Chapter 9 • Unit IV

Exercise 9.2: Classical and Axiomatic Approach to Probability

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

9.2 Classical and Axiomatic Approach to Probability

In Applied Mathematics, insurance risk valuation, betting odds calculation, and quality sampling, probability translates intuition into an exact numeric scale bounded between 0 and 1. While the classical approach relies on equally likely symmetry, Kolmogorov's Axiomatic Framework provides the modern mathematical foundation that applies universally across finite and continuous probability models.

Fundamental Definitions & Kolmogorov's Three Axioms

1. Classical (Laplacian) Probability

If a random experiment has \(n(S) = N\) mutually exclusive, exhaustive, and equally likely elementary outcomes, and \(n(E) = m\) of these outcomes are favorable to event \(E\): \[ P(E) = \frac{n(E)}{n(S)} = \frac{m}{N} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}} \]

2. Kolmogorov's Axiomatic Definition

Let \(S\) be a sample space. A probability measure \(P\) is a real-valued function defined on all events \(E \subseteq S\) satisfying three axioms:

  • Axiom 1 (Non-Negativity): For any event \(E\): \[ 0 \le P(E) \le 1 \]
  • Axiom 2 (Total Certainty): The probability of the sure event \(S\) is unity: \[ P(S) = 1 \]
  • Axiom 3 (Additivity of Disjoint Events): If \(A\) and \(B\) are mutually exclusive events (\(A \cap B = \emptyset\)): \[ P(A \cup B) = P(A) + P(B) \] Generalized to any sequence of pairwise disjoint events: \(P\left(\bigcup_{i=1}^k E_i\right) = \sum_{i=1}^k P(E_i)\).

3. Immediate Theoretical Corollaries

  • Impossible Event: \(P(\emptyset) = 0\).
  • Complementary Event Rule: \[ P(E') = 1 - P(E) \iff P(E) + P(E') = 1 \]
  • Subset Monotonicity: \[ A \subseteq B \implies P(A) \le P(B) \quad \text{and} \quad P(B - A) = P(B) - P(A) \]
  • Sum of Elementary Probabilities: For any discrete sample space \(S = \{\omega_1, \omega_2, \dots, \omega_n\}\): \[ \sum_{i=1}^n P(\omega_i) = 1 \quad \text{where } 0 \le P(\omega_i) \le 1 \]

4. Odds in Favor and Odds Against

  • Odds in Favor of Event \(E\): \[ \text{Odds in Favor} = \frac{P(E)}{P(E')} = \frac{m}{n} \implies P(E) = \frac{m}{m + n} \]
  • Odds Against Event \(E\): \[ \text{Odds Against} = \frac{P(E')}{P(E)} = \frac{n}{m} \implies P(E) = \frac{m}{m + n} \]

Comprehensive Card Sampling Example: Hypergeometric Selection

Problem: Four cards are drawn simultaneously at random from a well-shuffled pack of 52 cards. What is the probability that:

  1. All four cards are Aces?
  2. All four cards are of the same suit?
  3. There is exactly one card of each suit?
  4. All four cards are red face cards?

Solution:

The total number of ways to choose 4 cards out of 52 is:

\[ n(S) = \binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270{,}725. \]
  1. All four cards are Aces: There are 4 Aces in the deck, so there is only \(\binom{4}{4} = 1\) favorable way: \[ P(\text{4 Aces}) = \frac{1}{270{,}725}. \]
  2. All four cards are of the same suit: Choose 1 of the 4 suits, and choose 4 cards from that suit's 13 cards: \[ n(E) = \binom{4}{1} \times \binom{13}{4} = 4 \times 715 = 2{,}860. \] \[ P(\text{Same Suit}) = \frac{2{,}860}{270{,}725} = \frac{44}{4{,}165} \approx 0.01056. \]
  3. One card of each suit: Choose 1 card from Hearts, 1 from Diamonds, 1 from Clubs, 1 from Spades: \[ n(E) = \binom{13}{1} \times \binom{13}{1} \times \binom{13}{1} \times \binom{13}{1} = 13^4 = 28{,}561. \] \[ P(\text{One of Each Suit}) = \frac{28{,}561}{270{,}725} = \frac{2{,}197}{20{,}825} \approx 0.1055. \]
  4. All four are red face cards: Red face cards (J, Q, K of Hearts and Diamonds) = 6 cards. We choose 4 from these 6: \[ n(E) = \binom{6}{4} = \binom{6}{2} = 15. \] \[ P(\text{4 Red Face Cards}) = \frac{15}{270{,}725} = \frac{3}{54{,}145} \approx 0.0000554. \]
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