**Definition**: If E and F are two events associated with the same sample space S of a random experiment, the conditional probability of event E given that F has already occurred is denoted as **P(E|F)** and is defined as:
**P(E|F) = P(E ∩ F) / P(F)** provided **P(F) ≠ 0**
This formula gives the probability of E occurring when we know F has already occurred, thus reducing the sample space from S to F.
**Intuitive Explanation**: When event F occurs, the sample space effectively reduces to only the outcomes in F. Among these outcomes, we count how many also satisfy event E (these are in E ∩ F). The conditional probability is the ratio of favorable outcomes to total outcomes in the reduced sample space.
**Example 1: Card Problem**
From a standard deck of 52 cards, given that a card drawn is a heart (event F), find the probability it is an ace (event E).
**Example 2: Two Children Problem**
A family has two children. Find P(both boys | at least one boy).
**Key Insight**: The conditional probability is only defined when P(F) ≠ 0 because we cannot condition on an impossible event.
**Property 1: P(S|F) = 1** and **P(F|F) = 1**
**Property 2: Addition Rule for Conditional Probability**
For any events A, B with P(F) ≠ 0:
**P((A ∪ B)|F) = P(A|F) + P(B|F) − P((A ∩ B)|F)**
If A and B are disjoint (mutually exclusive):
**P((A ∪ B)|F) = P(A|F) + P(B|F)**
This follows from: P((A ∪ B)|F) = P[(A ∪ B) ∩ F]/P(F) = P[(A ∩ F) ∪ (B ∩ F)]/P(F)
**Property 3: Complement Rule**
**P(E'|F) = 1 − P(E|F)**
Proof: Since E ∪ E' = S and P(S|F) = 1, and E and E' are disjoint:
P(E|F) + P(E'|F) = P((E ∪ E')|F) = P(S|F) = 1
**Example**: In a class, P(student is intelligent | student studies hard) = 0.8
Then P(student is not intelligent | student studies hard) = 1 − 0.8 = 0.2
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**Definition**: The multiplication theorem relates the probability of simultaneous occurrence of two events to conditional probability.
**Multiplication Rule for Two Events:**
**P(E ∩ F) = P(E) · P(F|E) = P(F) · P(E|F)**
provided P(E) ≠ 0 and P(F) ≠ 0
**Derivation**: From the definition of conditional probability:
**Multiplication Rule for Three Events:**
**P(E ∩ F ∩ G) = P(E) · P(F|E) · P(G|(E ∩ F))**
This extends naturally: find probability of E, then F given E, then G given both E and F have occurred.
**Example 1: Drawing Without Replacement**
An urn contains 10 black and 5 white balls. Two balls are drawn one after another without replacement. Find P(both black).
**Example 2: Three Cards**
From a shuffled deck of 52 cards, draw three cards without replacement. Find P(first two are kings and third is an ace).
**Key Points**:
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**Definition 1 (Using Conditional Probability)**: Events E and F are **independent** if:
**P(F|E) = P(F)** (provided P(E) ≠ 0) and **P(E|F) = P(E)** (provided P(F) ≠ 0)
This means occurrence of one event does not change the probability of the other.
**Definition 2 (Using Intersection)**: Events E and F are **independent** if:
**P(E ∩ F) = P(E) · P(F)**
This is equivalent to Definition 1 and often more practical to use.
**Intuitive Meaning**: Two events are independent if knowledge that one has occurred does not change the probability assessment of the other.
**Example 1: Card Drawing**
Draw one card from a deck. Let E = "card is a spade" and F = "card is an ace".
**Example 2: Dice Rolls**
Roll a fair die twice. Let E = "first die shows odd" and F = "second die shows even".
**Important Distinctions**:
**Mutual Independence of Three Events**:
Events A, B, C are **mutually independent** if **all** of the following hold:
If any one condition fails, the events are not mutually independent. Note: pairwise independence does not guarantee mutual independence.
**Example 3: Testing Independence**
Let A = "first die shows odd", B = "second die shows odd", C = "sum is even".
