Counting and Sample Spaces

Overview

The study of probability provides the mathematical framework for reasoning about uncertainty. Before we can assign a measure of likelihood to any occurrence, we must first possess a complete and rigorous description of all possibilities. This chapter is dedicated to establishing this fundamental groundwork. We begin by formalizing the notion of a random experiment and introducing the concepts of the sample space—the set of all possible outcomes—and events, which are specific subsets of this space. A precise understanding of these elements is the indispensable first step in the structured analysis of any probabilistic scenario.

Mastery of this foundational material is of paramount importance for the GATE examination, where problems are often constructed to test not merely computational skill but conceptual clarity. Once the structure of an experiment is defined, we are frequently faced with the challenge of enumeration: determining the number of outcomes that constitute an event or the total number of outcomes in the sample space. To this end, we shall develop a systematic toolkit of counting principles, including permutations and combinations. These are not simply formulas to be memorized, but powerful logical tools for dissecting complex problems. We will conclude by unifying these concepts under the three axioms of probability, which provide the definitive, formal rules that govern all of probability theory.

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Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Sample Space and Events | Defining outcomes of random experiments. |
| 2 | Counting Principles | Systematic methods for enumerating possible outcomes. |
| 3 | Probability Axioms | The fundamental rules of probability theory. |

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Learning Objectives

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We now turn our attention to Sample Space and Events...
## Part 1: Sample Space and Events

Introduction

The study of probability is fundamentally concerned with the analysis of random phenomena. Before we can assign probabilities to outcomes, we must first establish a rigorous mathematical framework for describing all possible results of an experiment. This framework is built upon the foundational concepts of the sample space and events. A clear understanding of how to define a sample space and identify events within it is the first and most critical step in solving any problem in probability theory. This chapter provides the essential vocabulary and structure required for this purpose.

We begin by formalizing the notion of a random experiment, which is any process with an uncertain outcome. From this, we define the sample space as the set of all possible outcomes. An event, then, is simply a subset of this sample space, representing a result or a collection of results that are of particular interest to us.

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Key Concepts

The entire structure of probability theory rests upon a few key definitions. Mastering these is non-negotiable for success in the subject.

1. Sample Space

The sample space is the cornerstone upon which we build our analysis.

A sample space can be either discrete or continuous.
* Discrete Sample Space: Contains a finite or countably infinite number of sample points. For example, the outcomes of a coin toss, $S = \{H, T\}$, or the number of emails received in an hour, $S = \{0, 1, 2, ...\}$.
* Continuous Sample Space: Contains an infinite number of sample points that form a continuum. For example, the height of a student, where $S = \{h \in \mathbb{R} \mid h > 0\}$.

Worked Example:

Problem: Define the sample space for the experiment of rolling two distinct six-sided dice simultaneously.

Solution:

Step 1: Identify the outcomes for each die.
For the first die, the outcomes are $\{1, 2, 3, 4, 5, 6\}$. For the second die, the outcomes are also $\{1, 2, 3, 4, 5, 6\}$.

Step 2: Represent an outcome as an ordered pair.
Let an outcome be represented by an ordered pair $(x, y)$, where $x$ is the result of the first die and $y$ is the result of the second die.

Step 3: Systematically list all possible pairs.
The sample space $S$ is the set of all such ordered pairs.

Step 4: Determine the size of the sample space.
The total number of sample points is the product of the number of outcomes for each die.

Answer: The sample space consists of 36 ordered pairs, representing all possible combinations of outcomes from the two dice.

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2. Event

An event is what we are typically interested in measuring the probability of.

* Simple Event: An event consisting of a single sample point.
* Compound Event: An event consisting of more than one sample point.
* Sure Event: The entire sample space $S$. This event is certain to occur.
* Impossible Event: The empty set $\emptyset$. This event can never occur.

3. Algebra of Events

Since events are sets, we can apply the operations of set theory to them. This "algebra of events" is crucial for manipulating and calculating probabilities.

* Union ($A \cup B$): Represents the occurrence of "at least one of the events $A$ or $B$".
* Intersection ($A \cap B$): Represents the "simultaneous occurrence of both events $A$ and $B$".
* Complement ($A'$ or $A^c$): Represents the "non-occurrence of event $A$".
* Mutually Exclusive Events: Two events $A$ and $B$ are mutually exclusive (or disjoint) if they cannot occur at the same time. Mathematically, this means their intersection is the empty set, $A \cap B = \emptyset$.

