Fundamentals of Probability

Overview

Welcome to the foundational chapter in your journey through Probability and Statistics for the Masters in Data Science program. This chapter, "Fundamentals of Probability," is designed to equip you with the essential language and concepts necessary to quantify uncertainty. We will delve into the core definitions of probability, exploring how to define outcomes, events, and sample spaces, and how to calculate the likelihood of various occurrences. A robust grasp of these elementary principles is not just academic; it is the bedrock upon which all advanced statistical inference and machine learning algorithms are built.

For your CMI examinations, a precise understanding of elementary probability is non-negotiable. This chapter lays the groundwork for subsequent topics like conditional probability, random variables, and probability distributions, which are frequently tested and crucial for interpreting data-driven insights. Expect questions that assess your ability to correctly identify sample spaces, compute probabilities for simple and compound events, and apply basic probability rules with accuracy. Mastering these fundamentals ensures you can approach more complex problems with confidence and a solid conceptual framework.

Ultimately, probability is the mathematical framework for dealing with uncertainty, a constant companion in the world of data science. From predicting customer behavior to evaluating model performance, the ability to reason probabilistically is a core competency. This chapter will solidify your foundational knowledge, preparing you not only for exam success but also for the practical application of probabilistic thinking in real-world data science challenges.

Chapter Contents

| # | Topic | What You'll Learn |
|---|-------|-------------------|
| 1 | Elementary Probability | Define events, sample spaces, calculate probabilities. |

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Learning Objectives

❗ By the End of This Chapter

After studying this chapter, you will be able to:

Define fundamental probabilistic terms such as experiment, outcome, event, and sample space.

Calculate the probability of simple and compound events using various methods.

Apply basic probability rules, including the complement rule and the addition rule for mutually exclusive events.

Represent events and their relationships using set notation and Venn diagrams.

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Now let's begin with Elementary Probability...
## Part 1: Elementary Probability

Introduction

Probability theory is the mathematical framework for quantifying uncertainty. In the realm of Data Science, understanding probability is fundamental for statistical inference, machine learning algorithms, and decision-making under uncertainty. This unit covers the foundational concepts of probability, including sample spaces, events, axioms of probability, and various rules for calculating probabilities of simple and compound events. A strong grasp of these fundamentals is critical for CMI, as it forms the bedrock for more advanced topics like random variables, distributions, and stochastic processes. We will delve into both discrete and continuous probability spaces, with a particular focus on combinatorics for discrete spaces and geometric approaches for continuous ones, as these are frequently tested.

📖 Experiment, Sample Space, and Events

An experiment is any process that generates a well-defined outcome.
The sample space $S$ of an experiment is the set of all possible outcomes.
An event $E$ is any subset of the sample space $S$ .

The impossible event is the empty set $\emptyset$ .
The sure event is the entire sample space $S$ .

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Key Concepts

1. Sample Spaces and Events

The first step in any probability problem is to clearly define the sample space and the events of interest.

Types of Sample Spaces:
* Discrete Sample Space: A sample space that contains a finite or countably infinite number of outcomes.
Example:* Tossing a coin three times: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Example:* Number of cars passing a point until the first red car: $S = \{1, 2, 3, \ldots\}$
* Continuous Sample Space: A sample space that contains an uncountably infinite number of outcomes, typically represented by an interval or region in $\mathbb{R}^n$ .
Example:* The time a component functions before failing: $S = [0, \infty)$
Example:* The exact arrival time of a bus between 8:00 AM and 8:30 AM: $S = [8:00, 8:30]$

📖 Set Operations for Events

Given two events $A$ and $B$ from a sample space $S$ :

Union ( $A \cup B$ ): The event that $A$ occurs or $B$ occurs (or both).

Intersection ( $A \cap B$ ): The event that both $A$ and $B$ occur.

Complement ( $A^c$ or $\bar{A}$ ): The event that $A$ does not occur.

Mutually Exclusive Events: Events $A$ and $B$ are mutually exclusive if $A \cap B = \emptyset$ . They cannot occur simultaneously.

Symmetric Difference ( $A \Delta B$ ): The event that exactly one of $A$ or $B$ occurs.

A \Delta B = (A \cap B^c) \cup (A^c \cap B)

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2. Axioms of Probability

The probability $P(E)$ of an event $E$ must satisfy the following three axioms (Kolmogorov Axioms):

📐 Axioms of Probability

For any event $E$ , the probability of $E$ is non-negative:

P(E) \ge 0

The probability of the sample space $S$ (the sure event) is 1:

P(S) = 1

For any sequence of mutually exclusive events $E_1, E_2, \ldots$ (i.e., $E_i \cap E_j = \emptyset$ for $i \ne j$ ), the probability of their union is the sum of their individual probabilities:

P\left(\bigcup_{i=1}^{\infty} E_i\right) = \sum_{i=1}^{\infty} P(E_i)

For a finite number of mutually exclusive events $E_1, E_2, \ldots, E_n$ :

P\left(\bigcup_{i=1}^{n} E_i\right) = \sum_{i=1}^{n} P(E_i)

Application: These axioms form the basis for deriving all other probability rules.

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#
## 3. Basic Rules of Probability

From the axioms, several important rules can be derived.

Rule 1: Probability of the Complement
The probability that an event $E$ does not occur is $1$ minus the probability that it does occur.

