Q.A and B are two events such that $P(A) \neq 0$. Find $P(B|A)$, if

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.If A and B are two events such that $P(B) \neq 0$ and $P(B) \neq 1$, then $P(\bar{A}|\bar{B})$ is (A) $1-P(A|B)$ (B) $1-P(\bar{A}|B)$ (C) $\frac{1-P(A\cup B)}{P(\bar{B})}$ (D) $\frac{P(\bar{A})}{P(\bar{B})}$

   ›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

   1. $P(\bar A|\bar B)=\dfrac{P(\bar A\cap\bar B)}{P(\bar B)}$.
   2. $\bar A\cap\bar B=\overline{A\cup B}$ (De Morgan).
   3. So $P(\bar A\cap\bar B)=1-P(A\cup B)$.
   4. Substituting: $P(\bar A|\bar B)=\dfrac{1-P(A\cup B)}{P(\bar B)}$.

   Common Mistakes

   • Confusing $\bar A\cap\bar B$ with $\overline{A\cap B}$ — these are different sets; De Morgan swaps union and intersection under complementation.

   ✓Final answer

   The correct option is (C) — $\dfrac{1-P(A\cup B)}{P(\bar B)}$.

   ANSWER: C

2. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If A,B are any two events of a random experiment and $P(B)\neq1$, then $P(A\mid B^c)=$ (A) $\dfrac{P(A)+P(A\cap B)}{1-P(B)}$ (B) $\dfrac{P(A)-P(A\cap B)}{1-P(B)}$ (C) $\dfrac{P(A)+P(A\cap B)}{1+P(B)}$ (D) $\dfrac{P(A)}{1+P(B)}$

   ›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

   1. $P(A\mid B^c)=\dfrac{P(A\cap B^c)}{P(B^c)}$ by definition (needs $P(B^c)\ne0$, i.e. $P(B)\ne1$, as given).
   2. $A=(A\cap B)\cup(A\cap B^c)$, a disjoint union, so $P(A)=P(A\cap B)+P(A\cap B^c)\Rightarrow P(A\cap B^c)=P(A)-P(A\cap B)$.
   3. $P(B^c)=1-P(B)$.
   4. Combine: $P(A\mid B^c)=\dfrac{P(A)-P(A\cap B)}{1-P(B)}$.

   Common Mistakes

   • Writing $P(A\cap B^c)=P(A)+P(A\cap B)$ instead of subtracting — a very common sign slip.
   • Forgetting the denominator must be $P(B^c)=1-P(B)$, not $1+P(B)$.

   ✓Final answer

   The correct option is (B) — $\dfrac{P(A)-P(A\cap B)}{1-P(B)}$.

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If two events A and B are such that $P(\bar A)=0.3$, $P(B)=0.4$ and $P(A\cap\bar B)=0.5$, then $P(B/(A\cup\bar B))=$ (A) 0.25 (B) 0.6 (C) 0.45 (D) 0.8

   ›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

   1. From $P(\bar A)=0.3$, get $P(A)=1-0.3=0.7$.
   2. $P(A\cap\bar B)=0.5$ is given directly, so $P(A\cap B)=P(A)-P(A\cap\bar B)=0.7-0.5=0.2$.
   3. 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 $P(A\cup\bar B)=0.7+0.6-0.5=0.8$.
   4. Compute the numerator $P\big(B\cap(A\cup\bar B)\big)$: distributing, $B\cap(A\cup\bar B)=(B\cap A)\cup(B\cap\bar B)$. Since $B\cap\bar B=\varnothing$, this simplifies to just $A\cap B$, with probability $0.2$.
   5. Therefore $P(B\mid A\cup\bar B)=\dfrac{0.2}{0.8}=0.25$.

   Common Mistakes

   • Forgetting that $B\cap\bar B$ is empty and mistakenly adding an extra term.
   • Using $P(A\cap B)$ directly as $P(B)-P(A\cap \bar B)$ instead of $P(A)-P(A\cap\bar B)$.

   ✓Final answer

   The correct option is (A) — 0.25.

   ANSWER: A

4. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.It is given that in a random experiment events A and B are such that $P(A) = \dfrac{1}{4}$, $P(A|B) = \dfrac{1}{2}$ and $P(B|A) = \dfrac{2}{3}$ then $P(B) =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{6}$

   ›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

   1. From $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac23$, and $P(A)=\dfrac14$: $P(A\cap B)=\dfrac23\times\dfrac14=\dfrac16$.
   2. 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

   • Confusing which conditional probability to use to find $P(A\cap B)$ first (must use the one whose conditioning event's total probability, $P(A)$, is already known).
   • Dividing instead of multiplying (or vice versa) when rearranging the conditional probability formula.

   ✓Final answer

   The correct option is (A) — $\dfrac13$.

