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WBB • Class XI • Mathematics • Ch 14
Estimated Time: 90 minutes
Study Progress: In Progress

Probability

Probability is the branch of mathematics that formalizes the quantification of uncertainty, randomness, and chance. In everyday life, events cannot always be predicted with absolute certainty; whether predicting tomorrow weather, evaluating financial market risk, assessing clinical trial outcomes, or playing games of chance, uncertainty is inherent. While lower secondary school introduced empirical probability based on observed relative frequencies of past trials, Higher Secondary Class 11 introduces a rigorous, axiomatic mathematical foundation developed by Andrey Kolmogorov in 1933. The theory begins with the rigorous definition of random experiments, sample spaces, and sample points. Events are defined as subsets of the sample space, establishing a direct correspondence between set theory and probability. Students master classical Laplace probability for equiprobable discrete outcomes, event algebra encompassing unions, intersections, complements, and differences, and advance to Kolmogorov three fundamental axioms. Crucially, students examine the addition theorems for two and three arbitrary events, De Morgan laws in probability, Boole inequality, and the exact relationship between probability and odds in favor or odds against. This curriculum provides the indispensable mathematical foundation for conditional probability, Bayes theorem, random variables, and probability distributions in Class 12.

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Why This Chapter Matters

Probability is the foundational language of modern science, data analytics, artificial intelligence, quantum physics, actuarial mathematics, and high-stakes decision theory. In machine learning and computer vision, Bayesian inference, generative neural networks, and stochastic gradient descent rely entirely on probability measures. In financial engineering and quantitative trading, probability governs Black-Scholes option pricing, credit default models, and algorithmic risk management. In quantum mechanics, physical reality itself is inherently probabilistic, governed by wave function probabilities. In epidemiology and genetics, probability models viral transmission rates, Mendelian inheritance ratios, and vaccine efficacy. For students preparing for WBCHSE Class 11 annual examinations, WBJEE, and JEE Advanced, Probability is a core, conceptually rich chapter that rewards logical precision, combinatorial fluency, and rigorous set-theoretic reasoning.

Before You Begin (Prerequisites)

  • Class 11 Set Theory: Subsets, Unions, Intersections, Complements, De Morgan Laws, and Cardinality
  • Class 11 Permutations and Combinations: Fundamental Principle of Counting, nPr arrangements, and nCr selections

Chapter Roadmap & Progression

1 Module 1: Random Experiments, Outco...
2 Module 2: Event Algebra & Types of...
3 Module 3: Classical Definition of P...
4 Module 4: Axiomatic Approach to Pro...
5 Module 5: Addition Theorems & Proba...
6 Module 6: Odds in Favor, Odds Again...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Random Experiments, Outcomes & Sample Spaces

The foundation of mathematical probability rests on the formal distinction between deterministic and non-deterministic phenomena.

1. Deterministic vs. Random Experiments

In classical physical sciences, experiments conducted under identical conditions invariably yield identical results; such experiments are called deterministic (e.g., measuring the acceleration due to gravity or heating pure water at 1 atm to 100°C). In contrast, an experiment is defined as a random (or non-deterministic) experiment if it satisfies two essential criteria:

  • It has more than one possible outcome.
  • It is impossible to predict the exact outcome in advance of any particular performance (trial), even though the complete set of all possible outcomes is fully known beforehand.
2. Trials, Outcomes & Sample Space

A single performance of a random experiment is termed a trial, and the result obtained is an outcome. The set of all possible distinct outcomes of a random experiment is defined as the Sample Space, customarily denoted by the symbol \(S\) (or \(\Omega\)). Each individual element \(s \in S\) is called a sample point or elementary outcome.

3. Common Discrete Sample Spaces

A sample space is classified as discrete finite if it contains a finite number of sample points, and discrete countably infinite if its outcomes can be placed in one-to-one correspondence with natural numbers.

Random Experiment Sample Space \(S\) Number of Points \(n(S)\)
Tossing a single fair coin \(S = \{H, T\}\) \(2^1 = 2\)
Tossing \(n\) fair coins simultaneously \(S = \{(x_1, x_2, \dots, x_n) : x_i \in \{H, T\}\}\) \(2^n\)
Rolling a standard six-faced die \(S = \{1, 2, 3, 4, 5, 6\}\) \(6^1 = 6\)
Throwing a pair of distinct dice \(S = \{(i, j) : 1 \le i, j \le 6\}\) \(6^2 = 36\)
Drawing one card from a standard deck 4 suits (♠, ♣, ♥, ♦) \(\times\) 13 denominations \(52\)
Standard Deck of 52 Cards Architecture:

A standard deck consists of 52 cards divided into 2 colors: 26 Red (Hearts ♥, Diamonds ♦) and 26 Black (Spades ♠, Clubs ♣). Each of the 4 suits contains 13 cards: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. The 12 cards depicting people (Jack, Queen, King across all 4 suits) are designated as Face Cards (or Court Cards). The 16 cards comprising Aces, Kings, Queens, and Jacks are designated as Honor Cards.