**Example 4: Checking Dependence**
A box has 4 cards: RR, RB, BR, BB (two-sided cards).
Draw one card. Let A = "side 1 is red", B = "side 2 is red".
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**Statement**: If E₁, E₂, ..., Eₙ are mutually exclusive and exhaustive events with P(Eᵢ) > 0 for all i, and F is any event with P(F) > 0, then:
**P(Eᵢ|F) = [P(Eᵢ) · P(F|Eᵢ)] / P(F)**
where **P(F) = Σⱼ P(Eⱼ) · P(F|Eⱼ)** (law of total probability)
**Detailed Formula for Two Events**:
If E and E' partition the sample space and F is any event:
**P(E|F) = [P(E) · P(F|E)] / [P(E) · P(F|E) + P(E') · P(F|E')]**
**Terminology**:
**Derivation**: From the multiplication rule:
P(Eᵢ ∩ F) = P(Eᵢ) · P(F|Eᵢ) = P(F) · P(Eᵢ|F)
Therefore: P(Eᵢ|F) = [P(Eᵢ) · P(F|Eᵢ)] / P(F)
By total probability: P(F) = Σⱼ P(Eⱼ ∩ F) = Σⱼ P(Eⱼ) · P(F|Eⱼ)
**Example 1: Medical Diagnosis**
A certain disease affects 1% of the population. A test has 99% accuracy for infected people and 98% accuracy for healthy people (i.e., 2% false positive rate). If a person tests positive, what is the probability they actually have the disease?
Let D = "person has disease", T⁺ = "person tests positive"
Find P(D|T⁺):
Despite testing positive, there's only about 1 in 3 chance of having the disease due to the low prevalence.
**Example 2: Defective Products**
A factory has three machines A, B, C producing 50%, 30%, 20% of items respectively. Defect rates are 3%, 4%, 5% respectively. An item is found defective. Find probability it came from machine A.
Let A, B, C = "item from machine A/B/C", D = "item defective"
Total probability of defect:
P(D) = P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C)
P(D) = 0.5(0.03) + 0.3(0.04) + 0.2(0.05) = 0.015 + 0.012 + 0.01 = 0.037
By Bayes' theorem:
P(A|D) = [P(A) · P(D|A)] / P(D) = (0.5 × 0.03) / 0.037 = 0.015 / 0.037 ≈ **0.405 or 40.5%**
**Example 3: Coin Selection**
There are three coins: one fair coin and two biased coins (both show heads with probability 0.7). A coin is selected randomly and tossed, resulting in heads. What is the probability the selected coin is fair?
Let F = "fair coin selected", B = "biased coin selected", H = "heads appears"
Total probability:
P(H) = P(F) · P(H|F) + P(B) · P(H|B)
P(H) = (1/3)(1/2) + (2/3)(0.7) = 1/6 + 1.4/3 = 1/6 + 14/30 = 5/30 + 14/30 = 19/30
By Bayes' theorem:
P(F|H) = [P(F) · P(H|F)] / P(H) = [(1/3) · (1/2)] / (19/30)
P(F|H) = (1/6) × (30/19) = 30/114 = **5/19**
**Key Applications**:
**Common Mistakes**:
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**Definition**: A **random variable** is a function that assigns a real number to each outcome in the sample space of a random experiment. Typically denoted by capital letters X, Y, Z, etc.
**Formal Definition**: Let S be a sample space. A random variable X is a function X: S → ℝ that associates a real number X(ω) with each outcome ω ∈ S.
**Example 1: Coin Toss**
Toss a coin three times. Let X = number of heads.
**Example 2: Dice Roll**
Roll a die. Let X = outcome on the die.
**Types of Random Variables**:
This chapter focuses on **discrete random variables**.
**Probability Distribution (Probability Mass Function)**:
The **probability distribution** of a discrete random variable X lists all possible values of X and their corresponding probabilities.
