P(A) = (6)/(11),P(B) = (5)/(11),P(A ∪ B) = (7)/(11) ⇒ P(A ∩ B) = (a) (4)/(11) (b) (5)/(11) (c) (7)/(11) (d) (9

Question of 165

P(A) = \frac{6}{11},\ P(B) = \frac{5}{11},\ P(A \cup B) = \frac{7}{11} \Rightarrow P(A \cap B) =

(a)

\frac{4}{11}

(b)

\frac{5}{11}

(c)

\frac{7}{11}

(d)

\frac{9}{11}

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Concept understanding — Inclusion Exclusion Principle

The Inclusion–Exclusion Principle

Thirty students play cricket and twenty-five play football — so does "cricket or football" total $55$ ? Not if some play both. If ten play both, they have been counted twice, so the real total is $30 + 25 - 10 = 45$ . That single correction is the whole principle: add everything, then subtract what you double-counted.

Why "Inclusion" and "Exclusion"

Include every set by adding its size.
Exclude the overlaps counted more than once by subtracting the intersections.

For three or more sets the signs keep alternating, because each correction slightly over-corrects and must itself be fixed.

The General Statement

For finite sets $A_1, A_2, \dots, A_n$ ,

\left| \bigcup_{i=1}^n A_i \right| = \sum_{i} |A_i| - \sum_{i<j} |A_i \cap A_j| + \sum_{i<j<k} |A_i \cap A_j \cap A_k| - \cdots + (-1)^{n+1}\,|A_1 \cap \cdots \cap A_n|.

Add all singles, subtract all pairwise intersections, add all triples, and so on — the sign alternates, starting with $+$ .

Why It Works

An element lying in exactly $m$ of the sets is counted $\binom{m}{1}$ times among the singles, subtracted $\binom{m}{2}$ times among the pairs, added $\binom{m}{3}$ times among the triples, and so on. Its net count is

\binom{m}{1} - \binom{m}{2} + \binom{m}{3} - \cdots + (-1)^{m+1}\binom{m}{m} = 1,

using $(1-1)^m = 0$ from the binomial theorem. So every element is counted exactly once.

A Three-Set Example

Of $100$ students: $40$ take Physics, $30$ Chemistry, $25$ Biology; $15$ take P and C, $10$ P and B, $8$ C and B; $5$ take all three.

|P \cup C \cup B| = (40+30+25) - (15+10+8) + 5 = 95 - 33 + 5 = 67.

So $67$ take at least one subject and $33$ take none. …

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Rearranged addition rule gives $P(A\cap B)=\frac{4}{11}$ .

Addition rule: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ .

…

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

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution

The value $P=0.2$ is correctly computed, but the stated definition of "mutually exclusive" is wrong — that condition actually defines independence.

Assertion (A): For mutually exclusive events, $P(A\cup B)=P(A)+P(B)$ .

$0.6=0.4+P \Rightarrow P=0.2$ . So A is true.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): If n(X) = 17, n(Y) = 23 and n(X \cup Y) = 38, then n(X \cap Y) = 38. Reason (R): n(A \cup B) = n(A) + n(B) - n(A \cap B).$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
›Reveal solutionSolution

Using the correct formula n(X∩Y) = n(X)+n(Y)−n(X∪Y) = 2, not 38 — so the Assertion is false while the Reason (the formula itself) is true.

Checking the Assertion (A): Using the addition rule for sets:

$n(X\cup Y) = n(X)+n(Y)-n(X\cap Y)$

$n(X\cap Y) = n(X)+n(Y)-n(X\cup Y) = 17+23-38 = 2$

So the correct value is $n(X\cap Y)=2$ , not 38 as claimed in Assertion (A). Assertion (A) is false.

Note on the source paper: the printed English version of this paper states the conclusion as "n(X∩Y) = 38", but the correct mathematics (and the Hindi version of this same question paper) gives $n(X\cap Y)=2$ . This is treated here as a genuine printing discrepancy in the English text — the correct mathematical reading is used for grading.

…

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CBSE 2025Set E1 markMCQ
$Q. Addition theorem of probability is$
(a) $P(A \cup B) = P(A) + P(B)$
(b) $P(A \cup B) = P(A) + P(B) + P(A \cap B)$
(c) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
(d) $P(A \cup B) = P(A) \cdot P(B)$
›Reveal solutionSolution

The addition theorem is $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ .

Adding $P(A)$ and $P(B)$ counts the overlap $A\cap B$ twice, so it must be subtracted once:

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

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CBSE 2025Set ANNUAL1 markMCQ
$A$
(a) $0.2$
(b) $0.1$
(c) $\dfrac{4}{5}$
(d) none of these
›Reveal solutionSolution

Mutually exclusive events have no overlap, so their union probability is a simple sum.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): If S and T are two sets such that S has 21 elements, T has 32 elements and S \cap T has 11 elements, then S \cup T will have 42 elements. Reason (R): n(A \cup B) = n(A) + n(B) - n(A \cap B)$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
›Reveal solutionSolution

Plugging the given values into the stated formula reproduces the assertion exactly, so the reason is the correct explanation.