A ∩ B ≠ ∅


A
B
Not Mutually Exclusive

A ∩ B = ∅


A
B
Mutually Exclusive

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Problem-Solving Strategies

A common source of error in probability is an incorrectly defined sample space. A systematic approach is essential.

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="An experiment consists of tossing a fair coin four times. What is the total number of sample points in the sample space?" options=["8", "12", "16", "24"] answer="16" hint="Each toss has 2 possible outcomes. Use the multiplication principle for the sequence of four tosses." solution="
Step 1: Identify the number of outcomes for a single toss.
For one coin toss, there are 2 outcomes: Heads (H) or Tails (T).

Step 2: Apply the multiplication principle for a sequence of independent trials.
Since the coin is tossed four times, the total number of possible outcomes is the product of the number of outcomes at each step.

Step 3: Calculate the final value.

Result:
The sample space contains 16 sample points.
Answer: \boxed{16}
"
:::

:::question type="NAT" question="Two distinct six-sided dice are rolled. Let event A be that the sum of the numbers is a prime number. How many outcomes are in event A?" answer="15" hint="List all 36 outcomes in a grid. Identify the prime sums (2, 3, 5, 7, 11) and count the pairs that produce them." solution="
Step 1: Define the sample space $S$. The total number of outcomes is $|S|=6 \times 6 = 36$.

Step 2: Identify the possible prime sums. The minimum sum is $1+1=2$ and the maximum sum is $6+6=12$. The prime numbers in this range are 2, 3, 5, 7, and 11.

Step 3: List the outcomes for each prime sum.

• Sum = 2: $\{(1,1)\}$ (1 outcome)


• Sum = 3: $\{(1,2), (2,1)\}$ (2 outcomes)


• Sum = 5: $\{(1,4), (2,3), (3,2), (4,1)\}$ (4 outcomes)


• Sum = 7: $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$ (6 outcomes)


• Sum = 11: $\{(5,6), (6,5)\}$ (2 outcomes)

Step 4: Sum the number of outcomes for event A.
The total number of outcomes in event A is the sum of the counts from Step 3.

Result:
There are 15 outcomes in event A.
Answer: \boxed{15}
"
:::

:::question type="MSQ" question="From a standard deck of 52 playing cards, one card is drawn at random. Let event A be 'the card is a King' and event B be 'the card is a Spade'. Let event C be 'the card is a Heart'. Which of the following statements is/are true?" options=["Events A and B are mutually exclusive.", "Events B and C are mutually exclusive.", "Events A and C are not mutually exclusive.", "The sample space size is 52."] answer="Events B and C are mutually exclusive.,Events A and C are not mutually exclusive.,The sample space size is 52." hint="Check the intersection of each pair of events. A card can be a King and a Spade (King of Spades), but a card cannot be both a Spade and a Heart." solution="

Events A and B: Event A is drawing a King. Event B is drawing a Spade. The intersection $A \cap B$ is the event 'the card is the King of Spades'. Since $A \cap B \neq \emptyset$, events A and B are not mutually exclusive. So, the first option is false.

Events B and C: Event B is drawing a Spade. Event C is drawing a Heart. A single card cannot be both a Spade and a Heart, as these are distinct suits. Therefore, $B \cap C = \emptyset$. Events B and C are mutually exclusive. So, the second option is true.

Events A and C: Event A is drawing a King. Event C is drawing a Heart. The intersection $A \cap C$ is the event 'the card is the King of Hearts'. Since $A \cap C \neq \emptyset$, events A and C are not mutually exclusive. So, the third option is true.

Sample Space Size: A standard deck has 52 unique cards. Therefore, the size of the sample space for drawing one card is indeed 52. So, the fourth option is true.

Answer: \boxed{Events B and C are mutually exclusive., Events A and C are not mutually exclusive., The sample space size is 52.}
"
:::

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Summary

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What's Next?

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Part 2: Counting Principles

Introduction

In the study of probability, our primary objective is often to determine the likelihood of an event. This fundamentally requires us to quantify the number of possible outcomes, both for the event in question and for the entire experiment. The field of mathematics concerned with counting is known as combinatorics. While counting may seem elementary, the complexity of scenarios encountered in data analysis and computer science necessitates a systematic approach.

The foundational principles of counting—namely, the Addition and Multiplication Principles—provide the essential tools for enumerating outcomes in a structured manner. A mastery of these rules is not merely an academic exercise; it forms the bedrock upon which the entire edifice of discrete probability is built. These principles allow us to determine the size of sample spaces, which is the first and most critical step in calculating probabilities.