**Definition**: For a discrete random variable X taking values x₁, x₂, x₃, ..., the probability distribution is given by:
**Properties of Probability Distribution**:
1. **p(xᵢ) ≥ 0** for all i (probabilities are non-negative)
2. **Σᵢ p(xᵢ) = 1** (probabilities sum to 1)
**Representation**: Usually presented as a table or list:
| X | x₁ | x₂ | x₃ | ... |
|---|----|----|----|----|
| P(X) | p₁ | p₂ | p₃ | ... |
**Example 1: Three Coins, Number of Heads**
Toss three fair coins. X = number of heads.
Probability distribution:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | 1/8 | 3/8 | 3/8 | 1/8 |
Check: 1/8 + 3/8 + 3/8 + 1/8 = 8/8 = 1 ✓
**Example 2: Two Dice, Sum of Numbers**
Roll two fair dice. X = sum of numbers appearing.
Possible values: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Total outcomes: 36
Sum: 1+2+3+4+5+6+5+4+3+2+1 = 36/36 = 1 ✓
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**Definition**: The **mean** or **expected value** of a discrete random variable X, denoted as **E(X)** or **μ**, is:
**E(X) = Σᵢ xᵢ · P(X = xᵢ)**
This is the weighted average of all possible values, where weights are probabilities.
**Physical Interpretation**: If the experiment is repeated many times, the average of all outcomes approaches E(X).
**Example 1: Three Coins, Number of Heads**
From previous example:
E(X) = 0 · (1/8) + 1 · (3/8) + 2 · (3/8) + 3 · (1/8)
E(X) = 0 + 3/8 + 6/8 + 3/8 = 12/8 = **3/2 or 1.5**
This makes intuitive sense: with three fair coins, we expect 1.5 heads on average (by symmetry, exactly 3/2).
**Example 2: Fair Die**
Roll a fair die. X = number on die.
E(X) = 1 · (1/6) + 2 · (1/6) + 3 · (1/6) + 4 · (1/6) + 5 · (1/6) + 6 · (1/6)
E(X) = (1+2+3+4+5+6)/6 = 21/6 = **3.5**
**Example 3: Profit Distribution**
A company's daily profit X (in thousands of rupees) has distribution:
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | 0.2 | 0.3 | 0.4 | 0.1 |
E(X) = 0(0.2) + 1(0.3) + 2(0.4) + 3(0.1)
E(X) = 0 + 0.3 + 0.8 + 0.3 = **1.4 thousand rupees or ₹1400**
Expected average daily profit is ₹1400.
**Properties of Expected Value**:
1. **E(c) = c** where c is a constant
2. **E(cX) = c · E(X)** (constant factor comes out)
3. **E(X + Y) = E(X) + E(Y)** (expectation of sum = sum of expectations)
4. **E(X + c) = E(X) + c** (adding constant adds to expectation)
5. **If X and Y are independent: E(XY) = E(X) · E(Y)**
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**Definition of Variance**: The **variance** of a random variable X, denoted as **Var(X)** or **σ²**, measures how spread out the values are around the mean:
**Var(X) = E[(X − μ)²] = Σᵢ (xᵢ − μ)² · P(X = xᵢ)**
where μ = E(X)
**Alternative Formula**:
**Var(X) = E(X²) − [E(X)]²**
Proof:
E[(X − μ)²] = E[X² − 2μX + μ²]
= E(X²) − 2μE(X) + μ²
= E(X²) − 2μ · μ + μ²
= E(X²) − μ²
**Definition of Standard Deviation**: The **standard deviation** is:
**σ = √Var(X)**
It is in the same units as X and gives a more interpretable measure of spread.