📐 Complement Rule

P(E^c) = 1 - P(E)

Variables:

$E^c$ = complement of event $E$

$P(E)$ = probability of event $E$

When to use: Often useful for calculating the probability of "at least one" event.

Rule 2: Addition Rule (for two events)
The probability of the union of two events $A$ and $B$ is given by:

📐 General Addition Rule

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

Variables:

$P(A \cup B)$ = probability of $A$ or $B$ or both

$P(A)$ = probability of $A$

$P(B)$ = probability of $B$

$P(A \cap B)$ = probability of $A$ and $B$

When to use: When calculating the probability of at least one of two events occurring, especially if they are not mutually exclusive.

If $A$ and $B$ are mutually exclusive events ( $A \cap B = \emptyset$ ), then $P(A \cap B) = 0$ , and the addition rule simplifies to:

P(A \cup B) = P(A) + P(B)

Rule 3: Addition Rule (for three events)
For three events $A, B, C$ :

📐 Addition Rule for Three Events

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

When to use: When calculating the probability of at least one of three events occurring.

Worked Example: (Based on PYQ 2, 3, 10 concepts)

Problem: Let $A$ and $B$ be events such that $P(A) = 0.6$ , $P(B) = 0.4$ , and $P(A \cap B) = 0.2$ . Find $P(A \cup B)$ , $P(A^c)$ , and $P(A \Delta B)$ .

Solution:

Step 1: Calculate $P(A \cup B)$ using the general addition rule.

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

Step 2: Substitute the given values.

P(A \cup B) = 0.6 + 0.4 - 0.2

Step 3: Simplify to get the result.

P(A \cup B) = 0.8

Step 4: Calculate $P(A^c)$ using the complement rule.

P(A^c) = 1 - P(A)

Step 5: Substitute the given value.

P(A^c) = 1 - 0.6

Step 6: Simplify to get the result.

P(A^c) = 0.4

Step 7: Calculate $P(A \Delta B)$ . Recall $A \Delta B = (A \cap B^c) \cup (A^c \cap B)$ .
Also, $P(A \Delta B) = P(A \cup B) - P(A \cap B)$ .

P(A \Delta B) = P(A \cup B) - P(A \cap B)

Step 8: Substitute the calculated and given values.

P(A \Delta B) = 0.8 - 0.2

Step 9: Simplify to get the result.

P(A \Delta B) = 0.6

Answer: $P(A \cup B) = \boxed{0.8}$ , $P(A^c) = \boxed{0.4}$ , $P(A \Delta B) = \boxed{0.6}$ .

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#
## 4. Conditional Probability and Independence

Conditional Probability:
The probability of event $A$ occurring given that event $B$ has already occurred is called the conditional probability of $A$ given $B$ .

📐 Conditional Probability

P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad \text{provided } P(B) > 0

Variables:

$P(A|B)$ = probability of event $A$ given event $B$

$P(A \cap B)$ = probability of both $A$ and $B$ occurring

$P(B)$ = probability of event $B$

When to use: When the occurrence of one event affects the probability of another.

Multiplication Rule:
Rearranging the conditional probability formula gives the multiplication rule.

📐 Multiplication Rule

P(A \cap B) = P(B)P(A|B) = P(A)P(B|A)

When to use: To find the probability of both events occurring, especially in sequential processes or when events are dependent.

For multiple events, e.g., $P(A \cap B \cap C) = P(A)P(B|A)P(C|A \cap B)$ .

Independent Events:
Two events $A$ and $B$ are said to be independent if the occurrence of one does not affect the probability of the other.

📖 Independent Events

Events $A$ and $B$ are independent if any of the following equivalent conditions hold:

$P(A|B) = P(A)$ (provided $P(B) > 0$ )

$P(B|A) = P(B)$ (provided $P(A) > 0$ )

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

When to use: If events are known to be independent, their joint probability is simply the product of their individual probabilities. This is a common assumption in many problems (e.g., repeated trials, system reliability).

❗ Mutually Exclusive vs. Independent

Mutually Exclusive means $P(A \cap B) = 0$ . If $P(A)>0$ and $P(B)>0$ , then they cannot be independent.
Independent means $P(A \cap B) = P(A)P(B)$ . If $P(A)>0$ and $P(B)>0$ , then they cannot be mutually exclusive.

Unless one of the events has probability 0, mutually exclusive events are dependent, and independent events are not mutually exclusive.

Worked Example: (Based on PYQ 9, 12 concepts)

Problem: A machine produces masks, and there is a $0.001$ probability that a mask is not effective. What is the probability that among the first $100$ masks, at least one is not effective, assuming mask effectiveness is independent?

Solution:

Step 1: Define the event of interest. Let $E$ be the event that at least one mask is not effective.
It is easier to calculate the complement event $E^c$ , which is that all $100$ masks are effective.

Step 2: Let $P(\text{ineffective}) = 0.001$ .
Then $P(\text{effective}) = 1 - P(\text{ineffective}) = 1 - 0.001 = 0.999$ .

Step 3: Since the effectiveness of each mask is independent, the probability that all $100$ masks are effective is the product of their individual probabilities.

P(E^c) = P(\text{1st effective}) \times P(\text{2nd effective}) \times \ldots \times P(\text{100th effective})

Step 4: Calculate $P(E^c)$ .

P(E^c) = (0.999)^{100}

Step 5: Use the complement rule to find $P(E)$ .

