Home / Mathematics / Chapter 13 — Probability
Class 12 Maths · Chapter 13 · Bihar Board

Probability

Master conditional probability, multiplication theorem, independent events, total probability, and Bayes' theorem — with visual diagrams, step-marked solutions, MCQs, A-R, and PYQs.

10 Visual Diagrams Bayes' Theorem Step Marking
6Core Concepts
20MCQs
10Assertion-Reason
8PYQs
Section 01

Basic Concepts

Understanding conditional probability and beyond

13.1 Introduction

In earlier classes, we studied probability as a measure of uncertainty of events in a random experiment. We discussed the axiomatic approach formulated by Russian Mathematician A.N. Kolmogorov.

Key Idea

In this chapter, we discuss conditional probability of an event given that another event has occurred — which leads to Bayes' theorem, multiplication rule, and independence of events.

13.2 Conditional Probability

Definition

If E and F are two events associated with the same sample space, the conditional probability of E given that F has occurred is:

$$P(E|F) = \frac{P(E \cap F)}{P(F)}, \text{ provided } P(F) \neq 0$$

Example: Three fair coins are tossed. E = "at least two heads", F = "first coin shows tail". Then $P(E|F) = 1/4$.

13.2.1 Properties of Conditional Probability

Property 1

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

Property 2

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

If A and B are disjoint: $P((A \cup B)|F) = P(A|F) + P(B|F)$

Property 3

$P(E'|F) = 1 - P(E|F)$

Section 02

🎨 Visual Diagrams

See probability concepts in action

V1 Conditional Probability — Venn Diagram

P(E|F) = P(E ∩ F) / P(F) — reduced sample space S E F E∩F P(E|F) = P(E∩F) / P(F) When F has occurred, sample space reduces to F
Conditional probability $P(E|F)$ is the probability of $E$ within the reduced sample space $F$.

V2 Multiplication Rule — Sequential Events

P(E ∩ F) = P(E) × P(F|E) = P(F) × P(E|F) Event E P(E) First event occurs → Event F given E P(F|E) Second event occurs = Both occur P(E∩F) Joint probability P(E ∩ F) = P(E) · P(F|E) Also: P(E ∩ F) = P(F) · P(E|F) — symmetric form
The multiplication rule gives the probability that both events occur, using conditional probability.

V3 Independent Events — P(E∩F) = P(E)·P(F)

Occurrence of one does not affect the other Event E P(E) = 1/2 Odd on first die × Event F P(F) = 1/2 Odd on second die = Both E and F P(E∩F) = 1/4 = P(E) × P(F) Check: P(E ∩ F) = P(E) · P(F)? 1/4 = 1/2 × 1/2 ✓ → E and F are independent Independent events: P(E|F) = P(E) and P(F|E) = P(F)
Two events are independent if knowing one occurred does not change the probability of the other.

V4 Tree Diagram — Sequential Experiment

Coin shows H → toss again ; Coin shows T → throw die Start 1/2 H 1/2 T 1/2 (H,H) 1/4 1/2 (H,T) 1/4 (T,1) 1/12 (T,2) 1/12 (T,3) 1/12 … (T,4), (T,5), (T,6) each 1/12 Total = 1/4 + 1/4 + 6×(1/12) = 1 ✓ Tree diagrams help organise multi-stage experiments
A tree diagram shows all possible outcomes of a multi-stage experiment along with their probabilities.

V5 Bayes' Theorem — Reverse Probability

From P(A|Eᵢ) to P(Eᵢ|A) — reverse the conditional Hypotheses E₁, E₂, …, Eₙ (partition of S) → Event A observed P(A|Eᵢ) known (likelihood) → Find P(Eᵢ|A) (posteriori) Bayes' Theorem P(Eᵢ|A) = [P(Eᵢ) · P(A|Eᵢ)] / Σⱼ P(Eⱼ) · P(A|Eⱼ) Prior × Likelihood ÷ Total Probability Used to find the probability of a cause, given an observed effect Also called the "formula for probability of causes"
Bayes' theorem lets us compute reverse probabilities: given an effect, what is the probability of each cause?

V6 Partition of a Sample Space

E₁, E₂, …, Eₙ are pairwise disjoint and exhaustive S E₁ E₂ E₃ … Eₙ E₁ ∪ E₂ ∪ … ∪ Eₙ = S ; Eᵢ ∩ Eⱼ = φ for i ≠ j
A partition divides the sample space into non-overlapping, exhaustive parts — one and only one must occur.