**Example 1: Three Coins, Number of Heads**
From earlier: E(X) = 3/2
First calculate E(X²):
E(X²) = 0² · (1/8) + 1² · (3/8) + 2² · (3/8) + 3² · (1/8)
E(X²) = 0 + 3/8 + 12/8 + 9/8 = 24/8 = 3
Using the alternative formula:
Var(X) = E(X²) − [E(X)]²
Var(X) = 3 − (3/2)² = 3 − 9/4 = 12/4 − 9/4 = **3/4**
Standard deviation:
σ = √(3/4) = √3/2 ≈ **0.866**
Using the definition directly (verification):
Var(X) = (0 − 3/2)² · (1/8) + (1 − 3/2)² · (3/8) + (2 − 3/2)² · (3/8) + (3 − 3/2)² · (1/8)
= (9/4) · (1/8) + (1/4) · (3/8) + (1/4) · (3/8) + (9/4) · (1/8)
= 9/32 + 3/32 + 3/32 + 9/32 = 24/32 = 3/4 ✓
**Example 2: Fair Die**
From earlier: E(X) = 3.5
E(X²) = 1² · (1/6) + 2² · (1/6) + ... + 6² · (1/6)
= (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6
Var(X) = 91/6 − (3.5)² = 91/6 − 12.25 = 91/6 − 49/4
= 182/12 − 147/12 = 35/12 ≈ **2.917**
σ = √(35/12) ≈ **1.708**
**Example 3: Profit Distribution**
From earlier, X has distribution with E(X) = 1.4
E(X²) = 0² · (0.2) + 1² · (0.3) + 2² · (0.4) + 3² · (0.1)
= 0 + 0.3 + 1.6 + 0.9 = 2.8
Var(X) = E(X²) − [E(X)]²
Var(X) = 2.8 − (1.4)² = 2.8 − 1.96 = **0.84**
σ = √0.84 ≈ **0.917**
**Properties of Variance**:
1. **Var(c) = 0** (constant has zero variance)
2. **Var(cX) = c² · Var(X)** (variance scales with square of constant)
3. **Var(X + c) = Var(X)** (adding constant doesn't change variance)
4. **If X and Y are independent: Var(X + Y) = Var(X) + Var(Y)**
5. **Var(X) ≥ 0** (variance is always non-negative)
**Interpretation**:
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**Definition**: A **Binomial Distribution** models a series of independent experiments (called Bernoulli trials) where each trial has exactly two outcomes: success (with probability p) or failure (with probability q = 1−p).
**Conditions for Binomial Distribution**:
1. Fixed number of trials: n (known in advance)
2. Each trial is independent of others
3. Each trial has exactly two possible outcomes (success/failure)
4. Probability of success p is constant across all trials
5. We count the total number of successes X in n trials
**Bernoulli Trial**: A single experiment with two outcomes, success with probability p and failure with probability q = 1−p.
**Notation**: If X follows a binomial distribution with parameters n and p, we write:
**X ~ B(n, p)** or **X ~ Binomial(n, p)**
**Probability Mass Function**:
For X = 0, 1, 2, ..., n:
**P(X = k) = ⁿCₖ · pᵏ · qⁿ⁻ᵏ**
where:
**Proof/Derivation**:
To get exactly k successes in n trials:
**Example 1: Coin Tosses**
Toss a fair coin 4 times. Find distribution of X = number of heads.
Here n = 4, p = 0.5 (probability of heads), q = 0.5
Q1. If P(A) = 3/7, P(B) = 5/7, and P(A∩B) = 2/7, then P(A|B) = ?
Answer: A — Using P(A|B) = P(A∩B)/P(B) = (2/7)/(5/7) = 2/5.
Q2. A family has two children. What is the probability both are boys given at least one is a boy?
Answer: A — Sample space with at least one boy: {(b,b), (g,b), (b,g)}, favorable outcome for both boys: {(b,b)}, so P = 1/3, not 1/2 (common error: P(b,b) from total space is 1/4).
Q3. If P(E|F) = 0.6 and P(E'|F) = x, then x = ?
Answer: B — By Property 3, P(E'|F) = 1 − P(E|F) = 1 − 0.6 = 0.4.
Q4. Ten numbered cards 1–10 are in a box. If drawn card is known to be > 5, probability it is even is?
Answer: C — Given event F: {6,7,8,9,10}, favorable event E∩F: {6,8,10}, so P(E|F) = 3/5.
Q5. Which statement about conditional probability is NOT correct?
Answer: B — P(E|F) ≠ P(F|E) in general; they are different conditional probabilities representing different ordered events.
