Q.A fair die is rolled once. Let $E$ be the event "the number is even" and $F$ be the event "the number is at least $4$". Find $P(E\mid F)$ and $P(F\mid E)$.

Showing the 12 most recent of 88 on this concept.

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.

   Watch out

   A common mistake is to forget that the condition reduces the sample space. Some students count primes among all six numbers ($3$ primes) and odd numbers among all six ($3$ odds), then incorrectly write $\frac{3}{3}=1$ or $\frac{3}{6} \div \frac{3}{6}=1$. The formula forces you to consider only the overlap, which correctly gives $\frac{2}{3}$.

   5. Check the truth of Reason (R).

      The formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$ is the standard definition of conditional probability (provided $P(B) \neq 0$). It is always true. So Reason (R) is true.

   6. Does Reason (R) correctly explain Assertion (A)?

      Yes — the assertion's value $\frac{2}{3}$ is obtained directly by substituting the probabilities into this formula. The reason is not just a true statement; it is the very tool used to verify the assertion.

   Tip

   When both assertion and reason are true, and the reason is the principle that justifies the assertion, the answer is option (A). If the reason were true but irrelevant to the assertion, it would be option (B).

   ✓Final answer

   The correct option is (A) — Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

2. 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 $P(F)\neq0$),

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

   The sample space $S$ contains every outcome, so $S\cap F=F$ and $P(S\cap F)=P(F)$. Therefore

   $P(S\mid F)=\frac{P(F)}{P(F)}=1.$

   ✓Final answer

   $1$

3. 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):

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

   Union (addition rule), with common denominator $39$:

   $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\dfrac{5}{39}+\dfrac{15}{39}-\dfrac{6}{39}=\dfrac{14}{39}.$

   ✓Final answer

   Option (d) $\;P(A\cup B)=\dfrac{14}{39}$.

4. 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:

   $P(A|B) = \frac{\frac{4}{13}}{\frac{9}{13}} = \frac{4}{13}\cdot\frac{13}{9} = \frac{4}{9}.$

   ✓Final answer

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

5. 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

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

   Substituting the given values $P(B)=0.5$ and $P(A\cap B)=0.32$:

   $P(A\mid B)=\frac{0.32}{0.5}=0.64=\frac{64}{100}=\frac{16}{25}$

   ✓Final answer

   The correct option is (b) $\frac{16}{25}$.

6. 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).

   Of these, both boys occurs only in $\{BB\}$ (1 outcome).

   $P(\text{both boys}\mid\text{at least one boy}) = \dfrac13$.

   ✓Final answer

   The probability is $1/3$.

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

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

   (Note: $P(B)=0.5$ is extra information not needed for this computation.)

   ✓Final answer

   (c) $0.32$.

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

   $P(A\cap B)=P(A)+P(B)-P(A\cup B)=\dfrac{3}{10}+\dfrac{4}{10}-\dfrac{6}{10}=\dfrac{1}{10}$

   $P(B/A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{1/10}{3/10}=\dfrac{1}{3}$

   ✓Final answer

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

9. 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).

   Let $L$ be the event that he is late. We are given $P(L|E_1)=\dfrac{1}{4}$, $P(L|E_2)=\dfrac{1}{3}$, $P(L|E_3)=\dfrac{1}{12}$, and "if he comes by other means, he will not be late", i.e. $P(L|E_4)=0$.

   So $P(\text{not late}\,|\,E_4)=1-P(L|E_4)=1-0=1$.

   ✓Final answer

   $P(\text{not late}\mid \text{other means}) = 1$

10. 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 $P(A\cup B) = P(A)+P(B)-P(A\cap B)$:

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

    Now, conditional probability:

    $P(A/B) = \frac{P(A\cap B)}{P(B)} = \frac{1/5}{3/8} = \frac15\times\frac83 = \frac{8}{15}$

    ✓Final answer

    $P(A/B) = \dfrac{8}{15}$ (option b).

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.Assertion (A): If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 then P(A∩B) = 0.32 Reason (R): Conditional Probability of 'B' when event A has occurred is given by P(B/A) = P(A∩B) / P(A)

    (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) Both Assertion (A) and Reason (R) are false.

    ›Reveal solutionSolution

    Applying the conditional-probability formula in R directly to the given numbers reproduces the value claimed in A.

    Reason (R): $P(B/A) = \dfrac{P(A\cap B)}{P(A)}$ is the standard definition of conditional probability — true.

    Checking Assertion (A): Given $P(A)=0.8,\ P(B/A)=0.4$. Rearranging R's formula:

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

    This matches A exactly, so A is true, and R is precisely the formula used to derive it.

    (Note: $P(B)=0.5$ is extra information, not needed for this particular computation.)

    ✓Final answer

    Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A. (Option a)

12. CBSE 2026Set ANNUAL1 markMCQ
    Q.If $P(A) = \dfrac{1}{2}$ and $P(B) = 0$, then $P(A|B)$ is

    (a) 0

    (b) $\dfrac{1}{2}$

    (c) 1

    (d) Not defined

    ›Reveal solutionSolution

    Conditional probability is defined as $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$, which requires dividing by $P(B)$ — impossible when $P(B)=0$.

    Given $P(A)=\dfrac12$, $P(B)=0$.

    By definition:

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

    Since $P(B)=0$, this involves division by zero, which is undefined regardless of the value of $P(A\cap B)$ (and note $P(A\cap B)\le P(B)=0$ forces $P(A\cap B)=0$ too, giving the indeterminate $\frac00$).

    $P(A\mid B) \text{ is not defined.}$

    ✓Final answer

    (d) Not defined