If $A$ and $B$ are mutually exclusive events, $P(A) = 0.35$ and $P(B) = 0.45$, find (a) $P(A')$ (b) $P(B')$ (c) $P(A \cup B)$ (d) $P(A \cap B)$ (e) $P(A \cap B')$ (f) $P(A' \cap B')$.

If A and B are mutually exclusive events, P(A) = 0.35 and P(B) = 0.45, find (a) P(A') (b) P(B') (c) P(A ∪ B) (

NCERT Exemplar · Q7

A

(a)

P(A')

(b)

P(B')

(c)

P(A \cup B)

(d)

P(A \cap B)

(e)

P(A \cap B')

(f)

P(A' \cap B')

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Concept understanding — Probability Addition Rule

The Intuition: "Or" Means We Add — But Carefully

Imagine you have a bag of 20 marbles: 5 red, 3 blue, and 12 green. You pick one marble at random.

What's the probability that the marble is red or blue?

Your instinct might be: count the red ones (5), count the blue ones (3), add them up (8), and divide by total marbles (20). That gives $\frac{8}{20} = 0.4$ .

That works perfectly here. Why? Because no marble is both red and blue. The events "red" and "blue" cannot happen at the same time — they are mutually exclusive.

Now change the problem. The bag has 20 marbles: 5 red, 3 blue, and 4 striped red-and-blue marbles (counted in both colours). The rest are plain green.

What's the probability of picking a marble that is red or blue?

If you just add red (5 + 4 striped = 9) and blue (3 + 4 striped = 7), you get 16. But that double-counts the 4 striped marbles — they are both red and blue. The correct count is: red-only (5) + blue-only (3) + striped (4) = 12. Probability = $\frac{12}{20} = 0.6$ .

The simple addition overcounts when events can happen together. That's the core problem the Addition Rule solves.

The Precise Statement

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

Where:

$P(A \cup B)$ = probability that A or B (or both) occur
$P(A \cap B)$ = probability that both A and B occur together

The subtraction of $P(A \cap B)$ removes the double-counted overlap.

Two Special Cases

Case 1: Mutually exclusive events (can't happen together)

If $A$ and $B$ cannot both occur, then $P(A \cap B) = 0$ , and the rule simplifies to:

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

This is the "red or blue marble" case — no overlap, so just add.

Case 2: Events that can overlap (general case)

You must subtract the overlap. This is the "striped marble" case.

Watch out

A common mistake: forgetting to subtract the overlap when events can happen together. Always ask: "Can both events occur at the same time?" If yes, you need the subtraction.

Why It Works — A Visual Argument

Draw a rectangle for all possible outcomes. Inside, draw two overlapping circles — one for event A, one for event B. The overlap region is $A \cap B$ .

$P(A)$ counts everything in circle A.
$P(B)$ counts everything in circle B.
Adding them counts the overlap twice.
Subtracting $P(A \cap B)$ once corrects that.

The result is exactly the area covered by either circle — which is $P(A \cup B)$ .

Worked Example

A class has 30 students. 18 play cricket, 15 play football, and 8 play both. One student is chosen at random.

Question: What's the probability the student plays cricket or football?

Let $C$ = plays cricket, $F$ = plays football.

$P(C) = \frac{18}{30}$ , $P(F) = \frac{15}{30}$ , $P(C \cap F) = \frac{8}{30}$

Using the rule: …

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For mutually exclusive events, the addition rule simplifies to $P(A \cup B) = P(A) + P(B)$ . Using complement rules and set identities, we find all six probabilities: $P(A') = 0.65$ , $P(B') = 0.55$ , $P(A \cup B) = 0.80$ , $P(A \cap B) = 0$ , $P(A \cap B') = 0.35$ , $P(A' \cap B') = 0.20$ .

The key idea here is the Probability Addition Rule for mutually exclusive events. Two events are mutually exclusive when they cannot happen at the same time — their intersection is empty. This means $P(A \cap B) = 0$ , which simplifies everything downstream.

When events are mutually exclusive, the probability that at least one occurs is just the sum of their individual probabilities. That’s the core intuition: no overlap means no double-counting to correct for.

Let’s work through each part systematically.

1. (a) $P(A')$ — the complement of $A$

The complement rule says $P(A') = 1 - P(A)$ . Since $P(A) = 0.35$ , we get:

P(A') = 1 - 0.35 = 0.65

This is the probability that $A$ does not occur.