Q6. Two events A and B satisfy: P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2. Then P(A|B) and P(B|A) are respectively:
Answer: A — P(A|B) = 0.2/0.4 = 0.5; P(B|A) = 0.2/0.5 = 0.4; they are unequal, illustrating order-dependence.
Q7. For mutually exclusive events A, B and event F, if P(F) ≠ 0, then P((A∪B)|F) equals:
Answer: B — For mutually exclusive events, P((A∩B)|F) = 0, so Property 2 simplifies to P((A∪B)|F) = P(A|F) + P(B|F).
Q8. A box contains 5 red and 3 blue balls. One red ball is drawn and not replaced. Probability next ball is red given first was red:
Answer: B — After drawing one red ball without replacement, 4 red and 3 blue remain out of 7 total, so P(second red | first red) = 4/7.
Q9. **Assertion**: P(E|F) = P(E∩F)/P(F) is valid for all events F. **Reason**: This formula defines conditional probability when P(F) = 0.
Answer: C — The assertion is true and valid only when P(F) ≠ 0; the reason is false because the formula is not defined when P(F) = 0.
Q10. In 100 students, 60 like Math and 45 like Science, 25 like both. If a student likes Math, probability they like Science is:
Answer: C — P(Science|Math) = P(both)/P(Math) = 25/60, using students liking Math as the reduced sample space.
What is conditional probability P(E|F)?
P(E|F) = P(E∩F)/P(F), the probability of event E occurring given that event F has already occurred, provided P(F) ≠ 0.
Why must P(F) ≠ 0 in conditional probability formula?
Because division by zero is undefined, and an event with zero probability (impossible event) cannot serve as a given condition to reduce the sample space.
State Property 1 of conditional probability.
P(S|F) = P(F|F) = 1, meaning the probability of the entire sample space or the event itself given that event has occurred is always 1.
State Property 3 of conditional probability.
P(E'|F) = 1 − P(E|F), the complement rule holds: conditional probability of complement equals 1 minus conditional probability of the event.
For disjoint events A and B, what is P((A∪B)|F)?
P((A∪B)|F) = P(A|F) + P(B|F), since the intersection P((A∩B)|F) = 0 for disjoint events.
Two cards drawn without replacement: first is red. How does this affect probability second is red?
The reduced sample space now has one fewer red card and one fewer total card, making P(second red | first red) = (red cards−1)/(total cards−1), which differs from initial probability.
In a conditional probability problem, what does E∩F represent?
E∩F represents outcomes where both events E and F occur simultaneously; these are the favorable outcomes for calculating P(E|F).
A die shows number > 3. What is sample space for conditional probability?
The reduced sample space is {4, 5, 6} instead of {1, 2, 3, 4, 5, 6}, containing only outcomes satisfying the given condition.
Prove: P(E|F) + P(E'|F) = 1
Since E and E' are disjoint and E∪E' = S, we have P((E∪E')|F) = P(E|F) + P(E'|F) = P(S|F) = 1 by Property 1.
If P(A|B) = P(A), what can you conclude about events A and B?
Events A and B are independent because knowing B occurred does not change the probability of A, meaning P(A∩B) = P(A)·P(B).
Define conditional probability and state the condition under which P(E|F) = P(E∩F)/P(F) is valid. Give one example. [2 marks]
State P(E|F) formula explicitly, require P(F) ≠ 0, provide any real example like coin/card/dice with two events and calculate using the formula.
Prove that P(E|F) + P(E'|F) = 1. Then use this to find P(E'|F) if P(E|F) = 7/13. [5 marks]
Start with P(S|F) = P(E∪E'|F) = 1 from Property 1; expand using Property 2 for disjoint events E and E'; substitute values to compute P(E'|F) = 6/13.
A die is thrown twice. Event A: sum of numbers is 8; Event B: first throw is at least 3. Find P(A|B) and verify using definition. [6 marks]
List all outcomes in B: (3,1) to (6,6) — 24 outcomes; find A∩B: {(3,5), (4,4), (5,3)} — 3 outcomes; calculate P(A|B) = 3/24 = 1/8; verify by computing P(A∩B)/P(B) = (3/36)/(24/36) = 3/24.
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