Module 2: Event Algebra & Types of Events

In modern probability theory, events are defined strictly through set theory as subsets of the sample space.

1. Formal Definition of an Event

Any subset \(E\) of the sample space \(S\) is called an event (\(E \subseteq S\)). Whenever the actual outcome of the experiment is an element of \(E\), we say that the event \(E\) has occurred.

2. Types of Events
  • Elementary (Simple) Event: An event containing exactly one sample point, \(E = \{s_i\}\).
  • Compound Event: An event containing two or more sample points. For example, getting an even number on a die roll: \(E = \{2, 4, 6\}\).
  • Impossible Event (\(\emptyset\)): The empty set containing no sample points. Its probability is always zero: \(P(\emptyset) = 0\). Example: rolling a 7 on an ordinary die.
  • Certain (Sure) Event (\(S\)): The entire sample space containing all possible outcomes. Its occurrence is guaranteed: \(P(S) = 1\). Example: rolling a number between 1 and 6.
  • Complementary Event (\(E'\) or \(\bar{E}\)): The event that \(E\) does not occur, defined as \(E' = S - E = \{s \in S : s \notin E\}\).
3. Set-Theoretic Event Algebra

The operations of set theory translate directly into logical relationships among events:

Verbal Description Set-Theoretic Notation Meaning in Probability
"Not A" \(A'\) or \(A^c\) or \(\bar{A}\) Event A does not occur
"A or B" (at least one) \(A \cup B\) At least one of the events A or B occurs
"A and B" (both) \(A \cap B\) Both events A and B occur simultaneously
"A but not B" \(A - B = A \cap B'\) Event A occurs while event B does not occur
"Neither A nor B" \(A' \cap B' = (A \cup B)'\) Neither event A nor event B occurs
"Exactly one of A or B" \((A \cap B') \cup (A' \cap B)\) Either A occurs alone or B occurs alone
4. Mutually Exclusive & Exhaustive Events

Two events \(A\) and \(B\) associated with a random experiment are defined as mutually exclusive (or disjoint / incompatible) if they cannot both occur simultaneously in any single trial. In set notation:

\[A \cap B = \emptyset \implies P(A \cap B) = 0\]

A collection of events \(E_1, E_2, \dots, E_k\) is said to be mutually exhaustive if at least one of them is guaranteed to occur whenever the experiment is performed, meaning their union covers the entire sample space:

\[\bigcup_{i=1}^k E_i = E_1 \cup E_2 \cup \dots \cup E_k = S\]

If a set of events is both pairwise mutually exclusive (\(E_i \cap E_j = \emptyset\) for all \(i \neq j\)) and mutually exhaustive, the collection constitutes a partition of the sample space \(S\).

Module 3: Classical Definition of Probability & Combinatorial Techniques

The classical or mathematical definition of probability was formulated by Pierre-Simon Laplace in 1812.

1. Laplace's Classical Definition

If a random experiment results in \(n\) exhaustive, mutually exclusive, and equally likely outcomes, and among these, \(m\) outcomes are favorable to the occurrence of an event \(E\), then the probability of the event \(E\), denoted \(P(E)\), is defined as the ratio of favorable outcomes to the total number of possible outcomes:

\[P(E) = \frac{n(E)}{n(S)} = \frac{m}{n} = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible exhaustive outcomes}}\]
2. Equally Likely Outcomes

Outcomes are designated as equally likely (equiprobable) if, taking all relevant circumstances into consideration, none of them can be expected to occur in preference to any other. For an unbiased coin, a fair die, or a well-shuffled deck of cards, symmetry guarantees equiprobability.

3. Combinatorial Calculations in Probability

When the sample space is large, counting outcomes directly requires combinatorial principles:

  • Selection of \(r\) objects from \(n\) distinct objects: The number of ways to choose \(r\) objects without regard to order is given by combinations: \[\binom{n}{r} = \frac{n!}{r!(n - r)!}\]
  • Urn Problems (Sampling Without Replacement): If an urn contains \(w\) white balls and \(b\) black balls (total \(N = w + b\)), and \(k\) balls are drawn at random without replacement, the probability of obtaining exactly \(x\) white balls and \(k - x\) black balls is given by the hypergeometric form: \[P(X = x) = \frac{\binom{w}{x} \binom{b}{k - x}}{\binom{w + b}{k}}\]
  • Playing Card Selections: When drawing a hand of \(r\) cards from 52 cards, total possible hands equal \(\binom{52}{r}\).
4. Limitations of the Classical Definition

Although intuitive, Laplace classical definition suffers from two critical limitations:

  1. Circularity: The definition requires outcomes to be "equally likely", but "equally likely" itself means "having equal probability", thus defining probability in terms of itself.
  2. Finite and Symmetric Restriction: It completely fails when outcomes are not equally likely (e.g., a loaded die or a defective coin) or when the sample space contains an infinite number of outcomes.

These severe theoretical limitations motivated the development of Kolmogorov axiomatic approach.

Module 4: Axiomatic Approach to Probability (Kolmogorov Foundation)

In 1933, the Russian mathematician Andrey Nikolayevich Kolmogorov placed probability theory on a rigorous axiomatic foundation using measure theory.

