Q.If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then (A) $P(B|A) = 1$ (B) $P(A|B) = 1$ (C) $P(B|A) = 0$ (D) $P(A|B) = 0$

Showing the 12 most recent of 24 on this concept.

1. GUJCET 2026Set x1 markMCQ
   Q.Let $A$ and $B$ be two events such that $P(A) = \frac{5}{11}$, $P(B) = \frac{2}{11}$ and $P(A \cup B) = \frac{3}{11}$, then $P(A'|B') =$ ______ (A) $\frac{8}{9}$ (B) $\frac{3}{5}$ (C) $\frac{1}{2}$ (D) $\frac{2}{9}$

   ›Reveal solutionSolution

   [!TLDR] $P(A'|B') = \dfrac{1-P(A\cup B)}{1-P(B)} = \dfrac{8}{9}$.

   Concept

   By De Morgan's law $A'\cap B' = (A\cup B)'$, so $P(A'\cap B') = 1 - P(A\cup B)$. Then the conditional probability is $P(A'|B') = \dfrac{P(A'\cap B')}{P(B')}$ with $P(B') = 1 - P(B)$.

   Solution …

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2. GUJCET 2026Set x1 markMCQ
   Q.If $A$ and $B$ are any two events such that $P(A) + P(B) - P(A \text{ and } B) = P(A)$, then ______ (A) $P(B|A') = 1$ (B) $P(B|A) = 0$ (C) $P(A|B) = 1$ (D) $P(A|B) = 0$

   ›Reveal solutionSolution

   Simplify the given relation to find $P(A\cap B)=P(B)$, i.e. $B\subseteq A$ effectively.

   Given $P(A)+P(B)-P(A\text{ and }B)=P(A)$, which reduces to:

   $P(B)-P(A\cap B)=0\Rightarrow P(A\cap B)=P(B).$ …

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3. GUJCET 2026Set x1 markMCQ
   Q.Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. The probability that first two cards are kings and the third card drawn is an ace is ______ (A) $\frac{1}{135200}$ (B) $\frac{2}{5525}$ (C) $\frac{3}{5525}$ (D) $\frac{3}{135200}$

   ›Reveal solutionSolution

   Multiply the successive conditional probabilities for drawing without replacement.

   First card king: $\frac{4}{52}$. Second card king (3 kings left of 51): $\frac{3}{51}$. Third card ace (4 aces still present of 50): $\frac{4}{50}$. …

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4. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.If $P(A/B) > P(A)$, then which of the following is true?

   (a) $P(B|A) < P(B)$

   (b) $P(A\cap B) < P(A)\cdot P(B)$

   (c) $P(B|A) > P(B)$

   (d) $P(B|A) = P(B)$

   ›Reveal solutionSolution

   Convert the given inequality into one about the joint probability, then flip it around to get $P(B|A)$.

   $P(A|B)>P(A)\Rightarrow\dfrac{P(A\cap B)}{P(B)}>P(A)\Rightarrow P(A\cap B)>P(A)P(B)$.

   …

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5. CA Foundation 2026Set jan-20261 markMCQ
   Q.If in a class, 50% of the student study mathematics and science and 70% of the student study mathematics, then the probability of a student studying science given that he/she is already studying mathematics is (A) $\frac{3}{7}$ (B) $\frac{6}{7}$ (C) $\frac{4}{7}$ (D) $\frac{5}{7}$

   ›Reveal solutionSolution

   Conditional probability $P(S\mid M) = \dfrac{P(M \cap S)}{P(M)}$.

   Step 1 — identify the probabilities

   50% study both maths and science, so $P(M \cap S) = 0.5$; 70% study maths, so $P(M) = 0.7$.

   Step 2 — apply the conditional-probability formula

   $P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$ …

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6. CA Foundation 2026Set jan-20261 markMCQ
   Q.If two dice are rolled, then the probability of getting a greater number on the first die than the one on the second, given that the sum should be equal to 7 is (A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   Conditional probability on a reduced sample space: $P(A\mid B) = \dfrac{n(A \cap B)}{n(B)}$.

   Step 1 — list the outcomes with sum 7.

   $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \Rightarrow n(B) = 6.$

   Step 2 — count first die greater than second, among those.

   $(4,3),(5,2),(6,1)$ → 3 outcomes.

