Q.A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Showing the 12 most recent of 24 on this concept.

1. 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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2. 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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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. GUJCET 2019Set 171 markMCQ
   Q.If $6P(A) = 8P(B) = 14P(A \cap B) = 1$, then $P\left(\dfrac{A'}{B}\right) = \underline{\hspace{2cm}}$ (A) $\dfrac{4}{7}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{3}{7}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   With $P(A)=\tfrac16,P(B)=\tfrac18,P(A\cap B)=\tfrac1{14}$: $P(A'|B)=1-\dfrac{P(A\cap B)}{P(B)}=\dfrac37$.

   Concept: From $6P(A)=8P(B)=14P(A\cap B)=1$: $P(B)=\tfrac18$, $P(A\cap B)=\tfrac1{14}$. …

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5. GUJCET 2021Set 151 markMCQ
   Q.If $P(A) = \frac{6}{11}$, $P(B) = \frac{5}{11}$ and $P(A \cup B) = \frac{7}{11}$, then $P\left(\frac{A}{B}\right) = \underline{\hspace{2cm}}$ (A) $\frac{4}{5}$ (B) $\frac{2}{3}$ (C) $\frac{4}{11}$ (D) $\frac{2}{11}$

   ›Reveal solutionSolution

   Get P(A and B) from inclusion-exclusion, then divide by P(B).

   Concept. $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ and $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$. …

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6. 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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7. 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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8. GUJCET 2020Set 071 markMCQ
   Q.If A and B are any two events such that $P(A) + P(B) - P(A \cap B) = P(A)$ then ________. (A) $P\left(\frac{B}{A}\right) = 1$ (B) $P\left(\frac{A}{B}\right) = 0$ (C) $P\left(\frac{B}{A}\right) = 0$ (D) $P\left(\frac{A}{B}\right) = 1$

   ›Reveal solutionSolution

   $P(A)+P(B)-P(A\cap B)=P(A)$ forces $P(A\cap B)=P(B)$, meaning $B$ is contained in $A$, so $P(A/B)=1$.

   Concept. Rearrange the given equation: $P(B)-P(A\cap B)=0 \Rightarrow P(A\cap B)=P(B)$.

   Step. Conditional probability: $P(A/B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(B)}{P(B)}=1.$ …

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9. 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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10. GUJCET 2021Set 151 markMCQ
    Q.If A and B are two events such that $P(A) \ne 0$ and $P\left(\frac{B}{A}\right) = 1$, then $\underline{\hspace{2cm}}$ (A) $B \subset A$ (B) $B = \varnothing$ (C) $A = \varnothing$ (D) $A \subset B$

    ›Reveal solutionSolution

    [!TLDR] Melatonin is a pineal hormone controlling circadian rhythm, not the menstrual cycle. Estrogen, progesterone and relaxin are all reproductive hormones linked to the ovarian cycle.

    Concept

    In NCERT/CBSE-aligned human reproduction, the menstrual cycle is regulated by pituitary hormones (FSH, LH) and ovarian hormones. Estrogen (secreted by the growing follicle) causes the proliferative phase; progesterone (from the corpus luteum) maintains the secretory endometrium; and relaxin, also a corpus-luteum/reproductive hormone, prepares reproductive tissues. Melatonin, secreted by the pineal gland, governs the day–night biological clock and is not a menstrual-cycle hormone.

    Solution

    Evaluate each option: …

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11. 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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12. 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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