RD Sharma new

Finding the prob of numbers Cards - replaced and not replaced Dice - unbiased and biased

Class 11

Equally likely events

assume that all outcomes have equal chances to occur
Called Classical theory of probability

Random Experiments

tossing a coin
rolling a die
Conditions

more than one possible outcome
not possible to predict the outcome in advance

Outcome

possible results of an event

Sample Space

all possible outcomes of an event

Sample point

Each outcome of the random experiment
(ie.) each element of the sample space

Ordered pairs

possible pair of outcomes for a single event

Events

an event is associated with a subset of sample space

Occurrence of an event

the event E of a sample space S is said to have occurred if the outcome ω of the experiment is such that ω ∈ E. If the outcome ω is such that ω that the event E has not occurred.

Types of events

Impossible event - φ (null) is called an impossible event.
Sure event - S, i.e., the whole sample space is called the sure event.

Simple event

If an occasion has just one sample point of a sample space, it's called an easy (or elementary) event.

Compound events

If an occasion has quite one sample point, it's called a Compound event.

Complementary events - not A or A dash

A or B - A union B

A and B - A intersection B

A but not B - A - B

Mutually exclusive events

if the occurrence of any one of the events, excludes the occurrence of the other events
they cannot occur simultaneously
sets are disjoint

Exhaustive Events

Events are said to be exhaustive if, when you perform an experiment, at least one of them must necessarily occur.
A set of events $E_1, E_2, ..., E_n$ associated with a sample space $S$ are exhaustive if their union equals the total sample space. $E_1 \cup E_2 \cup ... \cup E_n = S$

Mutually Exclusive and Exhaustive System

Rolling a die:

$A = \{Odd numbers\} = \{1, 3, 5\}$
$B = \{Even numbers\} = \{2, 4, 6\}$
$A \cup B = S$ (Exhaustive)
$A \cap B = \phi$ (Mutually Exclusive)

Foundation of Probability - A N Kolmogorov, Russian Mathematician, in 1933 Axiomatic Approach of Probability

Class 12

📘 Chapter 13: Probability – Topic & Sub‑Topic Outline

1. Introduction

Revision of basic probability (from Class 11)
Random experiments, sample space, events

2. Conditional Probability

Definition of conditional probability
Multiplication theorem of probability
Independent events
Examples illustrating conditional probability

3. Bayes’ Theorem

Statement of Bayes’ theorem
Applications in real‑life problems
Examples using Bayes’ theorem

4. Random Variables & Probability Distribution

Definition of random variable
Probability distribution of a random variable
Mean (expected value) of a distribution
Variance and standard deviation

5. Bernoulli Trials & Binomial Distribution

Bernoulli trials definition
Binomial distribution formula
Mean and variance of binomial distribution
Examples of binomial distribution

📝 Grouping of Examples & Exercises by Method of Solving

Group A: Conditional Probability & Multiplication Rule

Examples: 13.1, 13.2, 13.3
Exercises: Q1–Q6 Method: Use definition P(A∣B)=P(A∩B)P(B), apply multiplication theorem.

Group B: Independent Events

Examples: 13.4, 13.5
Exercises: Q7–Q10 Method: Check independence using P(A∩B)=P(A)⋅P(B).

Group C: Bayes’ Theorem

Examples: 13.6, 13.7
Exercises: Q11–Q15 Method: Apply Bayes’ theorem formula

P(Ei∣A)=P(Ei)⋅P(A∣Ei)∑jP(Ej)⋅P(A∣Ej)

Group D: Random Variables & Distributions

Examples: 13.8, 13.9
Exercises: Q16–Q20 Method: Construct probability distribution table, calculate mean and variance.

Group E: Binomial Distribution

Examples: 13.10, 13.11
Exercises: Q21–Q25 Method: Use binomial probability formula

P(X=r)=(nr)prqn−r

with q=1−p.

📌 Final Hint Sheet – Formulas to Remember

From Chapter

Conditional probability: P(A∣B)=P(A∩B)P(B)
Multiplication theorem: P(A∩B)=P(A)⋅P(B∣A)
Independence: P(A∩B)=P(A)⋅P(B)
Bayes’ theorem: P(Ei∣A)=P(Ei)⋅P(A∣Ei)∑jP(Ej)⋅P(A∣Ej)
Mean of distribution: E(X)=∑xiP(xi)
Variance: Var(X)=E(X2)−[E(X)]2
Binomial distribution: P(X=r)=(nr)prqn−r

Outside Chapter (Useful)

Addition theorem: P(A∪B)=P(A)+P(B)−P(A∩B)
Complementary rule: P(Ac)=1−P(A)
Total probability theorem: P(A)=∑iP(Ei)⋅P(A∣Ei)

✅ With this structure, you can study progressively:

Start with conditional probability basics.
Move to independence and Bayes’ theorem.
Then practice random variables and binomial distribution.

Great choice! Let’s carefully unpack Conditional Probability step by step, keeping in mind that you’re approaching this for the first time. I’ll explain definitions, theorems, independence, and examples in a structured way.

🌱 1. Definition of Conditional Probability

Idea: Sometimes we want the probability of an event given that another event has already occurred.
Notation: P(A∣B) means "Probability of event A given that event B has occurred."
Formula:

P(A∣B)=P(A∩B)P(B),provided P(B)≠0

Interpretation: We restrict the sample space to B, then measure how much of B also satisfies A.

Example: A card is drawn from a deck. Find the probability that it is a king given that it is a face card.

Total face cards = 12 (J, Q, K of each suit).
Kings among them = 4.

P(King | Face card)=412=13

📐 2. Multiplication Theorem of Probability

Statement:

P(A∩B)=P(A)⋅P(B∣A)=P(B)⋅P(A∣B)

Use: Helps calculate joint probability of two events.
Why important: Many problems in NCERT exercises rely on this theorem to connect conditional probability with intersections.

Example: Probability of drawing two aces in succession without replacement:

P(First ace)=452
P(Second ace | First ace)=351

P(Two aces)=452⋅351

🔗 3. Independent Events

Definition: Two events A and B are independent if occurrence of one does not affect the probability of the other.
Condition:

P(A∩B)=P(A)⋅P(B)

Note: Independence is stronger than disjointness. Disjoint events cannot occur together, but independent events can.

Example: Tossing a coin and rolling a die.

P(Head)=12
P(Even number)=36=12
Joint probability: P(Head and Even)=12⋅12=14. Since this equals product of individual probabilities, they are independent.

📊 4. Examples Illustrating Conditional Probability

Example 1: Cards

Find probability of drawing a red card given that the card is a king.

Kings = 4, Red kings = 2.

P(Red | King)=24=12

Example 2: Dice

Two dice are thrown. Find probability that sum is 8 given that first die shows 3.

Restrict sample space: first die = 3, second die can be 1–6.
Possible sums = 3+5 = 8 (only one case).

P(Sum = 8 | First die = 3)=16

Example 3: Multiplication Rule

Probability that a student passes both Maths and English:

P(Maths)=0.7, P(English | Maths)=0.9.

P(Both)=0.7⋅0.9=0.63

✨ Quick Recap for Beginners

Conditional probability narrows the sample space.
Multiplication theorem connects conditional probability with joint probability.
Independence means probabilities don’t change when one event occurs.
Examples show how to apply formulas step by step.