Q.(a) A pair of dice is thrown simultaneously. If $X$ denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of $X$.

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1. CBSE 2026Set 65/1/11 markMCQ
   Q.Assertion (A): In an experiment of throwing an unbiased die, the probability of getting a prime number given that the number appearing on the die is odd is $\dfrac{2}{3}$. Reason (R): For any two events $A$ and $B$, $P(A|B) = \dfrac{P(A \cup B)}{P(B)}$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true and Reason (R) is false. (D) Assertion (A) is false and Reason (R) is true.

   ›Reveal solutionSolution

   The assertion is true: given the outcome is odd, the probability it is a prime is $\frac{2}{3}$. The reason states the correct conditional probability formula. Since the reason directly justifies the calculation in the assertion, both are true and the reason is the correct explanation.

   Concept first — Conditional probability asks: If we already know that event $B$ has occurred, what is the probability that event $A$ also occurs? The sample space shrinks from all possible outcomes to just those in $B$. The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the precise way to compute this reduced probability.

   Here, the die is unbiased, so each face $\{1,2,3,4,5,6\}$ has probability $\frac{1}{6}$. The assertion involves two events:

   • $A$: the number is prime. On a die, the primes are $2,3,5$.
   • $B$: the number is odd. The odd numbers are $1,3,5$.

   The condition "given that the number is odd" means we restrict attention to $B = \{1,3,5\}$. Among these three equally likely outcomes, the primes are $3$ and $5$ — that's two out of three. So the conditional probability is $\frac{2}{3}$.

   Now let's verify step by step using the formula in Reason (R).

   1. Define the events precisely.

      $A = \{2,3,5\}$, $B = \{1,3,5\}$.

      The sample space $S = \{1,2,3,4,5,6\}$.

   2. Compute $P(B)$.

      $B$ has 3 outcomes, each with probability $\frac{1}{6}$, so $P(B) = \frac{3}{6} = \frac{1}{2}$.

   3. Compute $P(A \cap B)$.

      $A \cap B$ = numbers that are both prime and odd = $\{3,5\}$. That's 2 outcomes, so $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$.

   4. Apply the formula from Reason (R).

   $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}.$

   This matches the assertion exactly. …

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2. CBSE 2026Set 65/2/11 markMCQ
   Q.For two events $A$ and $B$ such that $P(A) \ne 0$ and $P(B) \ne 1$, $P(A'/B') =$ (A) $1 - P(A/B)$ (B) $1 - P(A'/B)$ (C) $\frac{1 - P(A \cap B)}{P(B')}$ (D) $\frac{1 - P(A \cup B)}{P(B')}$

   ›Reveal solutionSolution

   We need to find $P(A'|B')$. By applying the definition of conditional probability, De Morgan's Law, and the complement rule, we can express this as $\frac{1 - P(A \cup B)}{P(B')}$, which corresponds to option (D).

   Let's break down this problem by first understanding the core concepts involved: conditional probability and the complement rule.

   Conditional probability, $P(X|Y)$, represents the probability of event $X$ occurring given that event $Y$ has already occurred. Its definition is fundamental:

   $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$, provided $P(Y) \ne 0$.

   The complement rule states that the probability of an event not happening is $1$ minus the probability of it happening. If $X'$ denotes the complement of event $X$ (i.e., $X$ does not occur), then:

   $P(X') = 1 - P(X)$.

   We are asked to find $P(A'|B')$, which means "the probability that event $A$ does not occur, given that event $B$ does not occur."

   Now, let's work through the problem step-by-step.

   1. Apply the definition of conditional probability. Using the formula $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$, we replace $X$ with $A'$ and $Y$ with $B'$.

   $P(A'|B') = \frac{P(A' \cap B')}{P(B')}$

   The problem states $P(B) \ne 1$. This is important because it implies $P(B') = 1 - P(B) \ne 0$, ensuring that the denominator is not zero and the conditional probability is well-defined.
   

   2. Simplify the numerator using De Morgan's Law.

   The term $A' \cap B'$ represents the event where neither $A$ nor $B$ occurs. This is equivalent to the event that $A \cup B$ (either $A$ or $B$ or both occur) does not occur. This is a direct application of De Morgan's Law for sets:

   $(A \cup B)' = A' \cap B'$

   Therefore, we can rewrite the numerator:
   

   $P(A' \cap B') = P((A \cup B)')$

   3. Apply the complement rule to the numerator. Now we have $P((A \cup B)')$. Using the complement rule $P(X') = 1 - P(X)$, where $X$ is the event $(A \cup B)$:

   $P((A \cup B)') = 1 - P(A \cup B)$

   4. Substitute back into the conditional probability formula. Substitute the simplified numerator back into the expression from Step 1: …

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3. CBSE 2026Set V11 mark
   Q.Choose from $[0, 3, -1, 2, -2, 1]$. If $F$ is an event of a sample space $S$ then $P(S|F)=$ ____.

   ›Reveal solutionSolution

   Since $S\cap F=F$, the conditional probability $P(S\mid F)=1$.

   By the definition of conditional probability (with $P(F)\neq0$),

   $P(S\mid F)=\frac{P(S\cap F)}{P(F)}.$ …

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4. CBSE 2026Set CX1 markMCQ
   Q.If $3P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$, then $P(A\cup B)$ will be:

   (a) $\dfrac{20}{39}$

   (b) $\dfrac{16}{39}$

   (c) $\dfrac{11}{39}$

   (d) $\dfrac{14}{39}$

   ›Reveal solutionSolution

   Using $P(A\cap B)=P(A/B)P(B)$ and the addition rule gives $P(A\cup B)=\dfrac{14}{39}$ — option (d).

