Axiomatic Approach to Probability — Class 11 Statistics And Probability Notes

1. Introduction

The classical definition of probability (number of favorable outcomes / total number of outcomes) works perfectly when all outcomes are equally likely, like in a fair coin toss or a die roll. However, what if the coin is biased? Or what if the events are more complex? To handle all scenarios, we use a more formal and powerful approach developed by the Russian mathematician A.N. Kolmogorov, known as the axiomatic approach.

Think of the axioms of probability like the rules of a game, such as chess. ♟️ We don't prove the rules themselves (e.g., "a bishop moves diagonally"); we accept them as the foundation. Using these basic rules, we can then derive every complex strategy and theorem about the game. In this approach, probability is a function, $P$ , that assigns a real number to every event, and this function must follow three fundamental rules (axioms).

2. The Axioms of Probability

Let $S$ be a sample space and let $P$ be the probability function defined on the events of $S$ . The following three axioms must hold:

Axiom 1: Non-negativity For any event E, the probability of E is greater than or equal to zero. $P(E) \ge 0$ Explanation: This means a probability can be 0% (impossible) or 50% (even chance), but it can never be negative. A '-20% chance of rain' is meaningless.

Axiom 2: Certainty The probability of the sure event (the entire sample space) is 1. $P(S) = 1$ Explanation: This sets the scale for probability. The total probability of all possible outcomes combined must equal 1 (or 100%). Something within the sample space is guaranteed to happen.

Axiom 3: Additivity If E and F are two mutually exclusive events (meaning they cannot happen at the same time, i.e., $E \cap F = \emptyset$ ), then the probability that at least one of them occurs is the sum of their individual probabilities. $P(E \cup F) = P(E) + P(F) \quad \text{(for mutually exclusive events)}$ Explanation: When rolling a die, the events 'getting a 1' and 'getting a 2' are mutually exclusive. So, the probability of getting a 1 OR a 2 is simply $P(1) + P(2) = 1/6 + 1/6 = 2/6$ . This axiom extends to any number of mutually exclusive events.

3. Key Theorems and Consequences

From these three simple axioms, all other rules of probability can be derived as theorems.

Probability of the Impossible Event: The probability of the impossible event (empty set) is 0. $P(\emptyset) = 0$ Proof: The sure event S and the impossible event $\emptyset$ are mutually exclusive, and their union is S. By Axiom 3, $P(S \cup \emptyset) = P(S) + P(\emptyset)$ . Since $S \cup \emptyset = S$ , this means $P(S) = P(S) + P(\emptyset)$ , which implies $P(\emptyset)=0$ .
Range of Probability: For any event E, its probability is always between 0 and 1, inclusive. $0 \le P(E) \le 1$
Probability of the Complement (The 'Not' Rule): The probability of an event E not happening is 1 minus the probability that it does happen. $P(E') = 1 - P(E)$
The General Addition Rule (The 'Or' Rule): For any two events A and B (not necessarily mutually exclusive), the probability of A or B occurring is: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Explanation: We subtract the probability of the intersection, $P(A \cap B)$ , because it was counted twice (once in P(A) and once in P(B)).
Probability of 'A but not B': $P(A - B) = P(A \cap B') = P(A) - P(A \cap B)$

Example 1: Basic Probability

Question: A single fair die is rolled. What is the probability of rolling an odd number?

Solution: The sample space is $S = \{1, 2, 3, 4, 5, 6\}$ , so the total number of possible outcomes is $n(S) = 6$ .

The event of rolling an odd number is $E = \{1, 3, 5\}$ , so the number of favorable outcomes is $n(E) = 3$ .

Since all outcomes are equally likely, the probability is: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{n(E)}{n(S)} = \frac{3}{6} = 0.5$

Example 2: Mutually Exclusive Events

Question: A card is drawn from a standard 52-card deck. What is the probability that the card is a Jack or a Queen?

Solution: Let J be the event 'drawing a Jack' and Q be the event 'drawing a Queen'.

$P(J) = \frac{4}{52}$ and $P(Q) = \frac{4}{52}$ .

Since a card cannot be both a Jack and a Queen at the same time, the events are mutually exclusive.

Therefore, by Axiom 3, the probability of drawing a Jack or a Queen is the sum of their individual probabilities:

$P(J \cup Q) = P(J) + P(Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$ .

Example 3: The Complement Rule

Question: The probability of winning a game is 0.3. What is the probability of not winning?

Solution: Let W be the event 'winning the game'. We are given $P(W) = 0.3$ .

