$A$ and $B$ are two events such that $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$ and $P(A \cap B) = \dfrac{1}{4}$. Find: (i) $P(A \mid B)$ (ii) $P(B \mid A)$ (iii) $P(A' \mid B)$ (iv) $P(A' \mid B')$

A and B are two events such that P(A) = (1)/(2), P(B) = (1)/(3) and P(A ∩ B) = (1)/(4). Find: (i) P(A mid B) (

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A

(i)

P(A \mid B)

(ii)

P(B \mid A)

(iii)

P(A' \mid B)

(iv)

P(A' \mid B')

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Using the definition of conditional probability $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$ , we compute each required probability directly from the given values. The answers are: (i) $\frac{3}{4}$ ,

(ii) $\frac{1}{2}$ ,

(iii) $\frac{1}{4}$ ,

(iv) $\frac{5}{8}$ .

Why conditional probability works here

Conditional probability answers the question: If we know that event $B$ has happened, how does that change the chance of event $A$ ? The key idea is that knowing $B$ occurred shrinks the "universe" of possible outcomes from the whole sample space down to just $B$ . So the probability of $A$ given $B$ is the proportion of $B$ that also belongs to $A$ — that is, $P(A \cap B)$ divided by $P(B)$ .

The formula is symmetric in logic: $P(A|B) = \frac{P(A \cap B)}{P(B)}$ , provided $P(B) \neq 0$ . For the complement events, we use the fact that $P(A' \cap B) = P(B) - P(A \cap B)$ , since $A' \cap B$ is just the part of $B$ that lies outside $A$ .

Let's work through each part.

1. Finding $P(A \mid B)$

We have $P(A \cap B) = \frac{1}{4}$ and $P(B) = \frac{1}{3}$ .

P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{1}{4} \times \frac{3}{1} = \frac{3}{4}

So given that $B$ occurred, there is a $\frac{3}{4}$ chance that $A$ also occurred.

2. Finding $P(B \mid A)$

Now we condition on $A$ instead. $P(A) = \frac{1}{2}$ .

P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{1/2} = \frac{1}{4} \times \frac{2}{1} = \frac{1}{2}

Notice that $P(B|A) = \frac{1}{2}$ is larger than $P(B) = \frac{1}{3}$ , which tells us that $A$ and $B$ are positively associated — knowing $A$ happened makes $B$ more likely.

3. Finding $P(A' \mid B)$

Here we want the probability that $A$ does not occur, given that $B$ has occurred. The event $A' \cap B$ is the part of $B$ that is outside $A$ .

P(A' \cap B) = P(B) - P(A \cap B) = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}

Now divide by $P(B)$ :

P(A' \mid B) = \frac{P(A' \cap B)}{P(B)} = \frac{1/12}{1/3} = \frac{1}{12} \times \frac{3}{1} = \frac{3}{12} = \frac{1}{4}

Tip

A quicker way: since $A$ and $A'$ partition the sample space, $P(A'|B) = 1 - P(A|B) = 1 - \frac{3}{4} = \frac{1}{4}$ . This works because conditional probabilities sum to 1 when conditioning on the same event.

4. Finding $P(A' \mid B')$

Now we condition on $B$ not happening. First, find $P(B')$ : …

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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 69 on this concept.

AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$P(A) = \dfrac{1}{4}$
›Reveal solutionSolution
Using $P(B|A)$ to find $P(A\cap B)$ , then dividing by $P(A|B)$ , gives $P(B)=\dfrac13$ .

Concept and Intuition

The conditional probability definitions $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ and $P(B|A)=\dfrac{P(A\cap B)}{P(A)}$ share the common quantity $P(A\cap B)$ — computing it from one equation lets us solve the other for the unknown probability.

Step-by-Step Solution

From $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac23$ , and $P(A)=\dfrac14$ : $P(A\cap B)=\dfrac23\times\dfrac14=\dfrac16$ .

From $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac12$ : $P(B)=\dfrac{P(A\cap B)}{P(A|B)}=\dfrac{1/6}{1/2}=\dfrac13$ .

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$P(A) = 0.5, P(B) = 0.4, P(A \cap B) = 0.3$
›Reveal solutionSolution
Using De Morgan's law $A'\cap B'=(A\cup B)'$ and the conditional probability formula gives $P(A'|B')=\dfrac23$ .

Concept and Intuition

$P(A'|B')$ asks: given we're outside $B$ , what's the chance we're also outside $A$ ? The key trick is that $A'\cap B' = (A\cup B)'$ (De Morgan), which is easy to compute from $P(A\cup B)$ .

Step-by-Step Solution

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.3=0.6$ .

$P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)=1-0.6=0.4$ .

$P(B')=1-P(B)=1-0.4=0.6$ . …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$P(A|B)$
›Reveal solutionSolution
Counting the 36 equally likely dice outcomes directly gives $P(A|B)=\frac15$ and $P(B|A)=\frac12$ .

Concept and Intuition

With two fair dice there are $36$ equally likely outcomes, so every probability here reduces to simple counting: enumerate the outcomes in $A$ , in $B$ , and in $A\cap B$ , then apply the conditional probability formula $P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$ directly as a ratio of counts.

