Q.For any two events $A$ and $B$, if $P(A) = \frac{1}{2}$, $P(B) = \frac{2}{3}$ and $P(A \cap B) = \frac{1}{4}$, then $P(\bar{A}/\bar{B})$ equals:
(A) $\frac{3}{8}$
(B) $\frac{8}{9}$
(C) $\frac{5}{8}$
(D) $\frac{1}{4}$

Showing the 12 most recent of 88 on this concept.

1. CBSE 2025Set 65/1/11 markMCQ
   Q.If E and F are two independent events such that $P(E) = \frac{2}{3}$, $P(F) = \frac{3}{7}$, then $P(E/\bar{F})$ is equal to : (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) $\frac{2}{3}$ (D) $\frac{7}{9}$

   ›Reveal solutionSolution

   For independent events, conditioning on the complement of one event does not change the probability of the other. Since $E$ and $F$ are independent, $P(E \mid \bar{F}) = P(E) = \frac{2}{3}$, which corresponds to option (C).

   Why conditional probability and independence work together

   The notation $P(E / \bar{F})$ means $P(E \mid \bar{F})$ — the probability that $E$ occurs, given that $F$ does not occur. The natural instinct is to reach for the conditional probability formula:

   $P(E \mid \bar{F}) = \frac{P(E \cap \bar{F})}{P(\bar{F})}$

   But here’s the key: independence between $E$ and $F$ tells us something deeper. If two events are independent, then knowing whether $F$ happened gives you zero information about $E$. That intuition extends to the complement too — if $F$ doesn’t happen, it still tells you nothing about $E$.

   So before doing any heavy algebra, we can already guess: the answer should be exactly $P(E)$, unchanged.

   ---

   Step-by-step reasoning

   1. State what independence means mathematically. For independent events $E$ and $F$:

   $P(E \cap F) = P(E) \cdot P(F)$

   This is the definition. But independence also implies that $E$ is independent of $\bar{F}$ — because if $F$ gives no information about $E$, then not-$F$ also gives no information. We can prove this quickly.

   2. Find $P(E \cap \bar{F})$ using the complement relationship. Any event $E$ can be split into two disjoint parts: when $F$ happens and when $F$ does not happen.

   $E = (E \cap F) \cup (E \cap \bar{F})$

   Since these two are mutually exclusive:

   $P(E) = P(E \cap F) + P(E \cap \bar{F})$

   Substitute $P(E \cap F) = P(E)P(F)$:

   $\frac{2}{3} = \left(\frac{2}{3} \cdot \frac{3}{7}\right) + P(E \cap \bar{F})$

   $\frac{2}{3} = \frac{2}{7} + P(E \cap \bar{F})$

   $P(E \cap \bar{F}) = \frac{2}{3} - \frac{2}{7} = \frac{14 - 6}{21} = \frac{8}{21}$

   3. Find $P(\bar{F})$. …

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2. CBSE 2020Set 65/1/11 markMCQ
   Q.If A and B are two independent events, where $P (A) = \frac{1}{3}$ and $P (B) = \frac{1}{4}$, then $P(B'|A)$ is equal to (A) $\frac{1}{4}$ (B) $\frac{1}{3}$ (C) $\frac{3}{4}$ (D) 1

   ›Reveal solutionSolution

   For independent events, the occurrence of A gives no information about B, so $P(B'|A) = P(B') = 1 - P(B) = \frac{3}{4}$.

   The key here is conditional probability — the probability that B does not happen, given that A has already happened. Many students rush to plug numbers into the conditional probability formula without first checking whether the events are independent. That’s where the trap lies.

   When two events are independent, knowing that one has occurred tells you nothing about the other. So $P(B|A) = P(B)$. By the same logic, $P(B'|A) = P(B')$. The condition “given A” becomes irrelevant.

   Let’s walk through it formally.

   1. Recall the definition of conditional probability For any two events A and B (with $P(A) > 0$):

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

   This is always true. But we can simplify it if we know something about the relationship between A and B.

   2. Use the independence condition A and B are independent. That means:

   $P(A \cap B) = P(A) \cdot P(B)$

   Independence also extends to complements: if A and B are independent, then A and $B'$ are independent too.

   Why? Because:

   $P(A \cap B') = P(A) - P(A \cap B) = P(A) - P(A)P(B) = P(A)[1 - P(B)] = P(A)P(B')$

   So $A$ and $B'$ are independent.

   3. Apply independence to the conditional probability Since A and $B'$ are independent:

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

   This is the cleanest path — no messy fraction needed.

