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) $P(A \cap B) = 2 P(B)$

> [!TLDR] > 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: $$P(A \cap B) = \frac{1}{2} P(B)$$ …

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

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

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

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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.

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)}

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)

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)

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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 88 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\dfrac{2}{3}$
›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).

Define the events precisely.

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

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

Compute $P(B)$ .

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

Compute $P(A \cap B)$ .

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

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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CBSE 2026Set V11 mark
$[0, 3, -1, 2, -2, 1]$
›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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CBSE 2026Set CX1 markMCQ
$3P(A)=P(B)=\dfrac{5}{13}$
(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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CBSE 2026Set A1 markMCQ
$P(A) = \frac{7}{13},\ P(B) = \frac{9}{13},\ P(A \cap B) = \frac{4}{13} \Rightarrow P(A/B) =$
(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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CBSE 2026Set ANNUAL1 markMCQ
$P(B)=0.5$
(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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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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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = 0.8$
(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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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{3}{10}$
(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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CBSE 2026Set ANNUAL1 mark
$\dfrac{3}{10}, \dfrac{1}{5}, \dfrac{1}{10}$
›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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CBSE 2026Set ANNUAL1 markMCQ
$Q. If P(A) = 1/2, P(B) = 3/8 and P(A\cupB) = 27/40 then P(A/B) is equal to:$
(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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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A): If P(A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 then P(A\capB) = 0.32 Reason (R): Conditional Probability of 'B' when event A has occurred is given by P(B/A) = P(A\capB) / P(A)$
(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:

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

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CBSE 2026Set ANNUAL1 markMCQ
$P(A) = \dfrac{1}{2}$
(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:

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

…

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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) $P(A \cap B) = 2 P(B)$

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

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