1. The Axiomatic Framework

Let \(S\) be the sample space of a random experiment, and let \(\mathcal{F}\) be a \(\sigma\)-algebra of events (subsets of \(S\)). A probability measure \(P\) is a real-valued function that assigns to each event \(E \subseteq S\) a real number \(P(E)\), satisfying the following three fundamental axioms:

Kolmogorov's Three Axioms of Probability:
  • Axiom I (Axiom of Non-Negativity): For any event \(E \subseteq S\), \[P(E) \ge 0\]
  • Axiom II (Axiom of Certainty / Normalization): For the sure event \(S\), \[P(S) = 1\]
  • Axiom III (Axiom of Additivity): If \(A\) and \(B\) are any two mutually exclusive events (i.e., \(A \cap B = \emptyset\)), then \[P(A \cup B) = P(A) + P(B)\] More generally, for any finite or countably infinite sequence of pairwise disjoint events \(A_1, A_2, A_3, \dots\) where \(A_i \cap A_j = \emptyset\) for all \(i \neq j\): \[P\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty P(A_i)\]
2. Fundamental Theorems Deduced Directly from the Axioms

From these three simple axioms, all operational rules of probability can be rigorously proven:

  • Theorem 1: Probability of the Impossible Event is Zero.
    Since \(S \cup \emptyset = S\) and \(S \cap \emptyset = \emptyset\), Axiom III yields \(P(S \cup \emptyset) = P(S) + P(\emptyset)\). Thus \(P(S) = P(S) + P(\emptyset) \implies P(\emptyset) = 0\).
  • Theorem 2: Probability of the Complementary Event.
    For any event \(A\), since \(A \cup A' = S\) and \(A \cap A' = \emptyset\), Axiom III yields \(P(A \cup A') = P(A) + P(A') = P(S) = 1\). Therefore: \[P(A') = 1 - P(A) \quad \text{or} \quad P(A) + P(A') = 1\]
  • Theorem 3: Boundedness of Probability.
    Since \(P(A') \ge 0\) by Axiom I, \(1 - P(A) \ge 0 \implies P(A) \le 1\). Combining with Axiom I yields: \[0 \le P(A) \le 1 \quad \text{for every event } A \subseteq S\]
  • Theorem 4: Monotonicity Property.
    If \(A \subseteq B\), then \(B = A \cup (B - A)\) where \(A \cap (B - A) = \emptyset\). Thus \(P(B) = P(A) + P(B - A)\). Since \(P(B - A) \ge 0\), it follows that: \[A \subseteq B \implies P(A) \le P(B) \quad \text{and} \quad P(B - A) = P(B) - P(A)\]

Module 5: Addition Theorems & Probability of Compound Events

When events are not mutually exclusive, calculating the probability of their union requires the General Addition Theorem.

1. Addition Theorem for Any Two Events

For any two arbitrary events \(A\) and \(B\) belonging to a sample space \(S\):

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
Proof:

The union \(A \cup B\) can be decomposed into two mutually disjoint sets: \(A \cup B = A \cup (B - (A \cap B))\). By Axiom III: \[P(A \cup B) = P(A) + P(B - (A \cap B))\] Since \((A \cap B) \subseteq B\), from the difference property \(P(B - (A \cap B)) = P(B) - P(A \cap B)\). Substituting this yields: \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] When \(A\) and \(B\) are mutually exclusive, \(A \cap B = \emptyset \implies P(A \cap B) = 0\), reducing to \(P(A \cup B) = P(A) + P(B)\).

2. Addition Theorem for Three Arbitrary Events

For any three events \(A, B\), and \(C\) associated with a sample space \(S\):

\[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)\]
3. Key Derived Compound Event Formulas
Compound Event Probability Formula Application Context
Only Event A occurs (A but not B) \(P(A \cap B') = P(A - B) = P(A) - P(A \cap B)\) Subtracting overlapping region from P(A)
Only Event B occurs (B but not A) \(P(B \cap A') = P(B - A) = P(B) - P(A \cap B)\) Subtracting overlapping region from P(B)
Exactly one of A or B occurs \(P(A) + P(B) - 2P(A \cap B) = P(A \cup B) - P(A \cap B)\) Symmetric difference of events
Neither A nor B occurs (De Morgan I) \(P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)\) Complement of the union
Not both A and B occur (De Morgan II) \(P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)\) Complement of the intersection
4. Boole's & Bonferroni's Inequalities

From \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) and \(P(A \cap B) \ge 0\), we deduce Boole's Inequality:

\[P(A \cup B) \le P(A) + P(B) \quad \text{and} \quad P\left(\bigcup_{i=1}^n A_i\right) \le \sum_{i=1}^n P(A_i)\]

Furthermore, since \(P(A \cup B) \le 1\), substituting into the addition theorem yields Bonferroni's Inequality:

\[P(A \cap B) \ge P(A) + P(B) - 1\]

Module 6: Odds in Favor, Odds Against & Practical Applications

In decision analysis, gambling, actuarial science, and competitive sports, likelihood is frequently expressed in terms of odds rather than direct probabilities.

