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

If P(A ∪ B) = P(A ∩ B) for any two events A and B, – 1st PUC Mathematics

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P(A \cup B) = P(A \cap B)

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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 condition forces $P(A)=P(B)$ , so the answer is (A).

Solution

By the addition rule,

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

Given $P(A\cup B)=P(A\cap B)$ , substitute:

P(A\cap B)=P(A)+P(B)-P(A\cap B)\ \Rightarrow\ P(A)+P(B)=2\,P(A\cap B).

Now use the inclusion chain $A\cap B\subseteq A\subseteq A\cup B$ and $A\cap B\subseteq B\subseteq A\cup B$ , which gives …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set A-11 markMCQ
$w_1, w_2, w_3, w_4$
›Reveal solutionSolution

Use $a+b=\frac{1}{12}$ from the given equation, then get $P(w_3)$ from the fact that the five outcome-probabilities of a sample space must add to 1.

Step 1 — use the given relation.
$12a + 12b - 1 = 0 \;\Longrightarrow\; 12(a+b) = 1 \;\Longrightarrow\; a+b=\frac{1}{12}.$
Step 2 — the axiom of total probability.

For a random experiment with the five (mutually exclusive, exhaustive) outcomes $w_1,\dots ,w_5$ ,
$P(w_1)+P(w_2)+P(w_3)+P(w_4)+P(w_5)=1 .$
Step 3 — substitute the known probabilities.

The probabilities quoted in the stem are $\frac16$ , $a$ , $b$ and $\frac1{12}$ ; the one that is not quoted is the one being asked for, $P(w_3)$ . Hence
$\frac16 + a + b + \frac1{12} + P(w_3) = 1 .$
Step 4 — solve.

Using $a+b=\frac1{12}$ : …
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KCET 2019Set A-11 markMCQ
$3k^2$
›Reveal solutionSolution
For a valid probability distribution, the sum of all probabilities must equal 1. Solving the resulting quadratic in $k$ and discarding the invalid root gives $k = \frac{1}{5}$ .

The core idea here is simple: a probability distribution must satisfy two conditions — each probability is between 0 and 1, and the sum of all probabilities is exactly 1. The second condition gives us an equation to solve for $k$ .

Let’s write down the sum of all probabilities:
$(k-1) + 3k + k + 3k + 3k^2 + k^2 + (k^2 + k) = 1$
Now combine like terms carefully.

Collect the constant terms: The only constant is $-1$ from the first term.

Collect the $k$ terms: $k + 3k + k + 3k + k = 9k$ (the last term contributes $+k$ ).

Collect the $k^2$ terms: $3k^2 + k^2 + k^2 = 5k^2$ .

So the sum becomes:
$5k^2 + 9k - 1 = 1$
Bring the 1 on the right over:
$5k^2 + 9k - 2 = 0$
This is a quadratic in $k$ . Solve it.
$5k^2 + 9k - 2 = 0$
The factors: $(5k - 1)(k + 2) = 0$ .

So $k = \frac{1}{5}$ or $k = -2$ .

Watch out

A probability can never be negative. If $k = -2$ , then $P(X=1) = k-1 = -3$ , which is impossible. So $k = -2$ is rejected outright.

Now check $k = \frac{1}{5}$ against each probability:

$P(1) = \frac{1}{5} - 1 = -\frac{4}{5}$ — wait, that’s negative!

This is a critical moment. Let’s re-read the table carefully. The first entry is $k-1$ . If $k = \frac{1}{5}$ , then $k-1 = -\frac{4}{5}$ , which is invalid.

Important

A valid probability distribution requires every individual probability to be non-negative. A single negative value makes the distribution invalid, regardless of the sum.

So $k = \frac{1}{5}$ also fails the non-negativity condition.

Let’s check the other values for $k = \frac{1}{5}$ :

$P(2) = 3k = \frac{3}{5}$ (ok)

$P(3) = k = \frac{1}{5}$ (ok)

$P(4) = 3k = \frac{3}{5}$ (ok)

$P(5) = 3k^2 = \frac{3}{25}$ (ok)

$P(6) = k^2 = \frac{1}{25}$ (ok)

$P(7) = k^2 + k = \frac{1}{25} + \frac{1}{5} = \frac{6}{25}$ (ok)

But $P(1)$ is negative. So $k = \frac{1}{5}$ is not valid either. …
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X	1	2	3	4	5	6	7
P(X)	k-1	3k	k	3k	$3k^2$	$k^2$	$k^2$ +k

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

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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

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