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Key Concepts

The two most fundamental rules in combinatorics are the Rule of Sum and the Rule of Product. Understanding when to apply each is crucial for correctly solving counting problems.

1. The Addition Principle (Rule of Sum)

The Addition Principle applies when we must make a choice between two or more mutually exclusive options. If an event can occur in one of several distinct ways, we sum the number of ways for each option.

Consider two tasks, $T_1$ and $T_2$. If task $T_1$ can be performed in $n_1$ ways and task $T_2$ can be performed in $n_2$ ways, and the two tasks cannot be performed simultaneously (they are disjoint), then the number of ways to perform either $T_1$ or $T_2$ is the sum of the number of ways for each task.

Set A
($n_1$ ways)
Set B
($n_2$ ways)
Total ways = $n_1 + n_2$

Since A and B are disjoint ($A \cap B = \emptyset$)

Worked Example:

Problem: A university library has 40 textbooks on Data Structures and 30 textbooks on Algorithms. A student wishes to borrow exactly one textbook. How many choices does the student have?

Solution:

Step 1: Identify the disjoint tasks.
The student can choose a Data Structures book OR an Algorithms book. These are mutually exclusive choices.
Let $T_1$ be the task of choosing a Data Structures book, and $T_2$ be the task of choosing an Algorithms book.

Step 2: Determine the number of ways for each task.
Number of ways for $T_1$ is $n_1 = 40$.
Number of ways for $T_2$ is $n_2 = 30$.

Step 3: Apply the Addition Principle.
The total number of choices is the sum of the number of choices for each task.

Step 4: Compute the final answer.

Answer: The student has 70 choices.

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2. The Multiplication Principle (Rule of Product)

The Multiplication Principle applies when a procedure or task is composed of a sequence of steps or stages. If the overall task is completed by performing a series of sub-tasks one after another, we multiply the number of ways to do each sub-task.

Let us consider a procedure that can be broken down into a sequence of two tasks, $T_1$ and $T_2$. If task $T_1$ can be performed in $n_1$ ways, and for each of these ways, task $T_2$ can be performed in $n_2$ ways, then the total number of ways to perform the entire procedure is the product of the number of ways for each task.

Worked Example:

Problem: A user is creating a password that must consist of one uppercase letter followed by two digits. How many different passwords can be created? (Assume the English alphabet has 26 letters and digits are 0-9).

Solution:

Step 1: Decompose the procedure into a sequence of steps.
The procedure of creating a password has three steps:

Choose the first character (an uppercase letter).

Choose the second character (a digit).

Choose the third character (a digit).

Step 2: Determine the number of ways for each step.

• Number of ways to choose the first character (A-Z): $n_1 = 26$.


• Number of ways to choose the second character (0-9): $n_2 = 10$.


• Number of ways to choose the third character (0-9): $n_3 = 10$.

Step 3: Apply the Multiplication Principle.
The total number of possible passwords is the product of the number of options at each step.

Step 4: Compute the final answer.

Answer: There are 2600 different possible passwords.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="A restaurant offers 5 distinct appetizers and 8 distinct main courses. A customer wants to order either an appetizer or a main course, but not both. How many different choices does the customer have?" options=["13", "40", "3", "35"] answer="13" hint="The customer is choosing one item from two separate, mutually exclusive categories. Consider the keyword 'or'." solution="
Step 1: Identify the tasks.
The customer can either choose an appetizer OR a main course. These are disjoint choices.

• Task 1: Choose an appetizer. There are $n_1 = 5$ ways.


• Task 2: Choose a main course. There are $n_2 = 8$ ways.

Step 2: Apply the Addition Principle.
Since the choices are mutually exclusive (the customer orders one or the other), we add the number of options.

Result: The customer has 13 different choices.
Answer: $\boxed{13}$
"
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:::question type="NAT" question="A 4-digit PIN is to be formed using the digits {0, 1, 2, 3, 5, 8}. If repetition of digits is allowed, how many different PINs can be formed?" answer="1296" hint="Forming a PIN is a 4-step process. Determine the number of options available for each of the four positions." solution="
Step 1: Decompose the task.
The task is to form a 4-digit PIN. This can be broken down into 4 sequential steps: choosing the first digit, choosing the second, choosing the third, and choosing the fourth.

Step 2: Count the options for each step.
The set of available digits is {0, 1, 2, 3, 5, 8}, which contains 6 distinct digits.