1. Odds in Favor and Odds Against

Let \(E\) be an event associated with a random experiment having \(n\) equally likely outcomes, of which \(m\) outcomes are favorable to \(E\) and the remaining \(n - m\) outcomes are unfavorable to \(E\).

  • Odds in Favor of Event \(E\): The ratio of the number of favorable outcomes to the number of unfavorable outcomes: \[\text{Odds in favor of } E = \frac{m}{n - m} = \frac{P(E)}{P(E')} = \frac{P(E)}{1 - P(E)}\] Written as the ratio \(m : (n - m)\).
  • Odds Against Event \(E\): The ratio of the number of unfavorable outcomes to the number of favorable outcomes: \[\text{Odds against } E = \frac{n - m}{m} = \frac{P(E')}{P(E)} = \frac{1 - P(E)}{P(E)}\] Written as the ratio \((n - m) : m\).
2. Conversion Formulas between Odds and Probability

Converting between odds and classical probability is governed by exact algebraic relationships:

Given Odds Statement Probability of Occurrence \(P(E)\) Probability of Non-occurrence \(P(E')\)
Odds in favor of \(E\) are \(a : b\) \(P(E) = \frac{a}{a + b}\) \(P(E') = \frac{b}{a + b}\)
Odds against \(E\) are \(a : b\) \(P(E) = \frac{b}{a + b}\) \(P(E') = \frac{a}{a + b}\)
3. Problem-Solving Strategy for Probability Examinations

When solving WBCHSE Class 11 examination questions in probability, follow this structured five-step framework:

  1. Identify the Random Experiment & Sample Space: State what experiment is being performed. Determine \(n(S)\) using combinations \(\binom{n}{r}\) or fundamental counting principles.
  2. Define Events Symbolically: Clearly define letters \(A, B, C\) representing the events mentioned in the problem.
  3. Translate Language to Set Notation: Convert statements like "at least one", "neither", "only A", and "both" into set expressions (\(A \cup B\), \(A' \cap B'\), \(A \cap B'\), \(A \cap B\)).
  4. Check for Mutual Exclusiveness: Check whether \(A \cap B = \emptyset\). If yes, use \(P(A \cup B) = P(A) + P(B)\); if not, subtract \(P(A \cap B)\).
  5. Sanity Check Final Answer: Ensure the calculated value is a valid probability lying in \([0, 1]\).

Key Formulas, Identities & Theorems

Classical (Laplace) Probability
$$P(E) = \frac{n(E)}{n(S)} = \frac{m}{n}$$
Kolmogorov Axioms of Probability
$$P(E) \ge 0, \quad P(S) = 1, \quad P(A \cup B) = P(A) + P(B) \; (\text{if } A \cap B = \emptyset)$$
General Addition Theorem for Two Events
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Addition Theorem for Three Events
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$$
Complementary Events & De Morgan Laws
$$P(A') = 1 - P(A), \quad P(A' \cap B') = 1 - P(A \cup B), \quad P(A' \cup B') = 1 - P(A \cap B)$$
Odds in Favor and Odds Against Conversion
$$\text{Odds in Favor } = \frac{P(E)}{1 - P(E)} = \frac{m}{n - m}, \quad P(E) = \frac{a}{a + b} \; (\text{from } a:b)$$

Conceptual Solved Examples & Case Studies

Example 1
A card is drawn at random from a well-shuffled pack of 52 playing cards. Find the probability of drawing: (i) a red face card, and (ii) a black king or a red ace.
Step-by-Step Solution:

Step 1: Determine Total Sample Space

A standard deck contains \(n(S) = 52\) cards.

Step 2: Sub-problem (i) — Red Face Card

Face cards are Jacks, Queens, and Kings (3 denominations per suit). There are two red suits: Hearts (♥) and Diamonds (♦).

\[\text{Number of red face cards } m_1 = 2 \times 3 = 6 \; (\text{J♥, Q♥, K♥, J♦, Q♦, K♦})\] \[P(\text{Red Face Card}) = \frac{m_1}{n(S)} = \frac{6}{52} = \frac{3}{26}\]

Step 3: Sub-problem (ii) — Black King or Red Ace

Let \(A\) be the event of drawing a black king, and \(B\) be the event of drawing a red ace. There are 2 black kings (Spades ♠, Clubs ♣) and 2 red aces (Hearts ♥, Diamonds ♦). Since a card cannot be simultaneously a king and an ace, the events are mutually exclusive: \(A \cap B = \emptyset\).

\[n(A \cup B) = n(A) + n(B) = 2 + 2 = 4\] \[P(A \cup B) = \frac{4}{52} = \frac{1}{13}\]

Final Answer: (i) \(\frac{3}{26}\), (ii) \(\frac{1}{13}\).

Example 2
Two unbiased fair dice are thrown simultaneously. Find the probability of obtaining: (i) a doublet, (ii) a total sum greater than 9, and (iii) an even sum or a sum divisible by 3.
Step-by-Step Solution:

Step 1: Total Sample Space

Each die has 6 faces. The total number of outcomes is \(n(S) = 6 \times 6 = 36\).

