Q.Among 120 students, 60 study Accountancy, 50 study Business Mathematics, and 45 study Costing. Of these, 25 study both Accountancy and Business Mathematics, 20 study both Business Mathematics and Costing, 15 study both Accountancy and Costing, and 10 study all three subjects. Find how many students study at least one of the three subjects, and how many study none.

Showing the 12 most recent of 17 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): If the events $A$ and $B$ are mutually exclusive events such that $P(A) = 0.4$, $P(A \cup B) = 0.6$ and $P(B) = P$, then $P = 0.2$ Reason (R): Two events $A$ and $B$ are mutually exclusive events if $P(A \cap B) = P(A).P(B)$.

   (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, 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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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): If n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, then n(X ∩ Y) = 38. Reason (R): n(A ∪ B) = n(A) + n(B) - n(A ∩ 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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3. 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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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $A$ and $B$ are mutually exclusive events such that $P(A) = 0.4$, $P(B) = x$ and $P(A \cup B) = 0.5$ then $x =$

   (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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5. 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 ∩ T has 11 elements, then S ∪ T will have 42 elements. Reason (R): n(A ∪ B) = n(A) + n(B) − n(A ∩ 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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6. CBSE 2024Set D1 markMCQ
   Q.$P(A) = \frac{6}{11},\ P(B) = \frac{5}{11},\ P(A \cup B) = \frac{7}{11} \Rightarrow P(A \cap B) =$

   (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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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.If P(A) = 6/11, P(B) = 5/11 and P(A∪B) = 7/11, then P(A∩B) 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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8. CBSE 2023Set E1 markMCQ
   Q.$P(E)=\dfrac{3}{7},\ P(F)=\dfrac{5}{7},\ P(E\cup F)=\dfrac{6}{7}\Rightarrow P(E\cap F) =$

   (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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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.If $P(A) = \dfrac{3}{8}$, $P(B) = \dfrac{1}{3}$ and $P(A \cap B) = \dfrac{1}{4}$, then $P(A \cup B) =$

   (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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10. CBSE 2023Set ANNUAL1 mark
    Q.If $P(A) = \dfrac{6}{11}$, $P(B) = \dfrac{5}{11}$ and $P(A \cup B) = \dfrac{7}{11}$, find $P(A \cap B)$.

    ›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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11. CBSE 2022Set ANNUAL1 mark
    Q.If set S has 21 elements, T has 32 elements and S ∪ T has 42 elements, then S ∩ 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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12. 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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