   ANSWER: A

5. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$P(A|A\cap B) + P(B|A\cap B) =$ (A) $1$ (B) $P(A\cup B)$ (C) $P(A\cap B)$ (D) $2$

   ›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

   1. $P(A|A\cap B) = \dfrac{P\big(A\cap(A\cap B)\big)}{P(A\cap B)}$.
   2. Since $A\cap(A\cap B) = A\cap B$ (intersecting with $A$ again changes nothing, as $A\cap B\subseteq A$), this is $\dfrac{P(A\cap B)}{P(A\cap B)}=1$.
   3. Similarly, $P(B|A\cap B) = \dfrac{P\big(B\cap(A\cap B)\big)}{P(A\cap B)} = \dfrac{P(A\cap B)}{P(A\cap B)}=1$.
   4. Sum $=1+1=2$.

   Common Mistakes

   • Trying to expand this using Bayes' theorem unnecessarily — the direct subset observation is much faster.
   • Assuming the answer depends on the actual probabilities of $A,B$ (it doesn't — it's always $2$, as long as $P(A\cap B)>0$).

   ✓Final answer

   The correct option is (D) — $2$.

   ANSWER: D

6. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Let $A$ and $B$ be two events with $P(A) = \dfrac{1}{7}$, $P(A/B) = \dfrac{2}{5}$ and $P(B) = \dfrac{2}{7}$. Then the value $P(B/A)$ is (A) $\dfrac{1}{5}$ (B) $\dfrac{5}{49}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{3}{5}$

   ›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

   1. $P(A/B) = \dfrac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B) = P(A/B)\cdot P(B) = \dfrac{2}{5}\times\dfrac{2}{7} = \dfrac{4}{35}$.
   2. Now use $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$.
   3. Dividing by $1/7$ is the same as multiplying by $7$: $\dfrac{4}{35}\times 7 = \dfrac{28}{35} = \dfrac{4}{5}$.

   Common Mistakes

   • Mixing up which probability ($P(A)$ or $P(B)$) belongs in the denominator at each step — always match the given event in the conditioning.
   • Forgetting to simplify $28/35$ to $4/5$.

   ✓Final answer

   The correct option is (C) — $\dfrac{4}{5}$.

   ANSWER: C

7. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Given $P(A) = 0.5, P(B) = 0.4, P(A \cap B) = 0.3$ then $P(A'/B')$ is equal to (A) $\dfrac{1}{3}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{3}{4}$

   ›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

   1. $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.3=0.6$.
   2. $P(A'\cap B')=P((A\cup B)')=1-P(A\cup B)=1-0.6=0.4$.
   3. $P(B')=1-P(B)=1-0.4=0.6$.
   4. $P(A'|B')=\dfrac{P(A'\cap B')}{P(B')}=\dfrac{0.4}{0.6}=\dfrac23$.

   Common Mistakes

   • Trying to compute $P(A'\cap B')$ directly instead of via De Morgan's law and the union probability, leading to more error-prone arithmetic.

   ✓Final answer

   The correct option is (C) — $\dfrac23$.

   ANSWER: C

8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Two dice are rolled. If A denote the event that the same number shows on each die and B denote the event that the sum of the numbers on both dice is greater than 7, then $P(A|B)$ and $P(B|A)$ respectively are (A) $\dfrac{2}{5}, \dfrac{1}{4}$ (B) $\dfrac{1}{5}, \dfrac{1}{2}$ (C) $\dfrac{1}{5}, \dfrac{1}{4}$ (D) $\dfrac{1}{2}, \dfrac{3}{5}$

   ›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

   1. $A$ = "same number on each die": outcomes $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — $6$ outcomes, so $P(A)=\dfrac6{36}$.
   2. $B$ = "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 $P(B)=\dfrac{15}{36}$.
   3. $A\cap B$: 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, $P(A\cap B)=\dfrac3{36}$.
   4. $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{3/36}{15/36}=\dfrac3{15}=\dfrac15$.
   5. $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{3/36}{6/36}=\dfrac36=\dfrac12$.

   Common Mistakes

   • Miscounting the number of outcomes with sum $>7$ (forgetting sum $=8$ is included since $>7$ means $\ge8$).
   • Swapping $P(A|B)$ and $P(B|A)$ in the final answer.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{5}, \dfrac{1}{2}$.

   ANSWER: B

9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In a sample space, $E$ is an event associated with the events $A$ and $B$. If $P(A)P(E|A) = l$ and $P(B)P(E|B) = m$ then, $P(B|E) =$ (A) $\dfrac{m}{l+m}$ always (B) $\dfrac{l}{l+m}$ only when $P(A)+P(B)=1$ (C) $\dfrac{m}{l+m}$ only when $P(A)+P(B)=1$ (D) $\dfrac{l}{l+m}$ always

   ›Reveal solutionSolution

   This tests when the "total probability" decomposition $P(E)=P(A\cap E)+P(B\cap E)$ is legitimate — only when $A,B$ partition the sample space. Answer: $\dfrac{m}{l+m}$, and only under that condition.

   Concept and Intuition

   Bayes' rule always gives $P(B\mid E) = P(B\cap E)/P(E)$. The numerator is handed to us as $m$. The tricky part is the denominator: $P(E)$ can only be written as $P(A\cap E)+P(B\cap E)=l+m$ if every outcome of $E$ passes through either $A$ or $B$ and not both — that is exactly the statement that $\{A,B\}$ partitions the sample space, equivalent to $A\cap B=\varnothing$ and $P(A)+P(B)=1$.

