Q.If $P(A \cup B) = P(A \cap B)$ for any two events $A$ and $B$, then
(A) $P(A) = P(B)$
(B) $P(A) > P(B)$
(C) $P(A) < P(B)$
(D) none of these

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 $P(A)$.

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).

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.

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 $\frac{1}{6} + \frac{1}{6}$.

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 $P(\emptyset) = 0$. …