Given $n(S)=21$ , $n(T)=32$ , $n(S\cap T)=11$ .

Reason (R): $n(A\cup B)=n(A)+n(B)-n(A\cap B)$ — this is the standard inclusion–exclusion formula for two sets, which is true.

Applying it: $n(S\cup T) = 21+32-11 = 42$

…

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CBSE 2024Set D1 markMCQ
$P(A) = \frac{6}{11},\ P(B) = \frac{5}{11},\ P(A \cup B) = \frac{7}{11} \Rightarrow P(A \cap B) =$
(a) $\frac{4}{11}$
(b) $\frac{5}{11}$
(c) $\frac{7}{11}$
(d) $\frac{9}{11}$
›Reveal solutionSolution

Rearranged addition rule gives $P(A\cap B)=\frac{4}{11}$ .

Addition rule: $P(A\cup B) = P(A)+P(B)-P(A\cap B)$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. If P(A) = 6/11, P(B) = 5/11 and P(A\cupB) = 7/11, then P(A\capB) is :$
(a) 1/11
(b) 3/11
(c) 2/11
(d) 4/11
›Reveal solutionSolution

Using the addition rule P(A∪B) = P(A) + P(B) − P(A∩B), rearranged to solve for P(A∩B), gives 4/11.

The general addition rule of probability states: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ .

…

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CBSE 2023Set E1 markMCQ
$P(E)=\dfrac{3}{7},\ P(F)=\dfrac{5}{7},\ P(E\cup F)=\dfrac{6}{7}\Rightarrow P(E\cap F) =$
(a) $\dfrac{4}{7}$
(b) $\dfrac{2}{7}$
(c) $\dfrac{1}{7}$
(d) $\dfrac{3}{7}$
›Reveal solutionSolution

Rearranging the addition rule gives $P(E\cap F)=\dfrac{2}{7}$ .

The addition rule of probability states

$P(E\cup F)=P(E)+P(F)-P(E\cap F).$

Solving for the intersection: …

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CBSE 2023Set ANNUAL1 markMCQ
$P(A) = \dfrac{3}{8}$
(a) $\dfrac{2}{3}$
(b) $\dfrac{1}{3}$
(c) $\dfrac{1}{2}$
(d) none of these
›Reveal solutionSolution

Use the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and simplify with a common denominator.

$P(A\cup B) = \dfrac{3}{8}+\dfrac{1}{3}-\dfrac{1}{4}$ . Using LCD 24: $= \dfrac{9}{24}+\dfrac{8}{24}-\dfrac{6}{24} = \dfrac{11}{24}$ .

…

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CBSE 2023Set ANNUAL1 mark
$P(A) = \dfrac{6}{11}$
›Reveal solutionSolution

Rearrange the addition theorem $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .

By the addition theorem of probability,

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

Solving for $P(A\cap B)$ :

…

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CBSE 2022Set ANNUAL1 mark
$Q. If set S has 21 elements, T has 32 elements and S \cup T has 42 elements, then S \cap T has ............ elements.$
›Reveal solutionSolution

Rearranging the union-intersection formula gives $n(S \cap T) = 11$ .

For any two sets:

$n(S \cup T) = n(S) + n(T) - n(S \cap T)$

Given $n(S) = 21$ , $n(T) = 32$ , $n(S \cup T) = 42$ : …

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CBSE 2021Set ANNUAL1 mark
$Q. In a school there are 20 teachers who teach Mathematics or Physics. Of these 12 teach Maths and 4 teach Maths and Physics. The number of teacher who teach Physics are ............. .$
›Reveal solutionSolution

Solving $|M \cup P| = |M| + |P| - |M \cap P|$ for $|P|$ gives 12.

Let M = set of teachers who teach Maths, P = set of teachers who teach Physics.

Given: $|M \cup P| = 20$ , $|M| = 12$ , $|M \cap P| = 4$ .

By the formula for the union of two sets:

…

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Q.P(A)=611, P(B)=511, P(A∪B)=711⇒P(A∩B)=P(A) = \frac{6}{11},\ P(B) = \frac{5}{11},\ P(A \cup B) = \frac{7}{11} \Rightarrow P(A \cap B) =P(A)=116​, P(B)=115​, P(A∪B)=117​⇒P(A∩B)=

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Q. $P(A) = \frac{6}{11},\ P(B) = \frac{5}{11},\ P(A \cup B) = \frac{7}{11} \Rightarrow P(A \cap B) =$

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