   Step-by-Step Solution

   1. From the given data, $P(A\cap E)=l$ and $P(B\cap E)=m$.
   2. By definition, $P(B\mid E)=\dfrac{P(B\cap E)}{P(E)}=\dfrac{m}{P(E)}$.
   3. If (and only if) $A,B$ partition the sample space, $E = (E\cap A)\cup(E\cap B)$ with no overlap, so $P(E)=l+m$.
   4. Substituting, $P(B\mid E)=\dfrac{m}{l+m}$ — but this substitution is valid only under the partition condition $P(A)+P(B)=1$.
   5. Without that condition, $P(E)$ could be larger (if there's sample space outside $A\cup B$ contributing to $E$), so the formula does not hold in general.

   Common Mistakes

   • Assuming $P(E)=l+m$ "always" — this silently assumes $A,B$ exhaust and don't overlap, which isn't given by default.
   • Confusing which of $l,m$ belongs in the numerator for $P(B\mid E)$ (it's $m$, since $m=P(B\cap E)$).

   ✓Final answer

   The correct option is (C) — $\dfrac{m}{l+m}$ only when $P(A)+P(B)=1$.

   ANSWER: C

10. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.If $A$ and $B$ are mutually exclusive events with $P(A) = \dfrac{1}{4}$ and $P(B) = \dfrac{3}{7}$. Then what is the value of $P(A / A \cup B) =$ (A) $\dfrac{7}{19}$ (B) $\dfrac{12}{19}$ (C) $\dfrac{06}{19}$ (D) $\dfrac{13}{19}$

    ›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

    1. $P(A\cup B) = P(A)+P(B) = \dfrac14+\dfrac37$.
    2. Common denominator 28: $\dfrac{7}{28}+\dfrac{12}{28}=\dfrac{19}{28}$.
    3. $P(A\mid A\cup B) = \dfrac{P(A)}{P(A\cup B)} = \dfrac{1/4}{19/28} = \dfrac14\times\dfrac{28}{19} = \dfrac{7}{19}$.

    Common Mistakes

    • Adding probabilities incorrectly by using the wrong common denominator.
    • Forgetting that mutual exclusivity is what allows $P(A\cup B)=P(A)+P(B)$ (no overlap term to subtract).

    ✓Final answer

    The correct option is (A) — $\dfrac{7}{19}$.

    ANSWER: A

11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.If E and F are events such that $P(\bar{F}) = 0.7$ and $P(E \cap F) = 0.2$, then $P(E|F)$ is (A) $2/3$ (B) $1/3$ (C) $3/4$ (D) $1/4$

    ›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

    1. $P(\bar F) = 0.7 \Rightarrow P(F) = 1 - 0.7 = 0.3$.
    2. $P(E\cap F) = 0.2$ is given directly.
    3. $P(E|F) = \dfrac{P(E\cap F)}{P(F)} = \dfrac{0.2}{0.3} = \dfrac{2}{3}$.

    Common Mistakes

    • Forgetting to convert $P(\bar F)$ to $P(F)$ and dividing by $0.7$ instead of $0.3$.
    • Confusing $P(E|F)$ with $P(F|E)$, which would need $P(E)$ instead.

    ✓Final answer

    The correct option is (A) — $2/3$.

    ANSWER: A

12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.A die is thrown twice. Let A be the event of getting a prime number when the die is thrown first time and B be the event of getting an even number when the die is thrown second time. Then $P(A/\bar B)=$ (A) $\dfrac{1}{2}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

    ›Reveal solutionSolution

    Since $A$ and $B$ (hence $\bar B$) come from two independent throws of the die, conditioning on $\bar B$ does not change $P(A)$; the answer is $1/2$.

    Concept and Intuition

    When two events are determined by physically independent trials (first throw vs second throw of a die), any conditional probability between them collapses to the unconditional probability — conditioning on an independent event changes nothing.

    Step-by-Step Solution

    1. $A$: prime number on the first throw, i.e. outcome in $\{2,3,5\}$, so $P(A)=\dfrac{3}{6}=\dfrac{1}{2}$.
    2. $B$: even number on the second throw, i.e. outcome in $\{2,4,6\}$, so $P(B)=\dfrac{1}{2}$, and $P(\bar B)=\dfrac{1}{2}$ (odd on the second throw).
    3. Because the first and second throws are independent, $A$ (a first-throw event) and $\bar B$ (a second-throw event) are independent of each other.
    4. Hence $P(A\mid \bar B) = P(A) = \dfrac{1}{2}$.

    Common Mistakes

    • Assuming some dependence between the two throws and trying to compute a joint sample space unnecessarily.
    • Confusing $P(A/\bar B)$ with $P(\bar B/A)$ or with $P(A\cap \bar B)$.

    ✓Final answer

    The correct option is (A) — $\dfrac{1}{2}$.

    ANSWER: A