• Options for the 1st digit: 6


• Options for the 2nd digit: 6 (since repetition is allowed)


• Options for the 3rd digit: 6 (since repetition is allowed)


• Options for the 4th digit: 6 (since repetition is allowed)

Step 3: Apply the Multiplication Principle.
The total number of PINs is the product of the number of options for each position.

Step 4: Calculate the final value.

Result: A total of 1296 different PINs can be formed.
Answer: $\boxed{1296}$
"
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:::question type="MSQ" question="A committee of two is to be formed from a group of 3 computer science (CS) students and 4 electrical engineering (EE) students. The committee must consist of students from different departments. Which of the following statements is/are correct?" options=["The total number of ways to form the committee is 12.", "The problem can be solved by choosing one CS student AND one EE student.", "If the students had to be from the same department, the total number of ways would be 9.", "The problem can be solved by calculating (3+4) * (3+4-1) / 2."] answer="The total number of ways to form the committee is 12.,The problem can be solved by choosing one CS student AND one EE student." hint="The committee requires one student from each department. This is a sequence of choices. Also, evaluate the alternative scenario in the options." solution="
Solution:
The problem requires forming a committee of two students from different departments. This means we must select exactly one student from the CS group and one student from the EE group.

• Statement A: We analyze this using the Multiplication Principle.

- Step 1: Select one CS student from the available 3. This can be done in 3 ways. - Step 2: Select one EE student from the available 4. This can be done in 4 ways. - Total ways = (Ways to choose CS student) $\times$ (Ways to choose EE student) = $3 \times 4 = 12$. Thus, statement A is correct.

• Statement B: This statement accurately describes the logical procedure required to solve the problem: choosing one student from the first group AND one from the second. This aligns perfectly with the Multiplication Principle. Thus, statement B is correct.

• Statement C: This statement describes a different scenario where students must be from the same department. While the calculation for that scenario ($\binom{3}{2} + \binom{4}{2} = 3 + 6 = 9$) is correct, it does not pertain to the original problem statement. Therefore, it is not a correct statement about the given problem.

• Statement D: This calculation, $\frac{(3+4)(3+4-1)}{2} = \frac{7 \times 6}{2} = 21$, represents the total number of ways to choose any 2 students from the total of 7, without any restrictions. This is not what the problem asks for. Thus, statement D is incorrect.

The correct statements that pertain to the given problem are A and B. Answer: $\boxed{\text{Statements A and B}}$ " :::

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Summary

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What's Next?

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Part 3: Probability Axioms

Introduction

In our preliminary study of probability, we often rely on an intuitive understanding based on relative frequencies or symmetry. However, for a rigorous and consistent framework, we must establish a formal mathematical foundation. This is achieved through an axiomatic approach, which defines probability not by what it is, but by the rules it must obey.

The axiomatic approach, formulated by Andrey Kolmogorov, provides a set of fundamental rules, or axioms, from which all other properties of probability can be logically derived. This framework is essential for handling complex problems in data analysis, machine learning, and other advanced fields. For the GATE examination, a firm grasp of these axioms and their immediate consequences is critical for building a robust understanding of probability theory. We shall now explore these foundational principles.

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The Three Axioms of Probability

The entire structure of probability theory rests upon three deceptively simple axioms. Let $(\Omega, \mathcal{F}, P)$ be a probability space. The probability measure $P$ must satisfy the following rules for any event $A \in \mathcal{F}$.

1. Axiom 1: Non-negativity

The probability of any event is a non-negative real number.

2. Axiom 2: Normalization

The probability of the entire sample space is equal to 1.

3. Axiom 3: Additivity

For any countable collection of pairwise disjoint (mutually exclusive) events $A_1, A_2, \dots$, the probability of their union is the sum of their individual probabilities.

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Key Consequences of the Axioms

From these three axioms, we can derive several fundamental properties of probability. These are not new axioms but are logical consequences that are frequently used in problem-solving.

1. Probability of the Impossible Event

The probability of the empty set, $\emptyset$, which represents an impossible event, is zero.

Derivation:

Step 1: Consider the sample space $\Omega$ and the empty set $\emptyset$. These two events are mutually exclusive, as $\Omega \cap \emptyset = \emptyset$.

Step 2: We can write the union as $\Omega \cup \emptyset = \Omega$.

Step 3: Apply Axiom 3 (Additivity) to this union.

Step 4: Since $\Omega \cup \emptyset = \Omega$, we have $P(\Omega \cup \emptyset) = P(\Omega)$.

Step 5: By Axiom 2, $P(\Omega) = 1$.