Step-by-Step Solution

$A$ = "same number on each die": outcomes $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — $6$ outcomes, so $P(A)=\dfrac6{36}$ .

$B$ = "sum $>7$ ", i.e. sum $\in\{8,9,10,11,12\}$ . Counting pairs per sum: sum 8 has 5 pairs, 9 has 4, 10 has 3, 11 has 2, 12 has 1 — total $5+4+3+2+1=15$ outcomes, so $P(B)=\dfrac{15}{36}$ .

$A\cap B$ : doubles with sum $>7$ — check each double: $(4,4)$ sum 8 ✓, $(5,5)$ sum 10 ✓, $(6,6)$ sum 12 ✓; $(1,1),(2,2),(3,3)$ have sums $2,4,6$ , all $\le7$ — excluded. So $A\cap B$ has $3$ outcomes, $P(A\cap B)=\dfrac3{36}$ . …
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AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$P(B)\neq1$
›Reveal solutionSolution
Direct application of the conditional-probability definition to the complement event gives $P(A\mid B^c)=\dfrac{P(A)-P(A\cap B)}{1-P(B)}$ .

Concept and Intuition

$A\cap B^c$ is exactly the part of $A$ that does not overlap with $B$ , so its probability is $P(A)$ minus the overlapping piece $P(A\cap B)$ . Dividing by $P(B^c)=1-P(B)$ (valid since $P(B)\ne1$ ) gives the conditional probability.

Step-by-Step Solution

$P(A\mid B^c)=\dfrac{P(A\cap B^c)}{P(B^c)}$ by definition (needs $P(B^c)\ne0$ , i.e. $P(B)\ne1$ , as given).

$A=(A\cap B)\cup(A\cap B^c)$ , a disjoint union, so $P(A)=P(A\cap B)+P(A\cap B^c)\Rightarrow P(A\cap B^c)=P(A)-P(A\cap B)$ .

$P(B^c)=1-P(B)$ . …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$P(\bar A)=0.3$
›Reveal solutionSolution
This tests careful set manipulation of conditional probability with a compound conditioning event $A\cup\bar B$ . Answer: 0.25.

Concept and Intuition

Conditional probability $P(B\mid E) = \dfrac{P(B\cap E)}{P(E)}$ works exactly the same way when $E$ is a compound event like $A\cup\bar B$ — we just need to correctly compute $P(E)$ and $P(B\cap E)$ using set algebra (distributing intersection over union, and using the fact that $B$ and $\bar B$ are disjoint).

Step-by-Step Solution

From $P(\bar A)=0.3$ , get $P(A)=1-0.3=0.7$ .

$P(A\cap\bar B)=0.5$ is given directly, so $P(A\cap B)=P(A)-P(A\cap\bar B)=0.7-0.5=0.2$ .

Compute $P(A\cup\bar B)$ using inclusion-exclusion: $P(A\cup\bar B)=P(A)+P(\bar B)-P(A\cap\bar B)$ . Here $P(\bar B)=1-0.4=0.6$ , so $P(A\cup\bar B)=0.7+0.6-0.5=0.8$ . …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$A$
›Reveal solutionSolution
Mutually exclusive events make $P(A\cup B)$ a simple sum, and since $A$ is entirely contained in $A\cup B$ , the conditional probability reduces to $P(A)/P(A\cup B)$ .

Concept and Intuition

For mutually exclusive $A,B$ : $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ . Also $A\cap(A\cup B)=A$ always, so $P(A\mid A\cup B)=\dfrac{P(A\cap(A\cup B))}{P(A\cup B)}=\dfrac{P(A)}{P(A\cup B)}$ .

Step-by-Step Solution

$P(A\cup B) = P(A)+P(B) = \dfrac14+\dfrac37$ .

Common denominator 28: $\dfrac{7}{28}+\dfrac{12}{28}=\dfrac{19}{28}$ . …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$P(B) \neq 0$
›Reveal solutionSolution
Directly apply the definition of conditional probability together with De Morgan's law; the answer is $\frac{1-P(A\cup B)}{P(\bar B)}$ .

Concept and Intuition

Conditional probability is defined as $P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$ . Here $X=\bar A$ , $Y=\bar B$ . By De Morgan's law, $\bar A\cap\bar B = \overline{A\cup B}$ , so its probability is $1-P(A\cup B)$ .

Step-by-Step Solution

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

$\bar A\cap\bar B=\overline{A\cup B}$ (De Morgan).

So $P(\bar A\cap\bar B)=1-P(A\cup B)$ . …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$P(\bar{F}) = 0.7$
›Reveal solutionSolution
This tests the basic conditional-probability formula together with the complement rule. $P(E|F)=2/3$ .

Concept and Intuition

$P(E|F)$ asks: given that $F$ has already happened, what fraction of that reduced sample space does $E\cap F$ occupy? It is always $\dfrac{P(E\cap F)}{P(F)}$ , so first we need $P(F)$ itself, which is given indirectly through its complement.

Step-by-Step Solution

$P(\bar F) = 0.7 \Rightarrow P(F) = 1 - 0.7 = 0.3$ .