V7 Theorem of Total Probability

P(A) = Σ P(Eⱼ) · P(A|Eⱼ) E₁ E₂ E₃ … Eₙ A P(A) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + … + P(Eₙ)P(A|Eₙ) Total probability of A = sum over all partitions
The total probability of an event is the sum of its probabilities over all parts of a partition.

V8 Bayes' Theorem — Box Selection Example

Given a red ball is drawn, find P(from Bag II) Bag I 3 Red + 4 Black = 7 P(R|Bag I) = 3/7 P(Bag I) = 1/2 Bag II 5 Red + 6 Black = 11 P(R|Bag II) = 5/11 P(Bag II) = 1/2 Required P(Bag II | R) = ? (reverse) Bayes' Theorem P(Bag II | R) = (1/2 × 5/11) / (1/2 × 3/7 + 1/2 × 5/11) = 35/68 ≈ 0.515 Posterior probability that the red ball came from Bag II
Bayes' theorem updates the probability of a hypothesis after observing evidence.

V9 Independence — Properties & Consequences

If E and F are independent, so are many related events If E and F are independent P(E ∩ F) = P(E) · P(F) ↓ E and F' also independent P(E∩F')=P(E)P(F') E' and F also independent P(E'∩F)=P(E')P(F) E' and F' also independent P(E'∩F')=P(E')P(F') At least one P(A∪B) = 1 − P(A')P(B') Independence is preserved under complementation
Complementing one or both events preserves independence — a powerful computational tool.

V10 Formula Summary — Chapter 13 Probability

All key formulas at a glance Conditional P(E|F) = P(E∩F)/P(F) if P(F) ≠ 0 Multiplication P(E∩F)=P(E)·P(F|E) = P(F)·P(E|F) Independence P(E∩F)=P(E)·P(F) P(E|F)=P(E) Theorem of Total Probability P(A) = Σⱼ P(Eⱼ) · P(A|Eⱼ) for a partition {Eⱼ} Bayes' Theorem P(Eᵢ|A) = P(Eᵢ)P(A|Eᵢ) / Σⱼ P(Eⱼ)P(A|Eⱼ) Reverse conditional probability Properties of Conditional Probability P(S|F)=P(F|F)=1 ; P(E'|F)=1−P(E|F) P((A∪B)|F)=P(A|F)+P(B|F)−P((A∩B)|F)
Master these formulas — they cover the entire chapter. Practise identifying which formula to apply.
Section 03

🎯 Conditional Probability — Deep Dive

Probability of E given F has occurred

C1 Definition and Intuition

Definition

If E and F are events with $P(F) \neq 0$, then:

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

Intuition: When F is known to have occurred, the sample space reduces from S to F. The conditional probability is the fraction of F in which E also occurs.

C2 Example — Three Coins

Experiment: Toss three fair coins.

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

E = "at least two heads" = {HHH, HHT, HTH, THH}

F = "first coin shows tail" = {THH, THT, TTH, TTT}

$E \cap F = \{THH\}$

$P(E) = 1/2$, $P(F) = 1/2$, $P(E \cap F) = 1/8$

$$P(E|F) = \frac{1/8}{1/2} = \frac{1}{4}$$

C3 Properties of Conditional Probability

Property 1

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

Property 2

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

For disjoint events: $P((A \cup B)|F) = P(A|F) + P(B|F)$

Property 3

$P(E'|F) = 1 - P(E|F)$

Section 04

✖️ Multiplication Theorem on Probability

Probability of simultaneous occurrence

M1 Multiplication Rule

Two Events

$$P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F)$$

provided $P(E) \neq 0$ and $P(F) \neq 0$.

Three Events

$$P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|E \cap F)$$

Extension: The rule extends to four or more events naturally.

M2 Example — Drawing Balls Without Replacement

An urn contains 10 black and 5 white balls. Two balls are drawn without replacement.

P(both black) = $P(E) \cdot P(F|E) = \frac{10}{15} \times \frac{9}{14} = \frac{3}{7}$

M3 Example — Drawing Cards

Three cards drawn without replacement. P(King, King, Ace):

$P(K) = 4/52$, $P(K|K) = 3/51$, $P(A|KK) = 4/50$

$P(KKA) = \frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} = \frac{2}{5525}$

Section 05

🔗 Independent Events

Events that don't affect each other

I1 Definition of Independence

Definition

Two events E and F are said to be independent if:

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

Equivalently: $P(E|F) = P(E)$ and $P(F|E) = P(F)$ (when the conditionals exist).