Step 6: Solving for $P(\emptyset)$, we find:

2. Probability of the Complement

For any event $A$, the probability of its complement, $A^c$, is $1$ minus the probability of $A$.

3. The General Addition Rule

For any two events $A$ and $B$ (not necessarily disjoint), the probability of their union is given by the sum of their probabilities minus the probability of their intersection.

$\Omega$


A
B





$A \cap B$
When adding P(A) and P(B), the intersection P(A ∩ B) is counted twice.

Worked Example:

Problem: Let $P(A) = 0.5$, $P(B) = 0.4$, and $P(A \cap B) = 0.2$. Find the probability that either event A or event B occurs.

Solution:

Step 1: We are asked to find $P(A \cup B)$. The events are not mutually exclusive since $P(A \cap B) \ne 0$.

Step 2: Apply the General Addition Rule.

Step 3: Substitute the given values into the formula.

Step 4: Compute the final result.

Answer: The probability that either A or B occurs is $0.7$.

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Let $A$ and $B$ be two events in a sample space $\Omega$. If $P(A) = 0.6$ and $P(A \cup B) = 0.8$, and $A$ and $B$ are mutually exclusive, what is the value of $P(B)$?" options=["0.2","0.4","1.4","Cannot be determined"] answer="0.2" hint="Recall the addition rule for mutually exclusive events." solution="For mutually exclusive events, $A \cap B = \emptyset$. The additivity axiom simplifies to $P(A \cup B) = P(A) + P(B)$. We are given $P(A) = 0.6$ and $P(A \cup B) = 0.8$.

"
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:::question type="NAT" question="For two events $A$ and $B$, it is known that $P(A) = 0.7$ and $P(B^c) = 0.4$. If $P(A \cup B) = 0.8$, calculate the value of $P(A \cap B)$." answer="0.5" hint="First, find $P(B)$ from its complement. Then, use the general addition rule." solution="Step 1: Find $P(B)$ using the complement rule.

Step 2: Use the general addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.


Step 3: Solve for $P(A \cap B)$.

"
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:::question type="MSQ" question="Which of the following statements are direct consequences of the three axioms of probability for any events $A$ and $B$ from a sample space $\Omega$?" options=["If $A \subseteq B$, then $P(A) \le P(B)$","P($A \cap B$) = $P(A)P(B)$","P($A \cup A^c$) = 1","P($\emptyset$) = 0"] answer="If $A \subseteq B$, then $P(A) \le P(B)$,P($A \cup A^c$) = 1,P($\emptyset$) = 0" hint="Evaluate each statement based on the axioms. Note that one statement relates to independence, which is a separate definition, not an axiom." solution="

• If $A \subseteq B$, then $P(A) \le P(B)$: This is a correct consequence. We can write $B = A \cup (B \cap A^c)$, where $A$ and $(B \cap A^c)$ are disjoint. By Axiom 3, $P(B) = P(A) + P(B \cap A^c)$. By Axiom 1, $P(B \cap A^c) \ge 0$, so $P(B) \ge P(A)$.


• P($A \cap B$) = $P(A)P(B)$: This is the definition of statistical independence, not a direct consequence of the axioms for any two events. It only holds if the events are independent.


• P($A \cup A^c$) = 1: The union of an event and its complement is the entire sample space, $A \cup A^c = \Omega$. By Axiom 2, $P(\Omega) = 1$. So, $P(A \cup A^c) = 1$. This is correct.


• P($\emptyset$) = 0: As derived earlier, this is a direct consequence of Axioms 2 and 3. This is correct.

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Summary

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What's Next?

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="A committee of 4 is to be selected from a group of 6 computer science engineers and 5 electronics engineers. If the selection is made randomly, what is the probability that the committee consists of exactly 2 computer science engineers?" options=["5/11", "4/11", "1/3", "15/330"] answer="A" hint="The total number of ways to form the committee is the size of the sample space. The numerator is the number of ways to select 2 from each discipline." solution="
The problem requires us to find the probability of a specific composition for a committee selected from a larger group. This is a classic application of combinations, as the order of selection does not matter.

Step 1: Determine the size of the sample space, $|\Omega|$.
The sample space consists of all possible 4-member committees that can be formed from the total group of $6+5=11$ engineers. The number of ways to choose 4 people from 11 is given by the combination formula $C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step 2: Determine the size of the event space, $|E|$.
The event $E$ is the selection of a committee with exactly 2 computer science (CS) engineers. Since the committee must have 4 members, this implies we must also select exactly 2 electronics (EC) engineers.