   Step 3 — conditional probability.

   $P = \frac{3}{6} = \frac{1}{2}.$ …

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7. GUJCET 2025Set 031 markMCQ
   Q.Let $A$ and $B$ be two events such that $P(A) = \frac{3}{8}$, $P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$. Then $P(A'|B) - P(A|B) =$ _____. (A) $\frac{1}{5}$ (B) $\frac{3}{5}$ (C) $\frac{2}{5}$ (D) $\frac{4}{5}$

   ›Reveal solutionSolution

   $P(A\cap B)=P(A)+P(B)-P(A\cup B)$, then use conditional probabilities on $B$.

   $P(A\cap B)=\frac38+\frac58-\frac34=1-\frac34=\frac14.$ …

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8. GUJCET 2025Set 031 markMCQ
   Q.A man is known to speak truth 4 out of 5 times. He throws a die and reports that it is a six. The probability that actually there was a six is (A) $\frac{5}{9}$ (B) $\frac{4}{9}$ (C) $\frac{5}{35}$ (D) $\frac{4}{35}$

   ›Reveal solutionSolution

   $P(\text{six}\mid\text{reports six})=\dfrac{P(6)P(T)}{P(6)P(T)+P(\text{not }6)P(\text{lie})}$.

   Let $P(6)=\tfrac16$, $P(\text{not }6)=\tfrac56$, truth $=\tfrac45$, lie $=\tfrac15$. …

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9. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.If $P(A) = 0.8$, $P(B) = 0.5$, and $P(B/A) = 0.4$, then $P(A/B)$ = ____.

   (a) $0.16$

   (b) $0.64$

   (c) $0.32$

   (d) $0.40$

   ›Reveal solutionSolution

   Find $P(A\cap B)$ from $P(B/A)$ first, then use it to get $P(A/B)$.

   $P(A\cap B)=P(B/A)\cdot P(A)=0.4\times0.8=0.32$.

   …

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10. GUJCET 2024Set 131 markMCQ
    Q.If A and B are two events such that $P(B) \ne 0$ and $P(A \mid B) = 1$, then __________. (A) $B \ne \phi$ (B) $B \subset A$ (C) $A \ne \phi$ (D) $A \subset B$

    ›Reveal solutionSolution

    $P(A\mid B)=1$ forces $P(A\cap B)=P(B)$, meaning $B\subset A$.

    Concept. By definition $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$.

    Steps. Given $P(A\mid B)=1$ and $P(B)\ne 0$,

    $\frac{P(A\cap B)}{P(B)}=1 \;\Rightarrow\; P(A\cap B)=P(B).$ …

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11. CA Foundation 2023Set jun-20231 markMCQ
    Q.If $P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{4}, P(A/B)=\dfrac{1}{6}$, the probability $P\left(B/A\right)$ is (A) $\dfrac{1}{8}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    P(B/A) = P(A∩B)/P(A) = (1/24)/(1/3) = 1/8.

    Step 1 — Find the joint probability

    $P(A\cap B) = P(A/B)\,P(B) = \frac{1}{6}\times\frac{1}{4} = \frac{1}{24}$

    Step 2 — Apply the definition of conditional probability

    $P(B/A) = \frac{P(A\cap B)}{P(A)} = \frac{1/24}{1/3} = \frac{3}{24} = \frac{1}{8}$

    Watch out

    P(A/B) and P(B/A) are not equal — you must recompute the joint probability first, then divide by P(A), not P(B). …

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12. GUJCET 2022Set 081 markMCQ
    Q.Probability that A speaks truth is $\frac{4}{5}$. A coin is tossed. A reports that a head appears. The probability that actually there was head is ______. (A) $\frac{2}{5}$ (B) $\frac{4}{5}$ (C) $\frac{1}{5}$ (D) $\frac{1}{2}$

    ›Reveal solutionSolution

    A truthful reporter (p = 4/5) on a fair coin makes the reported head almost as reliable as their honesty.

    Concept. Bayes' theorem: $P(H\mid R)=\dfrac{P(R\mid H)P(H)}{P(R\mid H)P(H)+P(R\mid T)P(T)}$.

    Solution. $P(H)=P(T)=\tfrac12$. A reports head truthfully with prob $\tfrac45$ (if head) and lies with prob $\tfrac15$ (if tail). …

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