I2 Properties of Independent Events

  • If E and F are independent, then E and F' are independent.
  • If E and F are independent, then E' and F are independent.
  • If E and F are independent, then E' and F' are independent.
  • $P(\text{at least one of E, F}) = 1 - P(E') \cdot P(F')$

I3 Three Events — Mutual Independence

Condition

Three events A, B, C are mutually independent if:

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

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

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

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

All four conditions must hold.

I4 Independence vs Mutually Exclusive

Important Distinction

Independent ≠ Mutually Exclusive

  • Independent: Occurrence of one doesn't affect the other. Defined via probabilities.
  • Mutually Exclusive: Cannot occur together. Defined via sets: $E \cap F = \phi$
  • If two events with non-zero probabilities are mutually exclusive, they are dependent (not independent).
Section 06

➕ Theorem of Total Probability

Weighted sum over a partition

T1 Partition of Sample Space

Definition

Events $E_1, E_2, \ldots, E_n$ form a partition of S if:

  • $E_i \cap E_j = \phi$ for $i \neq j$ (pairwise disjoint)
  • $E_1 \cup E_2 \cup \ldots \cup E_n = S$ (exhaustive)
  • $P(E_i) > 0$ for all $i$

T2 Theorem of Total Probability

$$P(A) = \sum_{j=1}^{n} P(E_j) \cdot P(A|E_j)$$

Where $\{E_1, \ldots, E_n\}$ is a partition of S and A is any event.

T3 Example — Construction Job

P(strike) = 0.65, P(completed | no strike) = 0.80, P(completed | strike) = 0.32

$P(A) = 0.65 \times 0.32 + 0.35 \times 0.80 = 0.208 + 0.28 = 0.488$

Section 07

🔄 Bayes' Theorem

Reverse conditional probability

B1 Statement of Bayes' Theorem

Bayes' Theorem

If $E_1, E_2, \ldots, E_n$ are pairwise disjoint and exhaustive events (a partition of S), and A is any event with $P(A) \neq 0$, then for any $i$:

$$P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{\sum_{j=1}^{n} P(E_j) \cdot P(A|E_j)}$$

Terminology
  • Prior probability: $P(E_i)$ — before observing A
  • Likelihood: $P(A|E_i)$ — probability of A given hypothesis $E_i$
  • Posterior probability: $P(E_i|A)$ — after observing A

B2 Example — Bag Selection

Bag I: 3R, 4B ; Bag II: 5R, 6B. One ball drawn from randomly selected bag and is red.

$P(\text{Bag II} | R) = \frac{(1/2)(5/11)}{(1/2)(3/7) + (1/2)(5/11)} = \frac{35}{68}$

B3 Example — Machine Defect

Machines A, B, C produce 25%, 35%, 40% with defect rates 5%, 4%, 2%.

P(defective from B) = $\frac{0.35 \times 0.04}{0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02} = \frac{28}{69}$

Quick Reference

📐 All Important Formulas

Memorize these for board exams

F1 Conditional Probability

$$P(E|F) = \frac{P(E \cap F)}{P(F)}, \quad P(F) \neq 0$$
$$P(E'|F) = 1 - P(E|F)$$
$$P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$$

F2 Multiplication Rule

$$P(E \cap F) = P(E) \cdot P(F|E) = P(F) \cdot P(E|F)$$
$$P(E \cap F \cap G) = P(E) \cdot P(F|E) \cdot P(G|(E \cap F))$$

F3 Independence

$$P(E \cap F) = P(E) \cdot P(F)$$
$$P(E|F) = P(E) \text{ and } P(F|E) = P(F)$$
$$P(A \cup B) = 1 - P(A') \cdot P(B')$$ (if independent)

F4 Total Probability

$$P(A) = \sum_{j=1}^{n} P(E_j) \cdot P(A|E_j)$$

Where $\{E_1, \ldots, E_n\}$ is a partition of S.

F5 Bayes' Theorem

$$P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{\sum_{j=1}^{n} P(E_j) \cdot P(A|E_j)}$$

F6 Complement and Union

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

F7 Mutual Independence (3 events)

$$P(A \cap B) = P(A)P(B) \text{ AND } P(A \cap C) = P(A)P(C)$$
$$P(B \cap C) = P(B)P(C) \text{ AND } P(A \cap B \cap C) = P(A)P(B)P(C)$$
Shortcuts

💡 Tricks & Shortcuts

Speed techniques for probability problems

Trick 1 — Identify the Formula

Ask: What is given? What is asked?