Step 2: Sub-problem (i) — Getting a Doublet

A doublet means both dice show the same number: \(E_1 = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}\).

\[n(E_1) = 6 \implies P(E_1) = \frac{6}{36} = \frac{1}{6}\]

Step 3: Sub-problem (ii) — Sum Greater than 9

Possible sums greater than 9 are 10, 11, and 12:

  • Sum = 10: \(\{(4,6), (5,5), (6,4)\}\) (3 pairs)
  • Sum = 11: \(\{(5,6), (6,5)\}\) (2 pairs)
  • Sum = 12: \(\{(6,6)\}\) (1 pair)
\[n(E_2) = 3 + 2 + 1 = 6 \implies P(E_2) = \frac{6}{36} = \frac{1}{6}\]

Step 4: Sub-problem (iii) — Even Sum or Sum Divisible by 3

Let \(A\) = event that the sum is even, and \(B\) = event that the sum is divisible by 3.

  • Even sums (2, 4, 6, 8, 10, 12): There are exactly 18 even sum pairs \(\implies n(A) = 18, P(A) = \frac{18}{36}\).
  • Sums divisible by 3 (sums 3, 6, 9, 12):
    • Sum = 3: \(\{(1,2), (2,1)\}\) (2)
    • Sum = 6: \(\{(1,5), (2,4), (3,3), (4,2), (5,1)\}\) (5)
    • Sum = 9: \(\{(3,6), (4,5), (5,4), (6,3)\}\) (4)
    • Sum = 12: \(\{(6,6)\}\) (1)
    Total \(n(B) = 2 + 5 + 4 + 1 = 12 \implies P(B) = \frac{12}{36}\).
  • Intersection \(A \cap B\): Sum is both even AND divisible by 3, meaning the sum is a multiple of 6 (sum = 6 or sum = 12). \[n(A \cap B) = 5 + 1 = 6 \implies P(A \cap B) = \frac{6}{36}\]

By the General Addition Theorem:

\[P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{18}{36} + \frac{12}{36} - \frac{6}{36} = \frac{24}{36} = \frac{2}{3}\]

Final Answer: (i) \(\frac{1}{6}\), (ii) \(\frac{1}{6}\), (iii) \(\frac{2}{3}\).

Example 3
Given that P(A) = 3/8, P(B) = 1/2, and P(A ∩ B) = 1/4, calculate: (i) P(A ∪ B), (ii) P(A' ∩ B'), (iii) P(A ∩ B'), and (iv) P(A' ∪ B').
Step-by-Step Solution:

Step 1: Calculate P(A ∪ B)

By the General Addition Theorem:

\[P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{8} + \frac{1}{2} - \frac{1}{4}\] \[P(A \cup B) = \frac{3}{8} + \frac{4}{8} - \frac{2}{8} = \frac{5}{8}\]

Step 2: Calculate P(A' ∩ B') (Neither A nor B)

By De Morgan's First Law, \(A' \cap B' = (A \cup B)'\):

\[P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8}\]

Step 3: Calculate P(A ∩ B') (Only A occurs)

Using the probability difference theorem \(P(A - B) = P(A \cap B') = P(A) - P(A \cap B)\):

\[P(A \cap B') = \frac{3}{8} - \frac{1}{4} = \frac{3}{8} - \frac{2}{8} = \frac{1}{8}\]

Step 4: Calculate P(A' ∪ B') (Not both A and B)

By De Morgan's Second Law, \(A' \cup B' = (A \cap B)'\):

\[P(A' \cup B') = 1 - P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}\]

Final Answer: (i) \(\frac{5}{8}\), (ii) \(\frac{3}{8}\), (iii) \(\frac{1}{8}\), (iv) \(\frac{3}{4}\).

Example 4
A committee of 4 persons is to be chosen at random from a group of 5 men, 4 women, and 3 children. Find the probability that: (i) exactly 2 men are selected, (ii) at least 1 woman is selected, and (iii) no child is selected.
Step-by-Step Solution:

Step 1: Compute Total Sample Space n(S)

Total number of available persons = \(5 + 4 + 3 = 12\). The number of ways to select 4 persons from 12 is:

\[n(S) = \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495\]

Step 2: Sub-problem (i) — Exactly 2 Men Selected

To have exactly 2 men, we choose 2 men from 5, and the remaining \(4 - 2 = 2\) persons from the \(4 + 3 = 7\) non-men (women and children):

\[n(E_1) = \binom{5}{2} \times \binom{7}{2} = \left(\frac{5 \times 4}{2 \times 1}\right) \times \left(\frac{7 \times 6}{2 \times 1}\right) = 10 \times 21 = 210\] \[P(E_1) = \frac{210}{495} = \frac{14}{33}\]

Step 3: Sub-problem (ii) — At Least 1 Woman Selected

We use complementary counting: \(P(\text{at least 1 woman}) = 1 - P(\text{no woman})\). Selecting 4 persons with zero women means selecting all 4 from the \(5 + 3 = 8\) remaining people (men and children):

\[n(\text{no woman}) = \binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\] \[P(\text{no woman}) = \frac{70}{495} = \frac{14}{99}\] \[P(\text{at least 1 woman}) = 1 - \frac{14}{99} = \frac{85}{99}\]

Step 4: Sub-problem (iii) — No Child Selected

Selecting 4 persons with no children means choosing all 4 from the \(5 + 4 = 9\) adults (men and women):

\[n(E_3) = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126\] \[P(E_3) = \frac{126}{495} = \frac{14}{55}\]

Final Answer: (i) \(\frac{14}{33}\), (ii) \(\frac{85}{99}\), (iii) \(\frac{14}{55}\).