P(E) = 1 - P(E^c)

Step 6: Substitute the value of $P(E^c)$ .

P(E) = 1 - (0.999)^{100}

Answer: The probability that at least one mask is not effective is $\boxed{1 - (0.999)^{100}}$ .

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#
## 5. Total Probability Theorem

The law of total probability is a fundamental rule relating marginal probabilities to conditional probabilities.

📐 Total Probability Theorem

If $B_1, B_2, \ldots, B_n$ are mutually exclusive and exhaustive events (i.e., they form a partition of the sample space $S$ , so $\bigcup_{i=1}^n B_i = S$ and $B_i \cap B_j = \emptyset$ for $i \ne j$ ), then for any event $A$ :

P(A) = \sum_{i=1}^n P(A|B_i)P(B_i)

When to use: When you want to find the overall probability of an event $A$ , but you know its probability under different conditions ( $B_i$ ).

Worked Example: (Based on PYQ 1 concepts)

Problem: In a survey, $60\%$ of respondents were asked Question 1 and $40\%$ were asked Question 2. Let $V$ be the event that a respondent has voted. Question 1 asks "Have you voted?" (YES if $V$ , NO if $V^c$ ). Question 2 asks "Is it true that you have never voted?" (YES if $V^c$ , NO if $V$ ). If $54\%$ of all respondents replied YES, what is the probability $P(V)$ ?

Solution:

Step 1: Define events and probabilities.
Let $Q_1$ be the event a respondent was asked Question 1, and $Q_2$ be the event a respondent was asked Question 2.
Given $P(Q_1) = 0.6$ and $P(Q_2) = 0.4$ .
Let $Y$ be the event a respondent replied YES.

Step 2: Express $P(Y)$ using the Law of Total Probability.
The events $Q_1$ and $Q_2$ form a partition of the sample space.

P(Y) = P(Y|Q_1)P(Q_1) + P(Y|Q_2)P(Q_2)

Step 3: Determine $P(Y|Q_1)$ and $P(Y|Q_2)$ .
If asked $Q_1$ ("Have you voted?"), a YES means the respondent voted. So, $P(Y|Q_1) = P(V)$ .
If asked $Q_2$ ("Is it true that you have never voted?"), a YES means the respondent has not voted. So, $P(Y|Q_2) = P(V^c) = 1 - P(V)$ .

Step 4: Substitute these into the total probability formula.
Given $P(Y) = 0.54$ .

0.54 = P(V) \times 0.6 + (1 - P(V)) \times 0.4

Step 5: Solve for $P(V)$ .

0.54 = 0.6 P(V) + 0.4 - 0.4 P(V)

0.54 - 0.4 = 0.6 P(V) - 0.4 P(V)

0.14 = 0.2 P(V)

P(V) = \frac{0.14}{0.2}

P(V) = 0.7

Answer: \boxed{0.7}

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#
## 6. Combinatorics in Probability

Many probability problems involve counting the number of possible outcomes or favorable outcomes. This often requires using permutations and combinations.

📖 Permutations and Combinations

Permutation: An ordered arrangement of items. The number of permutations of $n$ distinct items taken $k$ at a time is:

P(n, k) = \frac{n!}{(n-k)!}

Combination: An unordered selection of items. The number of combinations of $n$ distinct items taken $k$ at a time is:

\binom{n}{k} = \frac{n!}{k!(n-k)!}

💡 When to use Permutations vs. Combinations

Permutations are used when the order matters. Think "arrangements", "sequences", "positions".
Combinations are used when the order does NOT matter. Think "selections", "groups", "subsets".

Worked Example: (Based on PYQ 6, 7, 8, 16 concepts)

Problem: A bag contains $5$ red balls, $3$ blue balls, and $2$ green balls. If $4$ balls are selected randomly without replacement, what is the probability that exactly $2$ red balls and $2$ blue balls are selected?

Solution:

Step 1: Determine the total number of balls.
Total balls $N = 5 + 3 + 2 = 10$ .

Step 2: Calculate the total number of ways to select $4$ balls from $10$ . Since the order of selection doesn't matter, use combinations.

\text{Total outcomes} = \binom{10}{4}

\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210

Step 3: Calculate the number of ways to select $2$ red balls from $5$ .

\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10

Step 4: Calculate the number of ways to select $2$ blue balls from $3$ .

\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3

Step 5: Calculate the number of favorable outcomes (selecting $2$ red and $2$ blue). This is the product of the individual selections because these are independent choices from distinct groups.

\text{Favorable outcomes} = \binom{5}{2} \times \binom{3}{2}

\text{Favorable outcomes} = 10 \times 3 = 30

Step 6: Calculate the probability.

P(\text{2 red and 2 blue}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}

P(\text{2 red and 2 blue}) = \frac{30}{210}

P(\text{2 red and 2 blue}) = \frac{1}{7}

Answer: \boxed{1/7}

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#
## 7. Geometric Probability

Geometric probability deals with continuous sample spaces where outcomes correspond to points in a geometric region (e.g., length, area, volume). The probability of an event is then the ratio of the "measure" of the favorable region to the "measure" of the entire sample space.

📖 Geometric Probability

If a sample space $S$ is a region in $\mathbb{R}^n$ with measure $M(S)$ (length, area, volume), and an event $E$ is a sub-region of $S$ with measure $M(E)$ , then the probability of $E$ is:

P(E) = \frac{M(E)}{M(S)}

Provided that all points in

S

are equally likely.