The event 'not winning' is the complement of W, denoted as W'.

Using the complement rule, $P(W') = 1 - P(W)$ . $P(W') = 1 - 0.3 = 0.7$ .

Example 4: The General Addition Rule

Question: A card is drawn from a standard 52-card deck. What is the probability that the card is a Spade or an Ace?

Solution: Let A be the event 'drawing a Spade' and B be the event 'drawing an Ace'. These events are not mutually exclusive because the 'Ace of Spades' is in both sets.

$P(A) = \frac{13}{52}$ (since there are 13 spades).
$P(B) = \frac{4}{52}$ (since there are 4 aces).
$P(A \cap B)$ is the probability of drawing the Ace of Spades, which is $\frac{1}{52}$ .

Using the general addition rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$ .

Example 5: Finding the Intersection

Question: Given $P(A) = 0.5$ , $P(B) = 0.4$ , and $P(A \cup B) = 0.7$ , find $P(A \cap B)$ .

Solution: We start with the general addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

We can rearrange this formula to solve for the intersection:

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

Substituting the given values:

$P(A \cap B) = 0.5 + 0.4 - 0.7 = 0.2$ .

Example 6: The 'A but not B' Rule

Question: Using the data from Example 5 ( $P(A)=0.5, P(A \cap B)=0.2$ ), find the probability that only event A occurs.

Solution: The event 'only A occurs' is the same as 'A but not B', which is written as $A - B$ .

The formula is $P(A - B) = P(A) - P(A \cap B)$ .

Substituting the values:

$P(A - B) = 0.5 - 0.2 = 0.3$ .

Example 7: Using De Morgan's Law

Question: Using the data from Example 5 ( $P(A \cup B)=0.7$ ), find the probability that neither A nor B occurs.

Solution: The event 'neither A nor B occurs' is the intersection of their complements: $A' \cap B'$ .

By De Morgan's Law of sets, we know that $A' \cap B' = (A \cup B)'$ .

Therefore, we need to find the probability of the complement of the union of A and B.

Using the complement rule:

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

$P(A' \cap B') = 1 - 0.7 = 0.3$ .

Example 8: Word Problem with Union

Question: In a group of 100 people, 60 own a car, 30 own a motorcycle, and 20 own both. What is the probability that a person chosen at random owns a car or a motorcycle?

Solution: Let C be the event 'owns a car' and M be the event 'owns a motorcycle'. We want to find $P(C \cup M)$ .

From the given data:

$P(C) = \frac{60}{100} = 0.6$ .
$P(M) = \frac{30}{100} = 0.3$ .
$P(C \cap M) = \frac{20}{100} = 0.2$ .

Using the general addition rule:

$P(C \cup M) = P(C) + P(M) - P(C \cap M) = 0.6 + 0.3 - 0.2 = 0.7$ .

Example 9: Exhaustive & Mutually Exclusive Events

Question: In a horse race with three horses (A, B, and C), the probability of A winning is 0.5 and the probability of B winning is 0.2. Assuming there are no ties, what is the probability of C winning?

Solution: The events 'A wins', 'B wins', and 'C wins' are:

Mutually Exclusive: Only one horse can win.
Exhaustive: One of the three horses must win.

When events are both mutually exclusive and exhaustive, the sum of their probabilities must equal 1.

$P(A) + P(B) + P(C) = 1$ .

$0.5 + 0.2 + P(C) = 1$ .

$0.7 + P(C) = 1$ .

$P(C) = 1 - 0.7 = 0.3$ .

Example 10: Range of Probability

Question: The probability of event E is 0.6 and the probability of event F is 0.7. Can E and F be mutually exclusive?

Solution: Assume that E and F are mutually exclusive. According to Axiom 3, the probability of their union would be: $P(E \cup F) = P(E) + P(F) = 0.6 + 0.7 = 1.3$ .

However, the probability of any event (including the union $E \cup F$ ) cannot be greater than 1. This violates the fundamental theorems derived from the axioms.

Since our assumption leads to a contradiction, the assumption must be false.

Therefore, E and F cannot be mutually exclusive.

Example 11: Addition Rule for Three Events

Question: For events A, B, and C, we have: $P(A)=0.3, P(B)=0.4, P(C)=0.5$ . $P(A \cap B)=0.1, P(A \cap C)=0.2, P(B \cap C)=0.2$ . $P(A \cap B \cap C)=0.05$ . Find the probability that at least one of the events occurs.

Solution: The event 'at least one of A, B, or C occurs' is their union, $A \cup B \cup C$ .

We use the addition rule for three events:

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