The probability of an occurrence of event A is .7 and that of the occurrence of event B is .3 and the probabil

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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 given probabilities violate the fundamental axiom that $P(A \cap B) \leq \min(P(A), P(B))$ , because $0.4 > 0.3$ . Therefore, no such probability assignment is possible — the data is inconsistent.

The problem gives you three numbers: $P(A) = 0.7$ , $P(B) = 0.3$ , and $P(A \cap B) = 0.4$ . At first glance, these look like ordinary probabilities. But something is off — and the key is to check whether they obey the basic rules that every probability must follow.

Probability is not just any set of numbers. It must satisfy three axioms (the Kolmogorov axioms):

Every probability is between 0 and 1.
The probability of the whole sample space is 1.
For mutually exclusive events, probabilities add.

A direct consequence of these axioms is that the probability of the intersection of two events can never exceed the probability of either event individually. Why? Because $A \cap B$ is a subset of $A$ and also a subset of $B$ . If one set is contained in another, its probability cannot be larger. Formally:

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

This is a non-negotiable property. Let's test the given numbers.

Check the minimum bound. $P(A) = 0.7$ and $P(B) = 0.3$ . The smaller of the two is $0.3$ . So we must have:

P(A \cap B) \leq 0.3

Compare with the given intersection probability.

The problem states $P(A \cap B) = 0.4$ . But $0.4 > 0.3$ , which directly violates the inequality above.
Interpret the violation. …

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Q.The probability of an occurrence of event A is .7.7.7 and that of the occurrence of event B is .3.3.3 and the probability of occurrence of both is .4.4.4.

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Q.The probability of an occurrence of event A is $.7$ and that of the occurrence of event B is $.3$ and the probability of occurrence of both is $.4$ .

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