  • Given P(F) and P(E∩F), find P(E|F) → use definition
  • Given P(E) and P(F|E), find P(E∩F) → use multiplication rule
  • Given P(A|Eⱼ) and P(Eⱼ), find P(Eᵢ|A) → use Bayes
Trick 2 — Multiplication Rule for Sequential Events

For "and" problems with sequential draws:

  • Multiply probabilities along the branches
  • Adjust denominator after each draw (without replacement)
  • Example: P(both black) = $\frac{10}{15} \times \frac{9}{14}$
Trick 3 — Independence Check

Compute P(E)·P(F) and compare with P(E∩F)

  • If equal → independent
  • If not equal → dependent
  • Quick, reliable method
Trick 4 — "At Least One" = 1 − "None"

Use the complement trick!

  • P(at least one) = 1 − P(none)
  • For independent events: P(at least one of A,B) = 1 − P(A')P(B')
  • Much faster than adding many cases
Trick 5 — Bayes' Setup

Identify partition first!

  • What are the "causes" (hypotheses)? → E₁, E₂, ..., Eₙ
  • What is the observed effect? → A
  • Write prior P(Eᵢ), likelihood P(A|Eᵢ)
  • Then apply formula
Trick 6 — Tree Diagram for Bayes

Draw a tree!

  • First branch: hypotheses E₁, E₂, ... with probabilities P(Eᵢ)
  • Second branch: event A with conditional probabilities P(A|Eᵢ)
  • Multiply along paths for joint probabilities
  • Bayes = (path of interest) / (sum of all A-paths)
Trick 7 — Independent vs Mutually Exclusive

Don't confuse them!

  • Independent: P(A∩B) = P(A)P(B)
  • Mutually exclusive: P(A∩B) = 0
  • If non-trivial events are mutually exclusive → they are not independent
Trick 8 — Conditional with Reduced Sample Space

Count directly when the sample space is small

  • P(E|F) = (favourable elements in F) / (total elements in F)
  • Faster than using the formula for small spaces
  • Example: P(E|F) = n(E∩F)/n(F)
Trick 9 — Symmetry in Conditional

Use symmetry to simplify

  • $P(E|F) + P(E'|F) = 1$
  • $P(E|F) \cdot P(F) = P(E \cap F)$
  • Sometimes it's easier to find the complement
Trick 10 — Sanity Check

Always verify:

  • Is the answer between 0 and 1?
  • Are probabilities summing to 1 over all cases?
  • For Bayes: posterior values over all hypotheses sum to 1
  • Did you correctly identify the given information?
Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

Q1. If $P(A) = 7/13$, $P(B) = 9/13$, and $P(A \cap B) = 4/13$, evaluate $P(A|B)$.

Step 1 — Use definition0.5 Mark

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

Step 2 — Substitute1 Mark

$= \frac{4/13}{9/13} = \frac{4}{9}$

Step 3 — Final0.5 Mark

$P(A|B) = 4/9$

$4/9$ ✓
📋 Total: 3 Marks

Q2. A family has two children. What is the probability that both are boys given that at least one is a boy?

Step 1 — Sample space0.5 Mark

S = {(b,b), (g,b), (b,g), (g,g)}

Step 2 — Events0.5 Mark

E = "both boys" = {(b,b)}

F = "at least one boy" = {(b,b), (g,b), (b,g)}

Step 3 — Compute probabilities1 Mark

$P(F) = 3/4$, $P(E \cap F) = 1/4$

Step 4 — Apply formula1 Mark

$P(E|F) = \frac{1/4}{3/4} = \frac{1}{3}$

$1/3$ ✓
📋 Total: 3 Marks

Q3. A die is thrown twice and the sum is 6. Find the conditional probability that 4 appears at least once.

Step 1 — Define events0.5 Mark

E = "4 appears at least once" ; F = "sum is 6"

Step 2 — List F and E∩F1 Mark

F = {(1,5), (2,4), (3,3), (4,2), (5,1)} → 5 outcomes

E ∩ F = {(2,4), (4,2)} → 2 outcomes

Step 3 — Compute probabilities0.5 Mark

$P(F) = 5/36$, $P(E \cap F) = 2/36$

Step 4 — Apply formula1 Mark

$P(E|F) = \frac{2/36}{5/36} = \frac{2}{5}$

$2/5$ ✓
📋 Total: 3 Marks

Q4. An urn contains 10 black and 5 white balls. Two balls are drawn without replacement. Find P(both black).