Worked Example: The Meeting Problem (Based on PYQ 11, 14 concepts)

Problem: Two friends, Alice and Bob, agree to meet at a café between 12:00 PM and 1:00 PM. Each will wait for the other for 15 minutes, but not longer than 1:00 PM. What is the probability they will meet?

Solution:

Step 1: Define the sample space.
Let $x$ be Alice's arrival time and $y$ be Bob's arrival time, both measured in minutes past 12:00 PM. So $0 \le x \le 60$ and $0 \le y \le 60$ .
The sample space is a square in the $xy$ -plane with side length $60$ .
The measure of the sample space $M(S)$ is the area of this square.

M(S) = 60 \times 60 = 3600 \text{ square minutes}

Step 2: Define the event of interest (meeting condition).
Alice and Bob meet if their arrival times are within 15 minutes of each other.
This can be expressed as $|x - y| \le 15$ .

Step 3: Graph the sample space and the favorable region.

0
60
0
60

Alice's arrival (x)
Bob's arrival (y)

$y-x \le 15$
$x-y \le 15$
Meeting Region

The condition $|x - y| \le 15$ can be broken down into two inequalities:

x - y \le 15 \implies y \ge x - 15

y - x \le 15 \implies y \le x + 15

The favorable region is the area between the lines $y = x - 15$ and $y = x + 15$ , within the square $[0, 60] \times [0, 60]$ .

Step 4: Calculate the measure of the favorable region $M(E)$ .
It's easier to calculate the area of the non-favorable region (the two triangles) and subtract it from the total area.
The two triangles are formed by the lines $y = x + 15$ , $x=0$ , $y=60$ (top-left) and $y = x - 15$ , $y=0$ , $x=60$ (bottom-right).
The vertices of the top-left triangle are $(0, 15)$ , $(0, 60)$ , $(45, 60)$ . Its base is $45$ and height is $45$ .
The vertices of the bottom-right triangle are $(15, 0)$ , $(60, 0)$ , $(60, 45)$ . Its base is $45$ and height is $45$ .

\text{Area of one triangle} = \frac{1}{2} \times 45 \times 45 = \frac{2025}{2} = 1012.5

\text{Total non-favorable area} = 2 \times 1012.5 = 2025

Alternatively, the favorable region is the area of the square minus the two corner triangles.
The non-meeting region is composed of two right triangles.
One triangle has vertices $(0,0), (0,15), (15,0)$ . This is incorrect.
The non-meeting region is composed of two triangles.
The first triangle has vertices $(0,0)$ , $(0,15)$ , $(15,0)$ .

\text{Area} = \frac{1}{2} \times 15 \times 15 = \frac{225}{2}

This represents

y < x-15

and

x < y-15

combined which is incorrect.

Let's use the correct area calculation for the favorable region.
The favorable region is a hexagon, but it's simpler to calculate the area of the square minus the two corner triangles that represent not meeting.
The condition for not meeting is $|x-y| > 15$ , which means $y > x+15$ or $y < x-15$ .
The region $y > x+15$ within the square is a triangle with vertices $(0,15), (0,60), (45,60)$ . Base = 45, Height = 45.

\text{Area} = \frac{1}{2} \times 45 \times 45 = 1012.5

The region

y < x-15

within the square is a triangle with vertices

(15,0), (60,0), (60,45)

. Base = 45, Height = 45.

\text{Area} = \frac{1}{2} \times 45 \times 45 = 1012.5

\text{Total non-favorable area} = 1012.5 + 1012.5 = 2025

M(E) = M(S) - \text{Non-favorable area} = 3600 - 2025 = 1575

Step 5: Calculate the probability.

P(\text{meet}) = \frac{M(E)}{M(S)} = \frac{1575}{3600}

Step 6: Simplify the fraction.

\frac{1575}{3600} = \frac{315}{720} = \frac{63}{144} = \frac{7}{16}

Answer: The probability they will meet is $7/16$ .

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Problem-Solving Strategies

💡 CMI Strategy: Complement Rule for 'At Least One'

When a question asks for the probability of "at least one" event occurring (e.g., at least one defect, at least one success), it is often much simpler to calculate the probability of the complement event (i.e., none of the events occur) and subtract it from 1. This is especially true when dealing with independent trials.

P(\text{at least one}) = 1 - P(\text{none})

💡 CMI Strategy: Visualizing Continuous Probability

For problems involving continuous random variables (especially two of them, like arrival times), sketching the sample space and the favorable region on a 2D coordinate plane (e.g., a square or rectangle) can simplify the problem significantly. The probability then becomes a ratio of areas.

💡 CMI Strategy: Breaking Down Complex Events

For complex scenarios involving sequences of selections or multiple conditions, break the problem into smaller, manageable events. Use the multiplication rule for sequences of dependent events and the addition rule for unions. For counting, decide carefully if order matters (permutations) or not (combinations).

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Common Mistakes

⚠️ Avoid These Errors

❌ Confusing Mutually Exclusive with Independent:

Mutually exclusive events (

A \cap B = \emptyset

) cannot happen at the same time. If

A

happens,

B

cannot. If

P(A)>0

and

P(B)>0

, they are dependent. ✅ Correct: Independent events (

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

) imply that the occurrence of one does not affect the other. They are distinct concepts. Only if

P(A)=0

P(B)=0

can events be both mutually exclusive and independent.