   4. Compute $P(B')$ Given $P(B) = \frac{1}{4}$: $P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4}$ …

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3. CBSE 20201 markMCQ
   Q.If A and B be two events such that $P(A) = 0.2$, $P(B) = 0.4$ and $P(A \cap B) = 0.08$, then $P(A|B)$ is (A) $0.02$ (B) $0.2$ (C) $0.4$ (D) $0.08$

   ›Reveal solutionSolution

   Conditional probability $P(A|B)$ is the probability of $A$ given $B$ has occurred. Using the formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$, we get $\frac{0.08}{0.4} = 0.2$. The correct option is (B).

   The core idea here is conditional probability — the chance that event $A$ happens, once we already know that event $B$ has happened. This isn't the same as the plain probability of $A$; knowing $B$ has occurred shrinks the "possible world" from everything to just the outcomes where $B$ is true.

   Think of it visually: imagine a rectangle representing all possible outcomes. $A$ and $B$ are overlapping circles inside it. $P(A|B)$ asks: out of the area of $B$, what fraction is also inside $A$? That fraction is exactly the overlap area $P(A \cap B)$ divided by the area of $B$, $P(B)$.

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

   This formula works only when $P(B) > 0$, which is true here since $P(B) = 0.4$.

   Now let's plug in the numbers.

   1. Identify the given values.

      $P(A) = 0.2$, $P(B) = 0.4$, and $P(A \cap B) = 0.08$.

      Notice that $P(A \cap B)$ is not zero — the events are not mutually exclusive. Also, $0.08 = 0.2 \times 0.4$, so $A$ and $B$ are actually independent events. But we don't need that fact here; the conditional formula works regardless.

   2. Apply the conditional probability formula.

   $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.08}{0.4}$

   3. Simplify the fraction. …

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4. CBSE 2023Set 65/1/11 markMCQ
   Q.If $P\left(\frac{A}{B}\right) = 0.3$, $P(A) = 0.4$ and $P(B) = 0.8$, then $P\left(\frac{B}{A}\right)$ is equal to:

   (a) $0.6$

   (b) $0.3$

   (c) $0.06$

   (d) $0.4$

   ›Reveal solutionSolution

   Use the definition of conditional probability to find $P(A \cap B)$ from the given $P(A|B)$, then apply it again to compute $P(B|A)$. The answer is $0.6$.

   Understanding Conditional Probability

   Conditional probability measures the likelihood of an event occurring given that another event has already occurred. The notation $P(A|B)$ reads as "the probability of $A$ given $B$" and is defined as:

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

   This formula tells us that to find the probability of $A$ happening when we know $B$ has happened, we look at the overlap between $A$ and $B$ relative to the size of $B$ itself.

   The key insight here is that both $P(A|B)$ and $P(B|A)$ depend on the same intersection $P(A \cap B)$, just normalized by different denominators. Once we know the intersection, we can compute either conditional probability.

   Solution

   1. Extract the intersection probability

   We're given $P(A|B) = 0.3$, $P(A) = 0.4$, and $P(B) = 0.8$. Using the definition of conditional probability:

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

   Substituting the known values:

   $0.3 = \frac{P(A \cap B)}{0.8}$

   Solving for $P(A \cap B)$:

   $P(A \cap B) = 0.3 \times 0.8 = 0.24$ …

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5. CBSE 2024Set 65/1/11 markMCQ
   Q.If $P(A|B) = P(A'|B)$, then which of the following statements is correct ? (A) $P(A) = P(A')$ (B) $P(A) = 2 P(B)$ (C) $P(A \cap B) = \frac{1}{2} P(B)$ (D) $P(A \cap B) = 2 P(B)$

   ›Reveal solutionSolution

   The condition $P(A|B) = P(A'|B)$ means that given $B$, events $A$ and $A'$ are equally likely. This forces $P(A \cap B) = P(A' \cap B)$, which simplifies to $P(A \cap B) = \frac{1}{2} P(B)$. The correct option is (C).

   The key here is to understand what conditional probability actually says. $P(A|B)$ is the probability that $A$ happens, given that $B$ has already occurred. So when we say $P(A|B) = P(A'|B)$, we are told that inside the world of $B$, the chance of $A$ happening is exactly the same as the chance of $A$ not happening. That means, within $B$, $A$ and its complement are equally likely — each has probability $\frac{1}{2}$ of occurring, conditional on $B$.

   Let’s translate that into algebra.

