.3$
• $P(B) = 0.4$
• Events $A$ and $B$ are defined as independent.

Step 3: Application of the Conditional Probability Formula

The general formula for conditional probability is given by [Per Bayes' Theorem/Conditional Probability Axioms]:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

Since events $A$ and $B$ are independent, the intersection probability is defined as the product of their individual probabilities [Per the Multiplication Rule for Independent Events]:

$P(A \cap B) = P(A) \times P(B)$

Step 4: Algebraic Substitution and Calculation

Substituting the product rule into the conditional probability formula:

$P(A|B) = \frac{P(A) \times P(B)}{P(B)}$

Assuming $P(B) \neq 0$, we perform the algebraic cancellation:

$P(A|B) = P(A)$

Substituting the given value $P(A) = 0.3$:

$P(A|B) = 0.3$

Visualization of Independence

Final Solution: $P(A|B) = 0.3$

Step 1: Defining the Sample Space and Event

When a pair of fair six-sided dice is rolled, 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 given by:

$n(S) = 6 \times 6 = 36$ [Fundamental Counting Principle]

Step 2: Identifying the Target Outcome

The question requires the probability of obtaining an "even prime number" on each die. First, we identify the set of prime numbers within the set of possible outcomes $\{1, 2, 3, 4, 5, 6\}$.

• The set of prime numbers is $\{2, 3, 5\}$.
• Among these, the only even number is $\{2\}$.

Therefore, the condition "even prime number" is satisfied if and only if the outcome on a die is $2$.

Step 3: Calculating Individual and Joint Probabilities

Let $E_1$ be the event that the first die shows an even prime (2) and $E_2$ be the event that the second die shows an even prime (2).

• $P(E_1) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{1}{6}$
• $P(E_2) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{1}{6}$

Since the roll of one die is independent of the roll of the other, the joint probability is calculated as the product of the individual probabilities [Per the Multiplication Rule for Independent Events]:

$P(E_1 \cap E_2) = P(E_1) \times P(E_2)$

Step 4: Algebraic Evaluation

$P(E_1 \cap E_2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Visualization of Sample Space

Final Solution: The probability of obtaining an even prime number on each die is $\frac{1}{36}$.

Step 1: Defining the Sample Space and Event

In a single toss of a fair six-faced die, the sample space $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes for one toss is $n(S) = 6$.

Let $E$ be the event of obtaining an odd number in a single toss. The odd numbers on a die are $\{1, 3, 5\}$.

Probability of getting an odd number in one toss: $P(E) = \frac{3}{6} = \frac{1}{2}$ [Per the Classical Definition of Probability].

Consequently, the probability of not getting an odd number (getting an even number) in one toss is $P(E^c) = 1 - P(E) = 1 - \frac{1}{2} = \frac{1}{2}$.

Step 2: Defining the Compound Event for Three Tosses

The experiment consists of tossing the die three times ($n = 3$). The tosses are independent events [Per the Multiplication Rule for Independent Events].

We are required to find the probability of getting an odd number at least once. Let $X$ be the random variable representing the number of odd numbers obtained. We seek $P(X \ge 1)$.

Using the Complement Rule: $P(X \ge 1) = 1 - P(X = 0)$, where $P(X = 0)$ is the probability of getting no odd numbers in all three tosses.

Step 3: Calculation of Complementary Probability

For $X = 0$, the outcome of every toss must be an even number. Since the tosses are independent, the joint probability is the product of individual probabilities:

$P(X = 0) = P(\text{Even}_1) \times P(\text{Even}_2) \times P(\text{Even}_3)$

$P(X = 0) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3$

$P(X = 0) = \frac{1}{8}$

Step 4: Final Probability Determination

Applying the Complement Rule to find the probability of at least one success:

$P(\text{At least one odd}) = 1 - \frac{1}{8}$

$P(\text{At least one odd}) = \frac{8 - 1}{8} = \frac{7}{8}$

Final Solution: The probability of getting an odd number at least once in three tosses is $\frac{7}{8}$.

Step 1: Definition of Events and Initial Data

Let $H$ be the event that a student reads the Hindi newspaper, and $E$ be the event that a student reads the English newspaper. Based on the provided data, the probabilities are defined as follows:

• $P(H) = 60\% = 0.6$
• $P(E) = 40\% = 0.4$
• $P(H \cap E) = 20\% = 0.2$

Step 2: Calculating the Probability of the Union of Events

To find the probability that a student reads at least one of the two newspapers, we employ the Addition Theorem of Probability [Per the Addition Rule for Sets]:

$P(H \cup E) = P(H) + P(E) - P(H \cap E)$

Substituting the given values:

$P(H \cup E) = 0.6 + 0.4 - 0.2$

$P(H \cup E) = 1.0 - 0.2 = 0.8$

Step 3: Calculating the Complement Event

The event "reads neither Hindi nor English newspapers" is represented by the complement of the union, denoted as $P((H \cup E)')$. [By De Morgan's Law and Complementary Probability Axiom]:

$P((H \cup E)') = 1 - P(H \cup E)$

$P((H \cup E)') = 1 - 0.8$

$P((H \cup E)') = 0.2$

Visual Representation (Venn Diagram)

Data Summary Table

Final Solution: The probability that a student reads neither Hindi nor English newspapers is 0.2 (or 20%).