Example 5
In a high school class of 60 students, 30 opted for NCC, 32 opted for NSS, and 24 opted for both NCC and NSS. If one student is selected at random, find the probability that: (i) the student opted for NCC or NSS, (ii) the student opted for neither NCC nor NSS, and (iii) the student opted for NSS but not NCC. Also, determine the odds in favor of and against a student opting for neither.
Step-by-Step Solution:

Step 1: Define Given Quantities in Set Notation

Let \(A\) = student opted for NCC, \(B\) = student opted for NSS. Total students \(n(S) = 60\).

\[n(A) = 30 \implies P(A) = \frac{30}{60} = \frac{1}{2}\] \[n(B) = 32 \implies P(B) = \frac{32}{60} = \frac{8}{15}\] \[n(A \cap B) = 24 \implies P(A \cap B) = \frac{24}{60} = \frac{2}{5}\]

Step 2: Sub-problem (i) — Opted for NCC or NSS

\[P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{30}{60} + \frac{32}{60} - \frac{24}{60} = \frac{38}{60} = \frac{19}{30}\]

Step 3: Sub-problem (ii) — Opted for Neither NCC nor NSS

By De Morgan's First Law:

\[P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{19}{30} = \frac{11}{30}\]

Step 4: Sub-problem (iii) — Opted for NSS but not NCC

\[P(B \cap A') = P(B - A) = P(B) - P(A \cap B) = \frac{32}{60} - \frac{24}{60} = \frac{8}{60} = \frac{2}{15}\]

Step 5: Odds in Favor and Against Opting for Neither

Let \(E\) be the event of opting for neither. Here \(P(E) = \frac{11}{30}\), so \(P(E') = 1 - \frac{11}{30} = \frac{19}{30}\).

\[\text{Odds in favor of } E = \frac{P(E)}{P(E')} = \frac{11/30}{19/30} = \frac{11}{19} \implies 11 : 19\] \[\text{Odds against } E = \frac{P(E')}{P(E)} = \frac{19/30}{11/30} = \frac{19}{11} \implies 19 : 11\]

Final Answer: (i) \(\frac{19}{30}\), (ii) \(\frac{11}{30}\), (iii) \(\frac{2}{15}\); Odds in favor: \(11 : 19\); Odds against: \(19 : 11\).

Example 6
Three candidates A, B, and C are contesting for a single scholarship nomination. The odds in favor of candidate A are 1 : 2, and the odds against candidate B are 3 : 2. Assuming only one candidate can win and the winning events are mutually exclusive: (i) Find the probabilities P(A) and P(B). (ii) If A, B, and C together form an exhaustive set of outcomes, find P(C) and the odds in favor of candidate C. (iii) If in another independent scenario P(C) is measured to be 1/10, prove whether the three candidates remain exhaustive or not.
Step-by-Step Solution:

Step 1: Determine P(A) from Odds in Favor

The odds in favor of candidate \(A\) winning are \(1 : 2\). By the odds conversion formula with \(a = 1, b = 2\):

\[P(A) = \frac{a}{a + b} = \frac{1}{1 + 2} = \frac{1}{3}\]

Step 2: Determine P(B) from Odds Against

The odds against candidate \(B\) winning are \(3 : 2\). By the odds conversion formula with \(a = 3, b = 2\):

\[P(B) = \frac{b}{a + b} = \frac{2}{3 + 2} = \frac{2}{5}\]

Step 3: Sub-problem (ii) — Finding P(C) Under Exhaustive Assumption

Since only one candidate can win, the events \(A, B, C\) are pairwise mutually exclusive: \(A \cap B = \emptyset, B \cap C = \emptyset, C \cap A = \emptyset\). Given that the candidates are exhaustive, their union covers the entire sample space: \(A \cup B \cup C = S\). By Kolmogorov Axioms II and III:

\[P(A \cup B \cup C) = P(A) + P(B) + P(C) = P(S) = 1\] \[\frac{1}{3} + \frac{2}{5} + P(C) = 1\]

Find the common denominator of \(1/3\) and \(2/5\) (which is 15):

\[\frac{5}{15} + \frac{6}{15} + P(C) = 1 \implies \frac{11}{15} + P(C) = 1 \implies P(C) = 1 - \frac{11}{15} = \frac{4}{15}\]