$P(E\cap F) = 0.2$ is given directly. …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$4/5$
›Reveal solutionSolution
Chain the given conditional probabilities to find $P(Q)$ and $P(P\cap Q)$ , then use the complement rule — the answer is (A) $4/5$ .

Concept and Intuition

Conditional probability definitions let us cross-multiply to recover the joint probability $P(P\cap Q)$ from either conditional. Once $P(P), P(Q), P(P\cap Q)$ are all known, De Morgan's law converts "neither event" into the complement of the union.

Step-by-Step Solution

Given: $P(P)=\tfrac14$ , $P(P|Q)=\tfrac12$ , $P(Q|P)=\tfrac13$ .

$P(P\cap Q) = P(Q|P)\cdot P(P) = \tfrac13\times\tfrac14 = \tfrac1{12}$ .

Also $P(P\cap Q) = P(P|Q)\cdot P(Q) \Rightarrow \tfrac1{12} = \tfrac12\cdot P(Q) \Rightarrow P(Q)=\tfrac16$ .

$P(P\cup Q) = P(P)+P(Q)-P(P\cap Q) = \tfrac14+\tfrac16-\tfrac1{12} = \tfrac{3}{12}+\tfrac{2}{12}-\tfrac{1}{12}=\tfrac{4}{12}=\tfrac13$ .

By De Morgan's law, "neither P nor Q applies" is the complement of $P\cup Q$ : $P(\overline P\cap\overline Q) = 1-\tfrac13=\tfrac23$ .

$P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$ . …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$A$
›Reveal solutionSolution
Chain the conditional-probability definition twice: first get $P(A\cap B)$ from $P(A/B)$ , then get $P(B/A)$ from that.

Concept and Intuition

Conditional probability is defined as $P(E/F)=\dfrac{P(E\cap F)}{P(F)}$ . Given one conditional probability, we can back out the joint probability $P(A\cap B)$ , and then use it (with the definition again, but conditioning the other way) to find the other conditional probability.

Step-by-Step Solution

$P(A/B) = \dfrac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B) = P(A/B)\cdot P(B) = \dfrac{2}{5}\times\dfrac{2}{7} = \dfrac{4}{35}$ .

Now use $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$ . …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\dfrac{1}{4}$
›Reveal solutionSolution
This is a classic conditional-probability trap: conditioning on "at least one girl" (not "the elder/first child is a girl") leaves 3 equally likely outcomes, of which 1 is both-girls, giving $\dfrac13$ .

Concept and Intuition

With two children, birth order matters for counting equally likely outcomes: $BB, BG, GB, GG$ , each with probability $1/4$ . The event "at least one girl" is a set of outcomes, not a statement about a specific (e.g. first) child, so it correctly removes only $BB$ and keeps three outcomes, not two. This distinguishes it from the simpler (and different) question "given the elder child is a girl," which would leave only $\{GB,GG\}$ and give probability $1/2$ .

Step-by-Step Solution

Sample space (ordered by birth, say elder-younger): $\{BB,BG,GB,GG\}$ , each with probability $\dfrac14$ .

Event $E$ = "at least one girl" = $\{BG,GB,GG\}$ , so $P(E)=\dfrac34$ . …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$P(A|A\cap B) + P(B|A\cap B) =$
›Reveal solutionSolution
Both conditional probabilities equal $1$ because $A\cap B$ is a subset of both $A$ and $B$ , so their sum is $2$ .

Concept and Intuition

Conditioning an event on its own subset (or superset relationship) often collapses to a probability of exactly $1$ : if $E\subseteq F$ , then $P(F|E)=1$ , since knowing $E$ occurred guarantees $F$ occurred too.

Step-by-Step Solution

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

Since $A\cap(A\cap B) = A\cap B$ (intersecting with $A$ again changes nothing, as $A\cap B\subseteq A$ ), this is $\dfrac{P(A\cap B)}{P(A\cap B)}=1$ .

Similarly, $P(B|A\cap B) = \dfrac{P\big(B\cap(A\cap B)\big)}{P(A\cap B)} = \dfrac{P(A\cap B)}{P(A\cap B)}=1$ .

Sum $=1+1=2$ . …
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Q. $A$ and $B$ are two events such that $P(A) = \dfrac{1}{2}$ , $P(B) = \dfrac{1}{3}$ and $P(A \cap B) = \dfrac{1}{4}$ . Find:

Concept understanding — Conditional Probability

Conditional Probability

The Idea: Shrink the Sample Space

The Definition

Two Cautions

Method: Computing a Family of Conditional Probabilities Efficiently

Steps

Common Mistakes

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Q.AAA and BBB are two events such that P(A)=12P(A) = \dfrac{1}{2}P(A)=21​, P(B)=13P(B) = \dfrac{1}{3}P(B)=31​ and P(A∩B)=14P(A \cap B) = \dfrac{1}{4}P(A∩B)=41​. Find:

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Q. $A$ and $B$ are two events such that $P(A) = \dfrac{1}{2}$ , $P(B) = \dfrac{1}{3}$ and $P(A \cap B) = \dfrac{1}{4}$ . Find:

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