❌ Incorrectly Applying Addition Rule: Using $P(A \cup B) = P(A) + P(B)$ when events are NOT mutually exclusive.

✅ Correct: Always use

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

. The simpler form is a special case.

❌ Misusing Combinations vs. Permutations: Using combinations when order matters, or permutations when order doesn't matter.

✅ Correct: If the problem asks for "arrangements," "sequences," or "positions," use permutations. If it asks for "selections," "groups," or "subsets," use combinations. Read the problem carefully for clues about order.

❌ Overlapping Regions in Geometric Probability: Incorrectly calculating the area of the favorable region by double-counting or missing parts.

✅ Correct: Carefully define the inequalities that describe the event. Sketch the region and consider using the complement rule (total area minus unfavorable area) if it simplifies the calculation.

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Practice Questions

:::question type="MCQ" question="Two fair six-sided dice are rolled. What is the probability that the sum of the numbers rolled is 7, given that the first die shows an odd number?" options=[" $1/6$ "," $1/3$ "," $1/2$ "," $2/3$ "] answer=" $1/6$ " hint="Define the sample space, the event of interest, and the given condition. Then apply the formula for conditional probability." solution="Step 1: Define the sample space $S$ .
The total number of outcomes when rolling two fair six-sided dice is $6 \times 6 = 36$ . Each outcome $(d_1, d_2)$ is equally likely.

Step 2: Define event $A$ : The sum of the numbers rolled is 7.
The outcomes for $A$ are: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ .
So, $P(A) = 6/36 = 1/6$ .

Step 3: Define event $B$ : The first die shows an odd number.
The odd numbers are $\{1, 3, 5\}$ .
The outcomes for $B$ are:
$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$
$(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$
$(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)$
So, there are $3 \times 6 = 18$ outcomes for $B$ .
$P(B) = 18/36 = 1/2$ .

Step 4: Define the intersection event $A \cap B$ : The sum is 7 AND the first die is odd.
The outcomes for $A \cap B$ from the list in Step 2: $(1,6), (3,4), (5,2)$ .
So, there are $3$ outcomes for $A \cap B$ .
$P(A \cap B) = 3/36 = 1/12$ .

Step 5: Apply the conditional probability formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ .

P(A|B) = \frac{1/12}{1/2}

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

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

Answer:

\boxed{1/6}

" :::

:::question type="NAT" question="A box contains 8 red balls and 4 blue balls. If 3 balls are drawn randomly without replacement, what is the probability that at least one blue ball is drawn? Express your answer as a decimal rounded to three significant figures." answer="0.745" hint="Consider using the complement rule to simplify the calculation." solution="Step 1: Define the event of interest.
Let $E$ be the event that at least one blue ball is drawn.
The complement event $E^c$ is that no blue balls are drawn, meaning all 3 balls drawn are red.

Step 2: Calculate the total number of ways to draw 3 balls from 12.
Total balls = $8 \text{ (red)} + 4 \text{ (blue)} = 12$ .

\text{Total outcomes} = \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220

Step 3: Calculate the number of ways to draw 3 red balls from 8 red balls (event $E^c$ ).

\text{Favorable outcomes for } E^c = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56

Step 4: Calculate the probability of $E^c$ .

P(E^c) = \frac{\text{Number of ways to draw 3 red balls}}{\text{Total number of ways to draw 3 balls}} = \frac{56}{220}

Step 5: Simplify $P(E^c)$ .

P(E^c) = \frac{14}{55}

Step 6: Use the complement rule to find $P(E)$ .

P(E) = 1 - P(E^c) = 1 - \frac{14}{55}

P(E) = \frac{55 - 14}{55} = \frac{41}{55}

Step 7: Convert to decimal and round to three significant figures.

P(E) \approx 0.74545...

Rounded to three significant figures, this is

0.745

.
Answer:

\boxed{0.745}

"
:::

:::question type="MSQ" question="Let $E$ and $F$ be two events such that $P(E) = 0.5$ , $P(F) = 0.6$ , and $P(E \cup F) = 0.8$ . Which of the following statements is/are true?" options=[" $P(E \cap F) = 0.3$ ","Events $E$ and $F$ are independent.","Events $E$ and $F$ are mutually exclusive."," $P(E^c \cap F^c) = 0.2$ "] answer="A,B,D" hint="Use the general addition rule to find $P(E \cap F)$ . Then check for independence and mutual exclusivity. Use De Morgan's laws for the complement of the union." solution="Step 1: Calculate $P(E \cap F)$ using the general addition rule.

P(E \cup F) = P(E) + P(F) - P(E \cap F)

0.8 = 0.5 + 0.6 - P(E \cap F)

0.8 = 1.1 - P(E \cap F)

P(E \cap F) = 1.1 - 0.8 = 0.3

So, option "

P(E \cap F) = 0.3

" is TRUE.

Step 2: Check for independence.
For $E$ and $F$ to be independent, $P(E \cap F)$ must equal $P(E)P(F)$ .

P(E)P(F) = 0.5 \times 0.6 = 0.3

Since

P(E \cap F) = 0.3

and

P(E)P(F) = 0.3

, the events

E

and

F

are independent.
So, option "Events

E

and

F

are independent." is TRUE.

Step 3: Check for mutual exclusivity.
For $E$ and $F$ to be mutually exclusive, $P(E \cap F)$ must be $0$ .
Since $P(E \cap F) = 0.3 \ne 0$ , the events $E$ and $F$ are not mutually exclusive.
So, option "Events $E$ and $F$ are mutually exclusive." is FALSE.