   Given: $3P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$.

   So $P(B)=\dfrac{5}{13}$ and $P(A)=\dfrac{1}{3}\cdot\dfrac{5}{13}=\dfrac{5}{39}$.

   Intersection (multiplication rule):

   $P(A\cap B)=P(A/B)\,P(B)=\dfrac{2}{5}\cdot\dfrac{5}{13}=\dfrac{2}{13}.$

   …

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5. CBSE 2026Set A1 markMCQ
   Q.$1 - P(A' \cap B') =$

   (a) $P(A \cap B)$

   (b) $P(A \cup B)$

   (c) $P(A)$

   (d) $P(B)$

   ›Reveal solutionSolution

   $1 - P(A'\cap B') = P(A\cup B)$.

   By De Morgan's law,

   $A' \cap B' = (A\cup B)'.$

   So …

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6. CBSE 2026Set A1 markMCQ
   Q.$P(A) = \frac{7}{13},\ P(B) = \frac{9}{13},\ P(A \cap B) = \frac{4}{13} \Rightarrow P(A/B) =$

   (a) $\frac{4}{9}$

   (b) $\frac{4}{7}$

   (c) $\frac{12}{13}$

   (d) $\frac{1}{6}$

   ›Reveal solutionSolution

   $P(A|B) = \frac{4}{9}$.

   Use the conditional-probability definition:

   $P(A|B) = \frac{P(A\cap B)}{P(B)}.$

   Substitute the given values: …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $P(B)=0.5$ and $P(A\cap B)=0.32$, then write the value of $P(A\mid B)$.

   (a) $\frac{15}{23}$

   (b) $\frac{16}{25}$

   (c) $\frac{16}{27}$

   (d) $\frac{16}{23}$

   ›Reveal solutionSolution

   By the definition of conditional probability, $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{16}{25}$.

   The conditional probability of $A$ given $B$ is defined as

   $P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B) eq0$

   …

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8. CBSE 2026Set ANNUAL1 mark
   Q.A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

   ›Reveal solutionSolution

   List the equally likely outcomes for two children, restrict to those with at least one boy, then find the fraction that are both boys.

   Sample space $=\{BB, BG, GB, GG\}$, each equally likely.

   Given at least one boy: reduced sample space $=\{BB,BG,GB\}$ (3 outcomes).

   …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $P(A) = 0.8$, $P(B) = 0.5$ and $P\left(\dfrac{B}{A}\right) = 0.4$ then $P(A \cap B) =$

   (a) 0.8

   (b) 0.5

   (c) 0.32

   (d) 0.4

   ›Reveal solutionSolution

   Use the multiplication rule of conditional probability: $P(A\cap B) = P(B\mid A)\cdot P(A)$.

   Given $P(A)=0.8$, $P(B\mid A)=0.4$.

   $P(A\cap B) = P(B\mid A)\cdot P(A) = 0.4 \times 0.8 = 0.32$.

   …

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10. CBSE 2026Set ANNUAL1 markMCQ
    Q.If $P(A) = \dfrac{3}{10}$, $P(B) = \dfrac{2}{5}$ and $P(A \cup B) = \dfrac{3}{5}$, then $P(B/A)$ is:

    (a) $\dfrac{1}{4}$

    (b) $\dfrac{1}{3}$

    (c) $\dfrac{5}{12}$

    (d) $\dfrac{7}{12}$

    ›Reveal solutionSolution

    Find $P(A\cap B)$ from the addition rule, then use the conditional probability formula.

    Given $P(A)=\dfrac{3}{10}$, $P(B)=\dfrac{2}{5}$, $P(A\cup B)=\dfrac{3}{5}$.

    …

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11. CBSE 2026Set ANNUAL1 mark
    Q.A doctor is to visit a patient. From past experience it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $\dfrac{3}{10}, \dfrac{1}{5}, \dfrac{1}{10}$ and $\dfrac{2}{5}$. The probabilities that he will be late are $\dfrac{1}{4}, \dfrac{1}{3}$ and $\dfrac{1}{12}$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. Probability that he will not be late, when he comes by other means of transport.

    ›Reveal solutionSolution

    The problem statement itself states that if the doctor comes by other means, he will never be late.

    Let $E_1,E_2,E_3,E_4$ denote the events that the doctor comes by train, bus, scooter, or other means, with $P(E_1)=\dfrac{3}{10}$, $P(E_2)=\dfrac{1}{5}$, $P(E_3)=\dfrac{1}{10}$, $P(E_4)=\dfrac{2}{5}$ (these sum to 1, confirming they form a complete set of cases).

    …

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.If P(A) = 1/2, P(B) = 3/8 and P(A∪B) = 27/40 then P(A/B) is equal to:

    (a) 2/5

    (b) 8/15

    (c) 2/3

    (d) 5/8

    ›Reveal solutionSolution

    First find $P(A\cap B)$ using the addition rule, then apply the conditional probability formula $P(A/B)=\frac{P(A\cap B)}{P(B)}$.

    Given $P(A) = \frac12$, $P(B)=\frac38$, $P(A\cup B) = \frac{27}{40}$.

    Using the addition rule $P(A\cup B) = P(A)+P(B)-P(A\cap B)$:

    $P(A\cap B) = P(A)+P(B)-P(A\cup B) = \frac{20}{40}+\frac{15}{40}-\frac{27}{40} = \frac{8}{40} = \frac15$

    …

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