Step 1 — Define events0.5 Mark

E = "first ball is black" ; F = "second ball is black"

Step 2 — First draw0.5 Mark

$P(E) = \frac{10}{15} = \frac{2}{3}$

Step 3 — Second draw given first black1 Mark

After one black ball is drawn: 9 black, 5 white, total 14

$P(F|E) = \frac{9}{14}$

Step 4 — Multiplication rule1 Mark

$P(E \cap F) = \frac{10}{15} \times \frac{9}{14} = \frac{3}{7}$

$3/7$ ✓
📋 Total: 3 Marks

Q5. Bag I: 3R, 4B ; Bag II: 5R, 6B. One bag chosen at random, a red ball is drawn. Find P(from Bag II).

Step 1 — Define events0.5 Mark

$E_1$ = Bag I chosen, $E_2$ = Bag II chosen, A = red drawn

$P(E_1) = P(E_2) = 1/2$

Step 2 — Conditionals0.5 Mark

$P(A|E_1) = 3/7$, $P(A|E_2) = 5/11$

Step 3 — Apply Bayes1 Mark

$P(E_2|A) = \frac{(1/2)(5/11)}{(1/2)(3/7) + (1/2)(5/11)}$

Step 4 — Simplify1 Mark

$= \frac{5/22}{3/14 + 5/22} = \frac{5/22}{68/154} = \frac{35}{68}$

$35/68$ ✓
📋 Total: 3 Marks

Q6. Machines A, B, C produce 25%, 35%, 40% of items with defect rates 5%, 4%, 2%. A defective item is found. Find P(from B).

Step 1 — Priors0.5 Mark

$P(B_1) = 0.25$, $P(B_2) = 0.35$, $P(B_3) = 0.40$

Step 2 — Likelihoods0.5 Mark

$P(E|B_1) = 0.05$, $P(E|B_2) = 0.04$, $P(E|B_3) = 0.02$

Step 3 — Apply Bayes1 Mark

$P(B_2|E) = \frac{0.35 \times 0.04}{0.25 \times 0.05 + 0.35 \times 0.04 + 0.40 \times 0.02}$

Step 4 — Simplify1 Mark

$= \frac{0.0140}{0.0345} = \frac{28}{69}$

$28/69$ ✓
📋 Total: 3 Marks

Q7. A man speaks truth 3 out of 4 times. He reports a six. Find P(it's actually a six).

Step 1 — Define hypotheses0.5 Mark

$S_1$ = six occurs, $S_2$ = six does not occur

$P(S_1) = 1/6$, $P(S_2) = 5/6$

Step 2 — Likelihoods0.5 Mark

E = "reports six"

$P(E|S_1) = 3/4$, $P(E|S_2) = 1/4$

Step 3 — Apply Bayes1 Mark

$P(S_1|E) = \frac{(1/6)(3/4)}{(1/6)(3/4) + (5/6)(1/4)}$

Step 4 — Simplify1 Mark

$= \frac{3/24}{3/24 + 5/24} = \frac{3}{8}$

$3/8$ ✓
📋 Total: 3 Marks

Q8. If A and B are independent with P(A) = 3/5 and P(B) = 1/5, find P(A∩B) and P(A∪B).

Step 1 — Independence formula0.5 Mark

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

Step 2 — Compute intersection1 Mark

$P(A \cap B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$

Step 3 — Union formula1 Mark

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

$= \frac{3}{5} + \frac{1}{5} - \frac{3}{25} = \frac{15+5-3}{25} = \frac{17}{25}$

Step 4 — Final0.5 Mark

$P(A \cap B) = 3/25$, $P(A \cup B) = 17/25$

$P(A \cap B) = 3/25$, $P(A \cup B) = 17/25$ ✓
Test Yourself

✍️ MCQ Practice — 20 Questions

Select your answer and click "Check Answers"

1. $P(E|F)$ is defined as:

Correct: B — Definition of conditional probability.

2. If A and B are independent events, then $P(A \cap B)$ equals:

Correct: C — Definition of independence.

3. If A and B are mutually exclusive, then $P(A \cap B)$ equals:

Correct: A — Mutually exclusive events cannot occur together.