Step 4: Calculate $P(E^c \cap F^c)$ .
Using De Morgan's Law, $E^c \cap F^c = (E \cup F)^c$ .

P(E^c \cap F^c) = P((E \cup F)^c) = 1 - P(E \cup F)

P(E^c \cap F^c) = 1 - 0.8 = 0.2

So, option "

P(E^c \cap F^c) = 0.2

" is TRUE.
Answer:

\boxed{\text{A, B, D}}

"
:::

:::question type="SUB" question="Three points $X, Y, Z$ are chosen independently and uniformly at random on the circumference of a circle. What is the probability that these three points lie on a semicircle?" answer=" $3/4$ " hint="Fix one point and consider the positions of the other two relative to it. Alternatively, use a symmetry argument or consider the arcs formed." solution="Step 1: Understand the condition "lie on a semicircle".
Three points $X, Y, Z$ on a circle lie on a semicircle if and only if the triangle formed by these points does not contain the center of the circle in its interior. These two events are complements of each other.

Step 2: Recall or derive the probability that the triangle formed by three random points on a circle contains the center.
This is a classic geometric probability problem. Let the circumference be $C$ .
A common argument (often proved using order statistics or integration) shows that the probability for the center to be inside the triangle is $1/4$ .

Step 3: Apply the complement rule.
Let $E$ be the event that the three points lie on a semicircle.
Let $F$ be the event that the triangle formed by the three points contains the center $O$ .
We know $P(F) = 1/4$ .
The events $E$ and $F$ are complementary. If the points lie on a semicircle, the center is not in the triangle. If the points do not lie on a semicircle, the center must be in the triangle.
Therefore, $P(E) = 1 - P(F)$ .

Step 4: Calculate the probability.

P(E) = 1 - \frac{1}{4} = \frac{3}{4}

Answer:

\boxed{3/4}

"
:::

:::question type="SUB" question="A company offers 10 different types of mobile phones. A customer randomly selects 3 different phones to compare. What is the probability that at least one of the selected phones is a specific model (e.g., 'Model X'), which is one of the 10 types?" answer=" $3/10$ " hint="Use the complement rule. Calculate the probability that 'Model X' is NOT selected." solution="Step 1: Define the sample space.
The total number of ways to select 3 different phones from 10 is given by combinations, as the order of selection doesn't matter.

\text{Total outcomes} = \binom{10}{3}

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120

Step 2: Define the event of interest and its complement.
Let $E$ be the event that at least one of the selected phones is 'Model X'.
The complement event $E^c$ is that 'Model X' is not selected.

Step 3: Calculate the number of ways to select 3 phones without 'Model X'.
If 'Model X' is not selected, then the 3 phones must be chosen from the remaining $10 - 1 = 9$ types.

\text{Favorable outcomes for } E^c = \binom{9}{3}

\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84

Step 4: Calculate the probability of $E^c$ .

P(E^c) = \frac{\text{Number of ways to select 3 phones without Model X}}{\text{Total number of ways to select 3 phones}}

P(E^c) = \frac{84}{120}

Step 5: Simplify $P(E^c)$ .

P(E^c) = \frac{42}{60} = \frac{21}{30} = \frac{7}{10}

Step 6: Use the complement rule to find $P(E)$ .

P(E) = 1 - P(E^c)

P(E) = 1 - \frac{7}{10}

P(E) = \frac{3}{10}

Answer:

\boxed{3/10}

"
:::

---

Summary

❗ Key Takeaways for CMI

Axioms and Basic Rules: Master the three axioms of probability and their direct implications: complement rule, addition rule (general and for mutually exclusive events), and multiplication rule.

Conditional Probability and Independence: Clearly distinguish between mutually exclusive events and independent events. Understand how to calculate conditional probabilities and use the multiplication rule for dependent events.

Combinatorics: Be proficient in using permutations (order matters) and combinations (order doesn't matter) for counting outcomes in discrete probability problems. This is crucial for problems involving selections without replacement.

Geometric Probability: For continuous sample spaces, especially those involving time intervals or points on a geometric figure, visualize the sample space and favorable region as areas or lengths. The probability is then a ratio of these measures.

Complement Rule for "At Least One": Always consider using the complement rule $P(E) = 1 - P(E^c)$ when the problem asks for the probability of "at least one" occurrence, as it often simplifies calculations significantly.

---

What's Next?

💡 Continue Learning

This topic connects to:

Random Variables and Probability Distributions: The concepts of sample spaces and events naturally lead into defining random variables and their associated probability distributions (e.g., binomial, Poisson, uniform, normal).

Expected Value and Variance: Understanding how to calculate probabilities is a prerequisite for computing expected values and variances of random variables, which are key measures in statistical analysis.

Bayesian Inference: The Total Probability Theorem and conditional probability are foundational for Bayes' Theorem, which is central to Bayesian statistics and machine learning.

Master these connections for comprehensive CMI preparation!

---

Chapter Summary

📖 Fundamentals of Probability - Key Takeaways

To excel in CMI, a solid grasp of probability fundamentals is non-negotiable. Ensure you thoroughly understand these core concepts:

Sample Space, Events, and Outcomes: The foundation of probability. Understand how to define the sample space ( $S$ ) for an experiment, identify elementary outcomes, and form events as subsets of $S$ .