   1. Write the definition of conditional probability for both sides:

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

   The given equality is:

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

   2. Since $P(B) > 0$ (otherwise conditional probability isn’t defined), we can multiply both sides by $P(B)$ and get:

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

   3. Now, note that $A \cap B$ and $A' \cap B$ are disjoint sets whose union is exactly $B$ (because every outcome in $B$ is either in $A$ or not in $A$). So:

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

   4. Since the two probabilities are equal, let each be $x$. Then:

   $x + x = P(B) \quad \Rightarrow \quad 2x = P(B) \quad \Rightarrow \quad x = \frac{1}{2} P(B)$

   But $x = P(A \cap B)$, so:

   $P(A \cap B) = \frac{1}{2} P(B)$ …

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6. CBSE 2024Set 65/3/11 markMCQ
   Q.Let E and F be two events such that $P(E) = 0.1$, $P(F) = 0.3$, $P(E \cup F) = 0.4$, then $P(F\,|\,E)$ is: (A) $0.6$ (B) $0.4$ (C) $0.5$ (D) $0$

   ›Reveal solutionSolution

   To find the conditional probability $P(F\,|\,E)$, we first determine the probability of the intersection $P(E \cap F)$ using the Addition Rule, and then divide by $P(E)$. The events E and F are mutually exclusive, leading to $P(E \cap F) = 0$, so $P(F\,|\,E) = \boxed{0}$.

   When we talk about $P(F\,|\,E)$, we are asking for the probability that event F occurs, given that event E has already occurred. This is called conditional probability. The key idea here is that the sample space for event F is no longer the entire original sample space, but rather it is restricted to only those outcomes where event E has happened.

   The formula for conditional probability is:

   $P(F\,|\,E) = \frac{P(F \cap E)}{P(E)}$

   This formula tells us that the probability of F given E is the probability of both F and E happening, divided by the probability of E happening. We need to find $P(F \cap E)$ first, as $P(E)$ is already given.

   Let's break down the solution step-by-step.

   1. Identify Given Information and What's Needed:

      We are given:

      • $P(E) = 0.1$
      • $P(F) = 0.3$
      • $P(E \cup F) = 0.4$

      We need to find $P(F\,|\,E)$. To use the conditional probability formula, we require $P(F \cap E)$ and $P(E)$. We already have $P(E)$.

   2. Find the Probability of the Intersection, $P(E \cap F)$:

      We can use the Addition Rule for probabilities, which relates the probabilities of the union, individual events, and their intersection:

   $P(E \cup F) = P(E) + P(F) - P(E \cap F)$

   We can rearrange this formula to solve for $P(E \cap F)$:
   

   $P(E \cap F) = P(E) + P(F) - P(E \cup F)$

   Now, substitute the given values:
   

   $P(E \cap F) = 0.1 + 0.3 - 0.4$

   $P(E \cap F) = 0.4 - 0.4$

   $$P(E \cap F) = 0$$ …
   

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7. CBSE 2020Set 65/1/11 mark
   Q.Two cards are drawn successively and without replacement from a well-shuffled deck of 52 cards. Find the probability that one card is red and the other is black.

   ›Reveal solutionSolution

   The probability that one card is red and the other is black when drawing two cards without replacement is $\frac{26}{51}$. This comes from the fact that the first card can be either colour, and the second card must be the opposite colour — the order doesn't matter because the two favourable sequences are mutually exclusive and symmetric.

   Why conditional probability is the natural tool here

   When we draw without replacement, the outcome of the second draw depends on what happened in the first. That's exactly the situation conditional probability handles: $P(A \cap B) = P(A) \cdot P(B \mid A)$.

   We want one red and one black. There are two ways this can happen:

   • First red, then black.
   • First black, then red.

   These two sequences are mutually exclusive (they can't both happen in the same draw), so we can add their probabilities.

   ---

   Step-by-step reasoning

   1. Probability of first red, then black

   • First card red: there are 26 red cards out of 52, so $P(\text{first red}) = \frac{26}{52} = \frac12$.
   • After removing one red, 51 cards remain, of which 26 are black. So $P(\text{second black} \mid \text{first red}) = \frac{26}{51}$.
   • Therefore:

   $P(\text{red then black}) = \frac12 \times \frac{26}{51} = \frac{26}{102} = \frac{13}{51}.$

   2. Probability of first black, then red

   • First card black: 26 black out of 52, so $P(\text{first black}) = \frac12$.
   • After removing one black, 51 cards remain, of which 26 are red. So $P(\text{second red} \mid \text{first black}) = \frac{26}{51}$.
   • Therefore:

   $P(\text{black then red}) = \frac12 \times \frac{26}{51} = \frac{13}{51}.$

   3. Add the two mutually exclusive cases

   $P(\text{one red, one black}) = \frac{13}{51} + \frac{13}{51} = \frac{26}{51}.$ …

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8. 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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9. 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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10. 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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11. 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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12. 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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