Step 1: Theoretical Foundation & Given Variables

We are given two independent events, $A$ and $B$, with the following probabilities:

• $P(A) = 0.3$
• $P(B) = 0.6$

Since $A$ and $B$ are independent events, the occurrence of one does not affect the probability of the occurrence of the other. Consequently, their complements $A^c$ and $B^c$ are also independent [By the Law of Independent Complements].

Step 2: Defining the Complementary Events

The probability of the complement of an event is defined as $P(E^c) = 1 - P(E)$. Therefore:

• $P(A^c) = 1 - P(A) = 1 - 0.3 = 0.7$
• $P(B^c) = 1 - P(B) = 1 - 0.6 = 0.4$

Step 3: Calculating "Neither A nor B"

The event "neither $A$ nor $B$" is represented set-theoretically as the intersection of the complements of $A$ and $B$: $A^c \cap B^c$.

Per De Morgan's Law, $A^c \cap B^c = (A \cup B)^c$. Given the independence of the events, the probability of the intersection is the product of the individual probabilities [By the Multiplication Rule for Independent Events]:

$P(A^c \cap B^c) = P(A^c) \times P(B^c)$

Step 4: Substitution and Final Calculation

Substituting the values derived in Step 2:

$P(\text{neither } A \text{ nor } B) = 0.7 \times 0.4$

$P(\text{neither } A \text{ nor } B) = 0.28$

Visual Representation of Sample Space

Final Solution: $P(\text{neither } A \text{ nor } B) = 0.28$

Step 1: Identifying the Sample Space and Event Probabilities

Let $H$ be the event that the student reads the Hindi newspaper, and $E$ be the event that the student reads the English newspaper. Based on the provided data, we assign the following probabilities:

• $P(H) = 60\% = 0.60$
• $P(E) = 40\% = 0.40$
• $P(H \cap E) = 20\% = 0.20$

Step 2: Formulating the Conditional Probability

We are required to find the probability that a student reads the English newspaper given that she already reads the Hindi newspaper. Mathematically, this is expressed as the conditional probability $P(E|H)$.

According to the Definition of Conditional Probability, for any two events $A$ and $B$ where $P(B) > 0$:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

[By the axioms of conditional probability theory]

Step 3: Calculating the Result

Substituting the values defined in Step 1 into the formula:

$P(E|H) = \frac{P(E \cap H)}{P(H)}$

$P(E|H) = \frac{0.20}{0.60}$

Simplifying the fraction:

$P(E|H) = \frac{2}{6} = \frac{1}{3}$

Visual Representation of Sets

Final Solution: The probability that the student reads the English newspaper given that she reads the Hindi newspaper is $\frac{1}{3}$ (or approximately 0.333).

Step 1: Given Variables & Theoretical Foundation

We are provided with the probabilities of two independent events, $A$ and $B$:

• $P(A) = 0.3$
• $P(B) = 0.6$

The events are stated to be independent, which implies that the occurrence of one does not affect the probability of the other [Definition of Independent Events].

Step 2: Applying the Addition Theorem of Probability

To find $P(A \text{ or } B)$, we must calculate the union of the two events, denoted as $P(A \cup B)$. The Addition Theorem of Probability states:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Step 3: Calculating the Intersection $P(A \cap B)$

Since $A$ and $B$ are independent events, the probability of both occurring simultaneously is the product of their individual probabilities [Multiplication Rule for Independent Events]:

$P(A \cap B) = P(A) \times P(B)$

$P(A \cap B) = 0.3 \times 0.6$

$P(A \cap B) = 0.18$

Step 4: Final Substitution into the Addition Formula

Substitute the known values into the expression derived in Step 2:

$P(A \cup B) = 0.3 + 0.6 - 0.18$

$P(A \cup B) = 0.9 - 0.18$

$P(A \cup B) = 0.72$

Visual Representation (Venn Diagram)

Final Solution: $P(A \text{ or } B) = 0.72$

Step 1: Define the Sample Space and Event Sets

The sample space $S$ consists of a well-shuffled standard deck of cards. The total number of outcomes is $|S| = 52$.