Axioms of Probability: The three fundamental rules:

$P(E) \ge 0$ for any event $E$ .

P(S) = 1

For mutually exclusive events $E_1, E_2, \dots$ ,
$P(E_1 \cup E_2 \cup \dots) = P(E_1) + P(E_2) + \dots$
Union and Intersection of Events: Key formulas for combining probabilities:

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

A

B

P(A \cap B) = 0

P(A \cup B) = P(A) + P(B)

Conditional Probability: The probability of event $A$ occurring given that event $B$ has already occurred, denoted

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

P(B) > 0

Independence of Events: Events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other. Mathematically,

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

P(A|B) = P(A)

P(B)>0

Total Probability Theorem: A powerful tool to calculate the probability of an event $B$ by partitioning the sample space into mutually exclusive and exhaustive events $A_1, A_2, \dots, A_n$ :

P(B) = \sum_{i=1}^n P(B|A_i)P(A_i)

Bayes' Theorem: Used to update the probability of a hypothesis (event $A_i$ ) given new evidence (event $B$ ):

P(A_i|B) = \frac{P(B|A_i)P(A_i)}{P(B)} = \frac{P(B|A_i)P(A_i)}{\sum_{j=1}^n P(B|A_j)P(A_j)}

---

Chapter Review Questions

:::question type="MCQ" question="Let $A$ and $B$ be two events in a sample space $S$ . Given $P(A) = 0.6$ , $P(B) = 0.5$ , and $P(A|B) = 0.7$ . What is the probability of the event $A' \cap B$ (i.e., $B$ occurs but $A$ does not)?" options=[" $0.15$ " " $0.25$ " " $0.35$ " " $0.45$ "] answer=" $0.15$ " hint="Recall the definition of conditional probability and how to express $P(A' \cap B)$ in terms of $P(B)$ and $P(A \cap B)$ ." solution="We are given $P(A) = 0.6$ , $P(B) = 0.5$ , and $P(A|B) = 0.7$ .

First, we use the definition of conditional probability to find $P(A \cap B)$ :

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

0.7 = \frac{P(A \cap B)}{0.5}

P(A \cap B) = 0.7 \times 0.5 = 0.35

Next, we want to find $P(A' \cap B)$ . The event $B$ can be partitioned into two mutually exclusive events: $A \cap B$ (both $A$ and $B$ occur) and $A' \cap B$ (B occurs but $A$ does not). Therefore,

P(B) = P(A \cap B) + P(A' \cap B)

Rearranging this equation to solve for

P(A' \cap B)

P(A' \cap B) = P(B) - P(A \cap B)

Substitute the known values:

P(A' \cap B) = 0.5 - 0.35 = 0.15

Thus, the probability of $A' \cap B$ is $0.15$ .
Answer: $\boxed{0.15}$ "
:::

:::question type="NAT" question="A diagnostic test for a rare disease has the following characteristics: If a person has the disease, the test is positive 98% of the time (sensitivity). If a person does not have the disease, the test is negative 95% of the time (specificity). The disease affects 0.1% of the population. If a randomly selected person tests positive, what is the probability that they actually have the disease? Round your answer to two decimal places." answer="0.02" hint="This is a classic application of Bayes' Theorem. Define events for 'has disease', 'does not have disease', 'test positive', and 'test negative'." solution="Let $D$ be the event that a person has the disease, and $D'$ be the event that a person does not have the disease.
Let $T^+$ be the event that the test is positive, and $T^-$ be the event that the test is negative.

We are given:
* $P(D) = 0.001$ (0.1% of the population has the disease)
* $P(D') = 1 - P(D) = 1 - 0.001 = 0.999$
* $P(T^+|D) = 0.98$ (sensitivity: test is positive if person has the disease)
* $P(T^-|D') = 0.95$ (specificity: test is negative if person does not have the disease)

From $P(T^-|D') = 0.95$ , we can find $P(T^+|D')$ (false positive rate):

P(T^+|D') = 1 - P(T^-|D') = 1 - 0.95 = 0.05

We want to find $P(D|T^+)$ , the probability that a person has the disease given that they tested positive. We use Bayes' Theorem:

P(D|T^+) = \frac{P(T^+|D)P(D)}{P(T^+)}

First, we need to calculate

P(T^+)

using the Law of Total Probability:

P(T^+) = P(T^+|D)P(D) + P(T^+|D')P(D')

P(T^+) = (0.98)(0.001) + (0.05)(0.999)

P(T^+) = 0.00098 + 0.04995

P(T^+) = 0.05093

Now, substitute this into Bayes' Theorem:

P(D|T^+) = \frac{0.00098}{0.05093}

P(D|T^+) \approx 0.01924

Rounding to two decimal places, $P(D|T^+) = 0.02$ .
Answer: $\boxed{0.02}$ "
:::

:::question type="MCQ" question="A bag contains 5 red balls and 7 blue balls. If 4 balls are drawn at random without replacement, what is the probability that exactly 2 of them are red?" options=[" $\frac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}}$ " " $\frac{\binom{5}{2}}{\binom{12}{4}}$ " " $\frac{\binom{5}{2} + \binom{7}{2}}{\binom{12}{4}}$ " " $\frac{\binom{5}{2}\binom{7}{4}}{\binom{12}{4}}$ "] answer=" $\frac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}}$ " hint="This problem involves combinations. Determine the total number of ways to draw 4 balls and the number of ways to draw exactly 2 red and 2 blue balls." solution="The total number of balls in the bag is $5 \text{ (red)} + 7 \text{ (blue)} = 12$ balls.
We are drawing 4 balls at random without replacement.