4. The multiplication rule states:

Correct: D — Multiplication rule of probability.

5. If P(E) = 0.6, P(F) = 0.3, P(E∩F) = 0.2, then P(E|F) =

Correct: B — $0.2/0.3 = 2/3$.

6. If A and B are independent, then P(A|B) equals:

Correct: A — Independence means P(A|B) = P(A).

7. Bayes' theorem is used to compute:

Correct: C — Bayes computes $P(E_i|A)$ from $P(A|E_i)$.

8. The theorem of total probability uses:

Correct: B — Requires a partition {E₁, …, Eₙ}.

9. For independent events A and B, P(at least one) equals:

Correct: D — Complement of "neither occurs".

10. If two events are mutually exclusive and have non-zero probabilities, then they are:

Correct: A — P(A∩B) = 0 ≠ P(A)P(B) > 0 → dependent.

11. $P(E'|F)$ equals:

Correct: C — Complement property of conditional probability.

12. In Bayes' theorem, $P(E_i)$ is called:

Correct: B — Prior probability of hypothesis $E_i$.

13. A partition of a sample space requires:

Correct: D — All three conditions must hold.

14. If E and F are independent, then which is TRUE?

Correct: A — Independence preserved under complementation.

15. $P((A \cup B)|F)$ equals:

Correct: C — Addition rule for conditional probability.

16. A family has two children. Given at least one is a boy, P(both boys) =

Correct: B — Sample space reduces to {BB, BG, GB}.

17. Three events A, B, C are mutually independent if:

Correct: D — All four conditions must hold.

18. Two cards are drawn without replacement. P(both kings) =

Correct: A — Without replacement changes denominator.

19. If P(A) = 1/2, P(B) = 0, then P(A|B) is:

Correct: C — Conditional undefined when P(B) = 0.

20. Bayes' theorem is also called the formula for:

Correct: B — It computes probability of a "cause" given an effect.
Score: 0 / 20
New Pattern

🎯 Assertion & Reasoning — 10 Questions

Each question has an Assertion (A) and a Reason (R). Choose the correct option.

📖 Standard Options

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is NOT the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

1
Assertion (A) Conditional probability $P(E|F)$ is defined when $P(F) \neq 0$.
Reason (R) $P(E|F) = P(E \cap F) / P(F)$ involves division by $P(F)$.
Correct: A — Both true. R explains why the condition is needed.
2
Assertion (A) If A and B are independent, then $P(A \cap B) = P(A) \cdot P(B)$.
Reason (R) Independence means occurrence of one does not affect the other.
Correct: A — Both true. R gives the intuitive meaning.
3
Assertion (A) Mutually exclusive events are always independent.
Reason (R) For mutually exclusive events, $P(A \cap B) = 0$.
Correct: D — A is false (they're usually dependent). R is true.
4
Assertion (A) Multiplication rule: $P(E \cap F) = P(E) \cdot P(F|E)$.
Reason (R) This follows from the definition of conditional probability.
Correct: A — Both true. R correctly explains the derivation.
5
Assertion (A) Bayes' theorem is used to find the probability of a cause given that an effect has been observed.
Reason (R) It computes $P(E_i|A)$ from prior probabilities $P(E_i)$ and likelihoods $P(A|E_i)$.
Correct: A — Both true. Bayes' theorem reverses the conditional: given the observed effect $A$, it computes the probability of each cause $E_i$ using priors $P(E_i)$ and likelihoods $P(A|E_i)$.
6
Assertion (A) $P((A \cup B)|F) = P(A|F) + P(B|F) - P((A \cap B)|F)$.
Reason (R) This is the addition rule for conditional probability.
Correct: A — Both true. R identifies the property.
7
Assertion (A) The theorem of total probability requires a partition of the sample space.
Reason (R) $\{E_1, E_2, \ldots, E_n\}$ must be pairwise disjoint and exhaustive.
Correct: A — Both true. R defines a partition.
8
Assertion (A) If A and B are independent, then $P(A \cap B) = 0$.
Reason (R) For independent events, $P(A \cap B) = P(A) \cdot P(B)$.
Correct: D — A is false ($P(A \cap B) = P(A)P(B)$, usually nonzero). R is true.
9
Assertion (A) $P(E'|F) = 1 - P(E|F)$.
Reason (R) $E$ and $E'$ partition the sample space given $F$.
Correct: A — Both true. R explains the complement property.
10
Assertion (A) Independence is preserved under complementation.
Reason (R) If A and B are independent, so are A' and B, A and B', A' and B'.
Correct: A — Both true. R lists the consequences.
Score: 0 / 10
Exam Focus