• Let $E$ be the event that the card drawn is a spade. In a standard deck, there are 13 spades. Thus, $|E| = 13$.
• Let $F$ be the event that the card drawn is an ace. In a standard deck, there are 4 aces. Thus, $|F| = 4$.

Step 2: Calculate Individual Probabilities

Using the classical definition of probability $P(A) = \frac{|A|}{|S|}$:

$P(E) = \frac{|E|}{|S|} = \frac{13}{52} = \frac{1}{4}$ [By the classical probability axiom]

$P(F) = \frac{|F|}{|S|} = \frac{4}{52} = \frac{1}{13}$ [By the classical probability axiom]

Step 3: Calculate the Probability of the Intersection

The intersection event $E \cap F$ represents the event that the card drawn is both a spade and an ace. There is exactly one card in the deck that satisfies both conditions: the Ace of Spades.

$|E \cap F| = 1$

$P(E \cap F) = \frac{|E \cap F|}{|S|} = \frac{1}{52}$

Step 4: Verification of Independence

Two events $E$ and $F$ are independent if and only if $P(E \cap F) = P(E) \cdot P(F)$ [By the definition of independent events].

Calculate the product of individual probabilities:

$P(E) \cdot P(F) = \left(\frac{1}{4}\right) \cdot \left(\frac{1}{13}\right) = \frac{1}{52}$

Compare with $P(E \cap F)$:

Since $P(E \cap F) = \frac{1}{52}$ and $P(E) \cdot P(F) = \frac{1}{52}$, the condition $P(E \cap F) = P(E) \cdot P(F)$ is satisfied.

Visual Representation: Venn Diagram of Events

Final Solution: Since $P(E \cap F) = P(E) \cdot P(F) = \frac{1}{52}$, the events E and F are independent.

Step 1: Theoretical Foundation

According to the axiomatic definition of probability, two events $A$ and $B$ are defined as independent events if and only if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, this condition is expressed via the Multiplication Rule for independent events:

$P(A \cap B) = P(A) \times P(B)$

Step 2: Given Parameters

The problem provides the following numerical values for the events $A$ and $B$:

• $P(A) = 0.3$
• $P(B) = 0.4$

Step 3: Algebraic Calculation

To determine the intersection probability $P(A \cap B)$, we substitute the given values into the derived formula from Step 1:

$P(A \cap B) = 0.3 \times 0.4$

$P(A \cap B) = 0.12$

Visual Representation of Independent Intersection

Data Summary Table

Final Solution: $P(A \cap B) = 0.12$

Step 1: Identifying the Mathematical Property of Independence

For any two events $A$ and $B$, the events are defined as independent if and only if the probability of both events occurring simultaneously is the product of their individual probabilities. This is expressed by the Multiplication Rule for independent events:

$P(A \cap B) = P(A) \times P(B)$

[Justification: By the definition of independent events in probability theory, the occurrence of $A$ does not affect the probability of $B$, and vice versa.]

Step 2: Assigning Given Values

From the problem statement, we are provided with the following initial parameters:

• $P(A) = 0.3$
• $P(B) = 0.6$

Step 3: Executing the Calculation

Substitute the given values into the formula derived in Step 1:

$P(A \cap B) = 0.3 \times 0.6$

$P(A \cap B) = 0.18$

Step 4: Geometric Representation (Venn Diagram)

In the diagram above, the overlapping region represents the intersection $P(A \cap B)$, which contains the probability of both events occurring simultaneously.

Final Solution: The probability of both events A and B occurring, $P(A \cap B)$, is $0.18$.

Given Variables & Initial Setup

The problem provides the probabilities of two independent events, $A$ and $B$:

• $P(A) = 0.3$
• $P(B) = 0.4$

Since events $A$ and $B$ are independent, the probability of their intersection is defined as [Per the definition of independent events]:

$P(A \cap B) = P(A) \times P(B)$

Step 1: Calculating the Intersection $P(A \cap B)$

Substituting the given values into the multiplication rule for independent events:

$P(A \cap B) = 0.3 \times 0.4$

$P(A \cap B) = 0.12$

Step 2: Applying the Addition Theorem of Probability

To find the union of two events $P(A \cup B)$, we utilize the General Addition Theorem of Probability [Per the Addition Law of Probability]:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Step 3: Final Algebraic Substitution

Insert the known values into the addition theorem formula:

$P(A \cup B) = 0.3 + 0.4 - 0.12$

$P(A \cup B) = 0.7 - 0.12$

$P(A \cup B) = 0.58$

Visual Representation of Sample Space

Final Solution: $P(A \cup B) = 0.58$