The odds in favor of candidate \(C\) are:

\[\text{Odds in favor of } C = \frac{P(C)}{1 - P(C)} = \frac{4/15}{11/15} = \frac{4}{11} \implies 4 : 11\]

Step 4: Sub-problem (iii) — Testing Exhaustiveness when P(C) = 1/10

Compute the total probability of the union under mutual exclusiveness:

\[P(A \cup B \cup C) = P(A) + P(B) + P(C) = \frac{1}{3} + \frac{2}{5} + \frac{1}{10}\]

The lowest common denominator of 3, 5, and 10 is 30:

\[P(A \cup B \cup C) = \frac{10}{30} + \frac{12}{30} + \frac{3}{30} = \frac{25}{30} = \frac{5}{6}\]

For a set of events to be exhaustive, Kolmogorov Axiom II requires \(P(A \cup B \cup C) = P(S) = 1\). Here, \(P(A \cup B \cup C) = \frac{5}{6} < 1\). There exists a non-empty residual event \(E_{\text{other}} = S - (A \cup B \cup C)\) with probability: \[P(E_{\text{other}}) = 1 - \frac{5}{6} = \frac{1}{6} > 0\] Therefore, the three candidates do not cover all possible outcomes; they are NOT exhaustive.

Final Answer: (i) \(P(A) = \frac{1}{3}\), \(P(B) = \frac{2}{5}\); (ii) \(P(C) = \frac{4}{15}\), Odds in favor of \(C = 4 : 11\); (iii) Not exhaustive because \(P(A \cup B \cup C) = \frac{5}{6} \neq 1\).

Common Misconceptions & Examiner Traps

Common Misconception

Scientific Reality & Correction

Mutually exclusive and independent are fundamentally different concepts. Mutually exclusive means events cannot happen together: A ∩ B = ∅, which implies P(A ∩ B) = 0. In contrast, two events are independent if the occurrence of one does not affect the occurrence of the other: P(A ∩ B) = P(A) × P(B). If two events have non-zero probabilities, they CANNOT be both mutually exclusive and independent simultaneously, because P(A ∩ B) cannot simultaneously equal 0 and a positive product P(A)P(B).

Common Misconception

Scientific Reality & Correction

Odds in favor is defined as favorable : unfavorable = m : (n - m) = a : b, giving P(E) = a / (a + b). Odds against is defined as unfavorable : favorable = (n - m) : m = a : b, giving P(E) = b / (a + b). Students frequently confuse the numerator and denominator, calculating a / (a + b) for odds against and causing catastrophic calculation errors.

Common Misconception

Scientific Reality & Correction

Writing P(A ∪ B) = P(A) + P(B) without checking whether the events are disjoint. When A ∩ B is non-empty, adding P(A) and P(B) counts the intersection twice, frequently yielding an absurd probability greater than 1.

Common Misconception

Scientific Reality & Correction

Students frequently translate "neither A nor B occurs" as 1 - P(A ∩ B). In truth, 1 - P(A ∩ B) = P(A' ∪ B'), which means "not both A and B occur". The phrase "neither A nor B" means A does not occur AND B does not occur: A' ∩ B' = (A ∪ B)', whose probability is 1 - P(A ∪ B).

Common Misconception

Scientific Reality & Correction

When drawing objects successively without replacement (e.g. cards from a pack or balls from an urn), students often divide every stage probability by the original total count n rather than decreasing counts (n, n - 1, n - 2).

Probability Foundations, Kolmogorov Axioms & Addition Theorems Architecture

P(E) Probability — Core Foundations, Axiomatic System & Addition Theorems WBCHSE Class 11 Mathematics • Unit V: Statistics & Probability 1. Sample Space & Event Algebra Sample Space S, Events E ⊆ S, Mutually Exclusive & Exhaustive S (Universal Space) A B A ∩ B Sample Space & Events S = set of all outcomes; Event E ⊆ S Mutually Exclusive Events A ∩ B = ∅ ⇒ Disjoint (cannot occur together) Exhaustive Events A ∪ B = S ⇒ Together cover entire space 2. Kolmogorov's Axiomatic System Non-negativity, Certainty P(S)=1, and Finite Additivity Classical / Laplace Definition: P(E) = n(E) / n(S) = m / n Kolmogorov's Three Axioms of Probability: 1 Non-Negativity Axiom 0 ≤ P(E) ≤ 1 for every event E 2 Certainty Axiom P(S) = 1 (Sure event) & P(∅) = 0 3 Additivity Axiom If A ∩ B = ∅ (Mutually Exclusive): P(A ∪ B) = P(A) + P(B) 3. Addition Theorems & Odds General Events, De Morgan's Laws & Odds in Favor/Against General Addition Theorem (Any Two Events): P(A ∪ B) = P(A) + P(B) - P(A ∩ B) P(A - B) = P(A ∩ B') = P(A) - P(A ∩ B) Three Events Addition Theorem: P(A∪B∪C) = ΣP(A) - ΣP(A∩B) + P(A ∩ B ∩ C) De Morgan's Laws in Probability: P(A' ∩ B') = 1 - P(A ∪ B) P(A' ∪ B') = 1 - P(A ∩ B) Odds in Favor & Odds Against: Odds in favor a:b ⇒ P(E) = a/(a+b) Odds against b:a ⇒ P(E') = b/(a+b) TargetExams Gold Standard Study Material • WBCHSE Class 11 Mathematics WBCHSE Class 11 • Ch 14