📖 Previous Year Questions

Frequently asked questions in board exams

2024

Q. If $P(A) = 7/13$, $P(B) = 9/13$, $P(A \cap B) = 4/13$, evaluate $P(A|B)$.

Step 1 — Definition0.5 Mark

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

Step 2 — Substitute1 Mark

$= \frac{4/13}{9/13} = 4/9$

Step 3 — Final0.5 Mark

$4/9$

$4/9$ ✓
2024

Q. A family has two children. Find P(both are boys) given at least one is a boy.

Step 1 — Sample space0.5 Mark

S = {BB, BG, GB, GG}

Step 2 — Events1 Mark

E = {BB} ; F = {BB, BG, GB}

$P(F) = 3/4$, $P(E \cap F) = 1/4$

Step 3 — Apply1 Mark

$P(E|F) = \frac{1/4}{3/4} = 1/3$

$1/3$ ✓
2023

Q. Bag I: 3R, 4B ; Bag II: 5R, 6B. One bag chosen at random, red ball drawn. Find P(from Bag II).

Step 1 — Setup0.5 Mark

$P(E_1) = P(E_2) = 1/2$ ; $P(A|E_1) = 3/7$, $P(A|E_2) = 5/11$

Step 2 — Bayes1.5 Marks

$P(E_2|A) = \frac{(1/2)(5/11)}{(1/2)(3/7)+(1/2)(5/11)} = \frac{35}{68}$

Step 3 — Final1 Mark

$35/68$

$35/68$ ✓
2023

Q. A die is thrown twice, sum is 6. Find P(4 appears at least once | sum = 6).

Step 1 — Events0.5 Mark

E = "4 appears at least once" ; F = "sum is 6"

Step 2 — Count1 Mark

F = {(1,5),(2,4),(3,3),(4,2),(5,1)} → 5 outcomes

E∩F = {(2,4),(4,2)} → 2 outcomes

Step 3 — Apply1 Mark

$P(E|F) = \frac{2/36}{5/36} = 2/5$

$2/5$ ✓
2022

Q. Machines A, B, C produce 25%, 35%, 40% with defect rates 5%, 4%, 2%. Defective found. Find P(from B).

Step 1 — Priors/Likelihoods0.5 Mark

$P(B_1)=0.25$, $P(B_2)=0.35$, $P(B_3)=0.40$

$P(E|B_1)=0.05$, $P(E|B_2)=0.04$, $P(E|B_3)=0.02$

Step 2 — Bayes1.5 Marks

$P(B_2|E) = \frac{0.0140}{0.0345} = \frac{28}{69}$

Step 3 — Final1 Mark

$28/69$

$28/69$ ✓
2022

Q. A man speaks truth 3/4 times. Reports a six. Find P(it's actually a six).

Step 1 — Hypotheses0.5 Mark

$P(S_1)=1/6$, $P(S_2)=5/6$

Step 2 — Likelihoods0.5 Mark

$P(E|S_1)=3/4$, $P(E|S_2)=1/4$

Step 3 — Bayes1 Mark

$P(S_1|E) = \frac{3/24}{3/24+5/24} = \frac{3}{8}$

Step 4 — Final1 Mark

$3/8$

$3/8$ ✓
2021

Q. If A and B are independent with P(A) = 3/5, P(B) = 1/5, find P(A∩B) and P(A∪B).

Step 1 — Intersection1 Mark

$P(A \cap B) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$

Step 2 — Union1.5 Marks

$P(A \cup B) = \frac{3}{5} + \frac{1}{5} - \frac{3}{25} = \frac{17}{25}$

Step 3 — Final0.5 Mark

$3/25$ and $17/25$

$3/25$ and $17/25$ ✓
2020

Q. An urn contains 10 black and 5 white balls. Two drawn without replacement. Find P(both black).

Step 1 — First draw0.5 Mark

$P(E) = \frac{10}{15}$

Step 2 — Second draw given first black1 Mark

$P(F|E) = \frac{9}{14}$

Step 3 — Multiply1 Mark

$\frac{10}{15} \times \frac{9}{14} = \frac{3}{7}$

Step 4 — Final0.5 Mark

$3/7$

$3/7$ ✓