Check whether the following probabilities P(A) and P(B) are consistently defined (i) P(A) = 0.5, P(B) = 0.7, P

Exercise 14.2 · Q12

Q. Check whether the following probabilities P(A) and P(B) are consistently defined

(i)

P(A) = 0.5

P(B) = 0.7

P(A \cap B) = 0.6

(ii)

P(A) = 0.5

P(B) = 0.4

P(A \cup B) = 0.8

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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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Probabilities are consistently defined when they obey the axioms of probability; we check whether $P(A \cap B) \leq \min\{P(A), P(B)\}$ and whether $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ yields valid probabilities in $[0,1]$ . (i) is inconsistent; (ii) is consistent.

Why consistency matters

When we assign probabilities to events, we cannot pick numbers arbitrarily. The axioms of probability impose strict constraints: every probability must lie in $[0, 1]$ , the intersection of two events cannot be more probable than either event alone, and the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ must hold. Violating any of these signals an impossible or contradictory probability model.

The key checks are:

Does $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$ ? (The intersection is a subset of each event.)
Does the addition rule produce a valid probability in $[0, 1]$ ?

Let's examine each case.

Case (i): $P(A) = 0.5$ , $P(B) = 0.7$ , $P(A \cap B) = 0.6$

Check the intersection constraint.

The intersection $A \cap B$ consists of outcomes that belong to both $A$ and $B$ . Since every outcome in $A \cap B$ is also in $A$ , we must have $P(A \cap B) \leq P(A)$ .

Here, $P(A \cap B) = 0.6$ but $P(A) = 0.5$ .

This says the probability of both events occurring together is larger than the probability of $A$ alone—an impossibility. If $A$ happens only 50% of the time, how can $A$ and $B$ together happen 60% of the time?
Verdict.

The probabilities violate the fundamental constraint $P(A \cap B) \leq \min\{P(A), P(B)\}$ .

Watch out

A common mistake is to forget that $A \cap B \subseteq A$ , so $P(A \cap B)$ can never exceed $P(A)$ or $P(B)$ . Always check this first.

Case (i) is inconsistent.

Case (ii): $P(A) = 0.5$ , $P(B) = 0.4$ , $P(A \cup B) = 0.8$

Use the addition rule to find $P(A \cap B)$ . The addition rule states: $P(A \cup B) = P(A) + P(B) - P(A \cap B).$ …

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Q.Check whether the following probabilities P(A) and P(B) are consistently defined

Why consistency matters

Case (i): P(A)=0.5P(A) = 0.5P(A)=0.5, P(B)=0.7P(B) = 0.7P(B)=0.7, P(A∩B)=0.6P(A \cap B) = 0.6P(A∩B)=0.6

Case (ii): P(A)=0.5P(A) = 0.5P(A)=0.5, P(B)=0.4P(B) = 0.4P(B)=0.4, P(A∪B)=0.8P(A \cup B) = 0.8P(A∪B)=0.8

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Case (i): $P(A) = 0.5$ , $P(B) = 0.7$ , $P(A \cap B) = 0.6$

Case (ii): $P(A) = 0.5$ , $P(B) = 0.4$ , $P(A \cup B) = 0.8$

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