Chapter Summary & 10 Key Takeaways

Takeaway 1
A random experiment is an experiment whose all possible outcomes are known in advance, but the exact outcome of any specific trial cannot be predicted with certainty.
Takeaway 2
The sample space S is the set of all possible outcomes of a random experiment; each individual element of S is called a sample point.
Takeaway 3
An event E is any subset of the sample space S (E ⊆ S); an elementary event contains exactly one sample point, while a compound event contains two or more.
Takeaway 4
Two events A and B are mutually exclusive (disjoint) if they cannot occur simultaneously: A ∩ B = ∅, which implies P(A ∩ B) = 0.
Takeaway 5
A collection of events is exhaustive if their union equals the entire sample space: A ∪ B ∪ ... ∪ K = S, which implies P(A ∪ B ∪ ... ∪ K) = 1.
Takeaway 6
Under the classical Laplace definition with equally likely outcomes, the probability of an event E is P(E) = n(E) / n(S) = m / n.
Takeaway 7
Kolmogorov Three Axioms state: (1) Non-negativity P(E) ≥ 0, (2) Certainty P(S) = 1, and (3) Additivity P(A ∪ B) = P(A) + P(B) whenever A ∩ B = ∅.
Takeaway 8
For any two arbitrary events, the General Addition Theorem is P(A ∪ B) = P(A) + P(B) - P(A ∩ B); if mutually exclusive, P(A ∪ B) = P(A) + P(B).
Takeaway 9
By De Morgan Laws, the probability of neither event occurring is P(A' ∩ B') = 1 - P(A ∪ B), and the probability of at most one event occurring is P(A' ∪ B') = 1 - P(A ∩ B).
Takeaway 10
If the odds in favor of an event E are a : b, then P(E) = a / (a + b); if the odds against E are a : b, then P(E) = b / (a + b).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
What is the probability that a leap year selected at random will contain 53 Sundays?
Reveal Answer & Explanation
Answer: A leap year contains 366 days = 52 weeks + 2 extra days. The 52 complete weeks guarantee 52 Sundays. The 2 extra days can be any of the 7 consecutive pairs: {(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}. Among these 7 equally likely pairs, exactly 2 pairs contain Sunday: (Sun, Mon) and (Sat, Sun). Hence, the probability of 53 Sundays is 2/7.
A leap year has 366 days = 52 complete weeks (364 days) plus 2 extra consecutive days. Write down the sample space of the 2 consecutive days.
2
If P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.8, find: (i) P(A ∩ B), and (ii) P(A' ∩ B).
Reveal Answer & Explanation
Answer: (i) P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.6 + 0.5 - 0.8 = 0.3. (ii) P(A' ∩ B) = P(B - A) = P(B) - P(A ∩ B) = 0.5 - 0.3 = 0.2.
Use P(A ∩ B) = P(A) + P(B) - P(A ∪ B) for part (i), and P(A' ∩ B) = P(B) - P(A ∩ B) for part (ii).
3
Two letters are chosen at random from the 11 letters of the word "PROBABILITY". What is the probability that both chosen letters are vowels?
Reveal Answer & Explanation
Answer: The word "PROBABILITY" contains 11 letters. The vowels are O, A, I, I (total 4 vowels). The total ways to choose 2 letters from 11 is C(11, 2) = (11 × 10) / 2 = 55. The number of ways to choose 2 vowels from 4 is C(4, 2) = (4 × 3) / 2 = 6. Therefore, the probability that both letters are vowels is 6/55.
Count total letters and vowels (O, A, I, I). Note that the letter I appears twice.
4
The odds against a candidate passing a civil services examination are 7 : 3. What is the probability that the candidate passes the examination?
Reveal Answer & Explanation
Answer: Given odds against passing are 7 : 3. Here unfavorable cases a = 7 and favorable cases b = 3. Total cases = 7 + 3 = 10. The probability of passing is P(Pass) = b / (a + b) = 3 / (7 + 3) = 3/10 = 0.3.
If odds against an event are a : b, then the probability of occurrence is b / (a + b).
5
Can two events A and B have P(A) = 0.7, P(B) = 0.8, and P(A ∩ B) = 0.4? Justify your answer using Kolmogorov axioms.
Reveal Answer & Explanation
Answer: By the addition theorem: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.7 + 0.8 - 0.4 = 1.1. According to Kolmogorov Axiom II and its direct corollary, the probability of any event cannot exceed 1 (P(E) ≤ 1 for all E ⊆ S). Since P(A ∪ B) = 1.1 > 1, this violates the fundamental axioms of probability. Therefore, such a probability assignment is impossible.
Compute P(A ∪ B) using the addition theorem and check whether it obeys Axiom II (P(E) ≤ 1).
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