Class 11 Maths Chapter 16 Probability Notes

Key Points to Remember

• Probability Range: The probability of any event E, denoted P(E), always satisfies 0 ≤ P(E) ≤ 1.
• Sure Event: The probability of a sure event (sample space S) is P(S) = 1.
• Impossible Event: The probability of an impossible event (null set φ) is P(φ) = 0.
• Complement of an Event: P(E') = 1 - P(E), where E' is the complement of event E.
• Mutually Exclusive Events: If A and B are mutually exclusive (disjoint), then P(A ∩ B) = 0.
• Addition Theorem: For any two events A and B, P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• For Mutually Exclusive Events: If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B).
• Exhaustive Events: If E₁, E₂, ..., Eₙ are mutually exclusive and exhaustive events, then P(E₁) + P(E₂) + ... + P(Eₙ) = 1.
• Classical Definition: For an event E in a finite sample space S with equally likely outcomes, P(E) = (Number of outcomes favorable to E) / (Total number of possible outcomes).

Core Definitions

Random Experiment

An experiment whose outcome cannot be predicted with certainty, but all possible outcomes are known, and it can be repeated under identical conditions.

Sample Space (S)

The set of all possible outcomes of a random experiment. It is denoted by 'S'.

Event

A subset of the sample space. An event is a collection of some or all possible outcomes.

Elementary Event

An event containing only one sample point (single outcome) of the sample space.

Compound Event

An event containing more than one sample point of the sample space.

Mutually Exclusive Events

Two events A and B are mutually exclusive if the occurrence of one precludes the occurrence of the other. In set notation, A ∩ B = φ (empty set).

Exhaustive Events

A set of events E₁, E₂, ..., Eₙ is exhaustive if their union is the entire sample space, i.e., E₁ ∪ E₂ ∪ ... ∪ Eₙ = S. This means at least one of the events must occur.

Complement of an Event (E')

The set of all outcomes in the sample space S that are not in event E. Also denoted as S - E.

The Axiomatic Approach to Probability

While the classical definition of probability (favorable outcomes over total outcomes) works well for experiments with equally likely outcomes and a finite sample space, it has limitations. It fails when outcomes are not equally likely or when the sample space is infinite. To address this, the axiomatic approach provides a more rigorous and general foundation for probability theory. This approach defines probability as a function P, which assigns a real number to each event E, satisfying specific axioms:

1. Axiom 1 (Non-negativity): For any event E, P(E) ≥ 0. The probability of an event cannot be negative.
2. Axiom 2 (Normalization): The probability of the sample space S is 1, i.e., P(S) = 1. This means that an outcome from the experiment must occur.
3. Axiom 3 (Additivity for Mutually Exclusive Events): If A and B are mutually exclusive events (i.e., A ∩ B = φ), then P(A ∪ B) = P(A) + P(B). This axiom can be extended to any finite number of mutually exclusive events: P(E₁ ∪ E₂ ∪ ... ∪ Eₙ) = P(E₁) + P(E₂) + ... + P(Eₙ) if Eᵢ ∩ Eⱼ = φ for i ≠ j.

These three axioms form the basis from which all other properties of probability can be derived, including formulas for complements, general addition rules, and probabilities involving conditional events (though conditional probability is typically a Class 12 topic, its foundation rests here).

Classical vs. Axiomatic Probability

Worked Examples

• {"title":"Tossing Three Coins","bodyMarkdown":"Q: Three unbiased coins are tossed. Find the probability of getting exactly two heads.\n\nA:\n1. Sample Space (S): {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Total outcomes n(S) = 2³ = 8.\n2. Event (E) - Exactly two heads: {HHT, HTH, THH}. Favorable outcomes n(E) = 3.\n3. Probability P(E): n(E) / n(S) = 3 / 8."}
• {"title":"Drawing a Card","bodyMarkdown":"Q: A card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is a King or a Spade?\n\nA:\n1. Total outcomes n(S): 52.\n2. Event K (King): There are 4 Kings. P(K) = 4/52.\n3. Event S (Spade): There are 13 Spades. P(S) = 13/52.\n4. Event K ∩ S (King of Spades): There is 1 King of Spades. P(K ∩ S) = 1/52.\n5. P(K ∪ S) = P(K) + P(S) - P(K ∩ S) (Addition Theorem)\n P(K ∪ S) = (4/52) + (13/52) - (1/52) = (17 - 1)/52 = 16/52 = 4/13."}

Exam Traps & Tips

1. Correct Sample Space: Always list out or correctly identify the entire sample space first. Many errors stem from an incomplete or incorrect S. For example, for tossing two coins, the outcomes are HH, HT, TH, TT (4 outcomes), not HH, HT, TT (3 outcomes) if you mistakenly assume HT and TH are the same.
2. 'And' vs. 'Or': Understand the difference between 'AND' (intersection, A ∩ B) and 'OR' (union, A ∪ B). 'And' implies both events must occur; 'Or' implies at least one must occur.
3. Mutually Exclusive vs. Independent: In Class 11, focus heavily on mutually exclusive events. Remember that mutually exclusive events cannot happen at the same time (A ∩ B = φ), while independent events (covered in Class 12) do not affect each other's occurrence. These are distinct concepts!
4. Complement Rule: Don't forget P(E') = 1 - P(E). This is often easier for 'at least one' type problems. For example, 'probability of at least one head' can be calculated as 1 - P(no heads).
5. Show Your Work: Even for simple problems, clearly state the sample space, favorable outcomes, and the formula used. This helps secure partial marks even if the final answer has a calculation error.

Practice Questions with Solutions

• Q: What is the sample space when a die is rolled twice? A: S = {(1,1), (1,2), ..., (6,6)}, total 36 outcomes.
• Q: If P(A) = 0.6, P(B) = 0.3, and P(A ∩ B) = 0.2, find P(A ∪ B). A: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.6 + 0.3 - 0.2 = 0.7.
• Q: Can the probability of an event be -0.5? Justify. A: No, probability must always be a non-negative value, i.e., 0 ≤ P(E) ≤ 1, according to Axiom 1 of the axiomatic approach.
• Q: A bag contains 3 red and 2 blue balls. A ball is drawn at random. What is the probability it is red? A: Total balls = 5. Favorable (red) = 3. P(Red) = 3/5.

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