• The number of ways to choose 2 CS engineers from 6 is:

• The number of ways to choose 2 EC engineers from 5 is:

By the multiplication principle, the total number of ways to form the desired committee is the product of these two values:

Step 3: Calculate the probability.
Using the formula for equally likely outcomes, $P(E) = \frac{|E|}{|\Omega|}$:

Thus, the correct option is A.
"
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:::question type="NAT" question="How many 5-digit numbers can be formed using the digits {0, 1, 2, 3, 4} without repetition, such that the resulting number is divisible by 2?" answer="60" hint="A number is divisible by 2 if its last digit is even. Consider the case where the last digit is 0 separately from the cases where it is 2 or 4, due to the restriction on the first digit." solution="
To solve this, we must count the number of valid arrangements (permutations) subject to two constraints: (1) it must be a 5-digit number (so the first digit cannot be 0), and (2) it must be divisible by 2 (the last digit must be even). The available even digits are {0, 2, 4}.

We use the addition principle by breaking the problem into disjoint cases based on the last digit.

Case 1: The last digit is 0.
Let the 5-digit number be represented by five slots: _ _ _ _ _.

• The 5th slot (units place) is fixed as 0. (1 way)


• The 1st slot can be filled by any of the remaining 4 digits {1, 2, 3, 4}. (4 ways)


• The 2nd slot can be filled by any of the remaining 3 digits. (3 ways)


• The 3rd slot can be filled by any of the remaining 2 digits. (2 ways)


• The 4th slot can be filled by the last remaining digit. (1 way)

Total numbers in this case = $4 \times 3 \times 2 \times 1 \times 1 = 24$.

Case 2: The last digit is 2.

• The 5th slot is fixed as 2. (1 way)


• The 1st slot cannot be 0 and cannot be 2. Thus, it can be filled by any of the digits {1, 3, 4}. (3 ways)


• The 2nd slot can now be filled by any of the remaining digits, including 0. We have used two digits, so there are 3 remaining. (3 ways)


• The 3rd slot can be filled by any of the remaining 2 digits. (2 ways)


• The 4th slot can be filled by the last remaining digit. (1 way)

Total numbers in this case = $3 \times 3 \times 2 \times 1 \times 1 = 18$.

Case 3: The last digit is 4.
This case is symmetric to Case 2.

• The 5th slot is fixed as 4. (1 way)


• The 1st slot cannot be 0 and cannot be 4. It can be filled by any of {1, 2, 3}. (3 ways)


• The remaining slots can be filled in $3 \times 2 \times 1$ ways.

Total numbers in this case = $3 \times 3 \times 2 \times 1 \times 1 = 18$.

Final Calculation:
By the addition principle, the total number of such 5-digit numbers is the sum of the counts from the disjoint cases.
Total = (Numbers ending in 0) + (Numbers ending in 2) + (Numbers ending in 4)
Total = $24 + 18 + 18 = 60$.
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:::question type="MCQ" question="For two events A and B in a sample space $\Omega$, we are given $P(A \cup B) = 0.7$, $P(A) = 0.4$, and $P(B) = 0.5$. What is the value of $P(A^c \cup B^c)$?" options=["0.2", "0.3", "0.8", "0.5"] answer="C" hint="Use De Morgan's laws to simplify the expression $P(A^c \cup B^c)$. You will first need to find $P(A \cap B)$ using the inclusion-exclusion principle." solution="
The problem asks for the probability of the union of two complementary events. We can solve this by first finding the probability of the intersection of the original events and then applying De Morgan's laws.

Step 1: Find $P(A \cap B)$ using the Principle of Inclusion-Exclusion.
The formula for the union of two events is:

We are given the values for $P(A \cup B)$, $P(A)$, and $P(B)$. We can rearrange the formula to solve for $P(A \cap B)$:

Substituting the given values:

Step 2: Simplify the target expression using De Morgan's Laws.
De Morgan's laws state that for any two sets (or events) A and B:

• $(A \cup B)^c = A^c \cap B^c$


• $(A \cap B)^c = A^c \cup B^c$

The expression we need to evaluate is $P(A^c \cup B^c)$. Using the second law, we can rewrite this as:

Step 3: Calculate the final probability using the complement rule.
The probability of the complement of an event $E$ is $P(E^c) = 1 - P(E)$. Applying this to our simplified expression:

We found $P(A \cap B) = 0.2$ in Step 1. Therefore:

Thus, the correct option is C.
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What's Next?