Q.A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Showing the 12 most recent of 24 on this concept.

1. 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 $=\tfrac15$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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). …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $6P(A)=8P(B)=14P(A\cap B)=1$: $P(B)=\tfrac18$, $P(A\cap B)=\tfrac1{14}$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $B$.

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

   🔒 Start your 14-day free trial to see the full solution →
6. 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): $\frac{4}{50}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 …

   🔒 Start your 14-day free trial to see the full solution →
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: $P(A/B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{P(B)}{P(B)}=1.$ …

   🔒 Start your 14-day free trial to see the full solution →
9. 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: …

   🔒 Start your 14-day free trial to see the full solution →
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 $P(B)\ne 0$,

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

    🔒 Start your 14-day free trial to see the full solution →
11. 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:

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

    🔒 Start your 14-day free trial to see the full solution →
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 $P(B|A)$.

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

    …

    🔒 Start your 14-day free trial to see the full solution →