The total number of ways to choose 4 balls from 12 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ :

\text{Total ways} = \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495

We want to find the number of ways to draw exactly 2 red balls. If 2 balls are red, then the remaining $4-2=2$ balls must be blue.
Number of ways to choose 2 red balls from 5 red balls:

\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10

Number of ways to choose 2 blue balls from 7 blue balls:

\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21

The number of ways to draw exactly 2 red balls and 2 blue balls is the product of these two combinations (by the multiplication principle):

\text{Favorable ways} = \binom{5}{2} \times \binom{7}{2} = 10 \times 21 = 210

The probability of drawing exactly 2 red balls is the ratio of favorable ways to total ways:

P(\text{exactly 2 red}) = \frac{\text{Favorable ways}}{\text{Total ways}} = \frac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}}

P(\text{exactly 2 red}) = \frac{210}{495}

This fraction can be simplified by dividing both numerator and denominator by 15:

P(\text{exactly 2 red}) = \frac{14}{33}

Answer:

\boxed{\frac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}}}

"
:::

:::question type="NAT" question="Let $A, B, C$ be three events in a sample space $S$ . Given the following probabilities:
$P(A) = 0.4$
$P(B) = 0.3$
$P(C) = 0.2$
$P(A \cap B) = 0.1$
$P(A \cap C) = 0.05$
$P(B \cap C) = 0.06$
$P(A \cap B \cap C) = 0.02$
Find the probability that none of the events $A, B,$ or $C$ occur. Round your answer to two decimal places." answer="0.29" hint="The event 'none of the events $A, B,$ or $C$ occur' is equivalent to $(A \cup B \cup C)'$ . Use the Principle of Inclusion-Exclusion for $P(A \cup B \cup C)$ ." solution="We want to find $P(A' \cap B' \cap C')$ . By De Morgan's Laws, this is equivalent to $P((A \cup B \cup C)')$ .
The probability of the complement of an event is $1$ minus the probability of the event:

P(A' \cap B' \cap C') = 1 - P(A \cup B \cup C)

Now, we need to calculate $P(A \cup B \cup C)$ using the Principle of Inclusion-Exclusion for three events:

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

Substitute the given values:

P(A \cup B \cup C) = 0.4 + 0.3 + 0.2 - 0.1 - 0.05 - 0.06 + 0.02

P(A \cup B \cup C) = (0.4 + 0.3 + 0.2) - (0.1 + 0.05 + 0.06) + 0.02

P(A \cup B \cup C) = 0.9 - 0.21 + 0.02

P(A \cup B \cup C) = 0.69 + 0.02

P(A \cup B \cup C) = 0.71

Finally, we can find the probability that none of the events occur:

P(A' \cap B' \cap C') = 1 - P(A \cup B \cup C) = 1 - 0.71 = 0.29

Answer:

\boxed{0.29}

"
:::

---

What's Next?

💡 Continue Your CMI Journey

You've just completed a crucial chapter: Fundamentals of Probability! This foundational knowledge is absolutely essential for almost every subsequent topic in probability and statistics for CMI.

Key connections:

* From Previous Learning: Your understanding of basic set theory (union, intersection, complement), logical reasoning, and elementary counting principles (permutations and combinations) from mathematics courses are directly applied here. This chapter provided the probabilistic framework for these concepts.

Building Blocks for Future Chapters: This chapter serves as the bedrock for more advanced topics. You will immediately build upon these concepts in:
Random Variables and Probability Distributions: The very definition of a random variable and the concept of its probability distribution (e.g., Bernoulli, Binomial, Poisson, Uniform, Normal) are direct extensions of defining events and assigning probabilities. Understanding discrete vs. continuous probability spaces begins here.
Expectation and Variance: Calculating the expected value and variance of random variables heavily relies on the probability values assigned to outcomes or events.
Joint Distributions and Conditional Expectations: When dealing with multiple random variables, concepts like joint probability, marginal probability, and conditional probability distributions are direct extensions of conditional probability for events.
Law of Large Numbers and Central Limit Theorem: These cornerstone theorems of statistics require a firm understanding of basic probability and the properties of random variables.
Statistical Inference (Estimation and Hypothesis Testing): All methods of statistical inference, from constructing confidence intervals to performing hypothesis tests, are built upon the theoretical framework of probability distributions and the likelihood of observing certain outcomes.

Mastering these fundamentals ensures you have the conceptual tools to tackle the more complex and abstract ideas that lie ahead in your CMI preparation. Keep practicing!

Fundamentals of Probability

Fundamentals of Probability

Overview

Chapter Contents

Learning Objectives

Introduction

Key Concepts

1. Sample Spaces and Events

2. Axioms of Probability

Problem-Solving Strategies

Common Mistakes

Practice Questions

Summary

What's Next?

Chapter Summary

Chapter Review Questions

What's Next?

🎯 Key Points to Remember

Related Topics in Probability and Statistics

Random Variables and Distributions

Data Interpretation and Summary Statistics

More Resources

Study Notes

Short Notes

Test Series

Mock Tests

Previous Year Papers

Chapter-wise PYQs

Chapter Practice

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