2. (b) $P(B')$ — the complement of $B$

Exactly the same logic:

P(B') = 1 - 0.45 = 0.55

3. (c) $P(A \cup B)$ — the union of $A$ and $B$

Because $A$ and $B$ are mutually exclusive, the addition rule becomes:

P(A \cup B) = P(A) + P(B) = 0.35 + 0.45 = 0.80

Watch out

A common mistake is to use the general addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and forget that $P(A \cap B) = 0$ here. That would give the same answer, but only because the subtraction term is zero — don’t mechanically subtract unless you’ve checked for overlap.

4. (d) $P(A \cap B)$ — the intersection of $A$ and $B$

Mutually exclusive means $A$ and $B$ cannot both occur. So:

P(A \cap B) = 0

5. (e) $P(A \cap B')$ — $A$ occurs but $B$ does not

Think of this as “only $A$ happens.” Since $A$ and $B$ are mutually exclusive, whenever $A$ occurs, $B$ automatically does not. So $A \cap B'$ is actually just $A$ itself. Let’s verify:

A \cap B' = A \cap (\text{not } B)

Because $A$ and $B$ share no outcomes, every outcome in $A$ is automatically outside $B$ . So: …

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

CBSE 2026Set ANNUAL1 markMCQ
$E$
(a) $\dfrac{3}{4}$
(b) $\dfrac{7}{8}$
(c) $\dfrac{3}{8}$
(d) $\dfrac{5}{8}$
›Reveal solutionSolution

"Not $E$ and not $F$ " is the complement of " $E$ or $F$ "; find $P(E\cup F)$ first via the addition rule, then subtract from 1.

By De Morgan's law, the event "not $E$ and not $F$ " is the complement of " $E$ or $F$ ": $(E \cup F)'$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If P(A) = 0.35, P(A \cap B) = 0.25, P(A \cup B) = 0.75, then P(B) is:$
(a) 0.65
(b) 0.40
(c) 0.10
(d) None of these
›Reveal solutionSolution

Rearrange the addition rule P(A∪B) = P(A) + P(B) − P(A∩B) to solve for P(B).

The addition theorem of probability (same structure as $n(A\cup B)=n(A)+n(B)-n(A\cap B)$ for sets) states:

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

Rearranging for $P(B)$ : …

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CBSE 2024Set ANNUAL1 markMCQ
$P(A)=\dfrac{1}{3}$
(a) $\dfrac{7}{15}$
(b) $\dfrac{8}{15}$
(c) $\dfrac{6}{11}$
(d) None of these
›Reveal solutionSolution

Use the addition rule of probability: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ , subtracting the overlap once so it isn't double-counted.

Given $P(A) = \dfrac13$ , $P(B) = \dfrac15$ , $P(A\cap B) = \dfrac{1}{15}$ .

Convert to a common denominator of $15$ :

$P(A) = \frac{5}{15}, \quad P(B) = \frac{3}{15}, \quad P(A\cap B) = \frac{1}{15}$

…

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CBSE 2022Set ANNUAL1 mark
$Q. If E and F are such that P(E) = 1/4, P(F) = 1/2 and P(E and F) = 1/8, then find P(E or F).$
›Reveal solutionSolution

Applying the addition theorem of probability gives $5/8$ .

Given $P(E) = \dfrac{1}{4}$ , $P(F) = \dfrac{1}{2}$ , $P(E \cap F) = \dfrac{1}{8}$ .

$P(E \cup F) = P(E) + P(F) - P(E \cap F) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8}$

…

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CBSE 2022Set ANNUAL1 mark
$E$
›Reveal solutionSolution

$P(E\text{ or }F)=\dfrac58$ .

…

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CBSE 2021Set ANNUAL1 mark
$Q. If A and B are two events such that P(A) = 1/2, P(B) = 7/10 and P(A\capB) = 3/5, then find P(A\cupB).$
›Reveal solutionSolution

Substituting the given probabilities into the addition rule gives $3/5$ .

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

Over a common denominator of 10: $P(A) = \dfrac{5}{10}$ , $P(B) = \dfrac{7}{10}$ , $P(A\cap B) = \dfrac{6}{10}$

…

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CBSE 2020Set ANNUAL1 markMCQ
$P(A) = \frac{3}{8}$
(a) $\frac{11}{24}$
(b) $\frac{1}{8}$
(c) $\frac{1}{3}$
(d) $\frac{13}{24}$
›Reveal solutionSolution

Apply the addition rule for probabilities: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .

Given $P(A)=\dfrac38$ , $P(B)=\dfrac13$ , $P(A\cap B)=\dfrac14$ .

Using LCD $24$ : $\dfrac38=\dfrac9{24}$ , $\dfrac13=\dfrac8{24}$ , $\dfrac14=\dfrac6{24}$ …

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Q.If $A$ and $B$ are mutually exclusive events, $P(A) = 0.35$ and $P(B) = 0.45$ , find

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