The probability of intersection of two events A and B is always less than or equal to those favourable to the

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Q. The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.

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Concept understanding — Probability Axioms

Probability Axioms: From Intuition to Precision

Imagine you're rolling a fair six-sided die. Before you throw it, you know a few things for certain: the result will be one of the numbers 1 through 6. You also know that some outcomes are equally likely — each face has a 1-in-6 chance. And you know that the chance of getting either a 1 or a 2 is simply the sum of their individual chances: $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ .

These three ideas — that probabilities are numbers between 0 and 1, that something must happen (total probability = 1), and that probabilities of non-overlapping events add — are the bedrock of all probability theory. They are so fundamental that we call them axioms: self-evident truths from which everything else is derived.

The Three Axioms (Kolmogorov's Axioms)

Let’s make this precise. We have a sample space $S$ — the set of all possible outcomes. An event $A$ is any subset of $S$ (like "rolling an even number" = $\{2,4,6\}$ ). The probability of an event $A$ is written $P(A)$ .

P(A) \geq 0 \quad \text{for every event } A

Axiom 1 (Non-negativity): A probability can never be negative. This matches our intuition: you can't have a "less than zero" chance of something happening. The smallest possible probability is 0 (an impossible event).

P(S) = 1

Axiom 2 (Normalization): The probability that some outcome in the sample space occurs is exactly 1. Something must happen. This is why we say "the die will show 1,2,3,4,5, or 6" with certainty.

A

Axiom 3 (Additivity): For events that don't overlap, the probability of "A or B" is just the sum of their individual probabilities. This is why the chance of rolling a 1 or a 2 is $\frac{1}{6} + \frac{1}{6}$ .

Watch out

This additivity only works for mutually exclusive events. If events can happen together (like "rolling an even number" and "rolling a number greater than 3"), you cannot simply add their probabilities — you'd double-count the overlap.

Why These Three Are Enough

From these three simple rules, we can derive everything else in probability. For example:

Complement rule: $P(\text{not } A) = 1 - P(A)$ . Why? Because $A$ and "not $A$ " are mutually exclusive and together cover the whole sample space. By Axiom 3: $P(A) + P(\text{not } A) = P(S) = 1$ .
Probability of an impossible event: $P(\emptyset) = 0$ . Since $S$ and $\emptyset$ are mutually exclusive and $S \cup \emptyset = S$ , we get $P(S) + P(\emptyset) = P(S)$ , so $P(\emptyset) = 0$ . …

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The intersection $A \cap B$ can never contain more outcomes than $A$ itself, so $P(A \cap B) \leq P(A)$ always holds. The statement is true.

Why this must be true

Probability rests on a simple counting principle: the probability of an event measures the "size" of the set of outcomes that make it happen. When we talk about $A \cap B$ , we're looking at outcomes that satisfy both $A$ and $B$ simultaneously. This set can only be as large as—or smaller than—the set of outcomes satisfying $A$ alone.

Think of it this way: if you have a basket of red balls (event $A$ ) and you ask "which of these are also large?" (event $B$ ), the red-and-large balls can't outnumber all the red balls. The intersection is always a subset.

The formal argument

Set relationship first By definition, $A \cap B$ is the set of all outcomes that belong to both $A$ and $B$ . This means every outcome in $A \cap B$ is automatically in $A$ . In set notation:

A \cap B \subseteq A

Probability preserves order

One of the fundamental axioms of probability states that if $E_1 \subseteq E_2$ , then $P(E_1) \leq P(E_2)$ . This is because probability is a measure—it assigns non-negative numbers to sets, and larger sets get larger (or equal) measures.
Apply to our case

Since $A \cap B \subseteq A$ , we immediately have:

P(A \cap B) \leq P(A)

The equality $P(A \cap B) = P(A)$ occurs precisely when $A \subseteq B$ , meaning every outcome in $A$ is also in $B$ .

Tip

This inequality is one half of a pair: we also have $P(A \cap B) \leq P(B)$ . In fact, $P(A \cap B) \leq \min\{P(A), P(B)\}$ . …

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Q.The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.

Why this must be true

The formal argument

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