Q.If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find

Showing the 12 most recent of 69 on this concept.

1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

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 2025Set eng-2025-05-24-FN1 markMCQ
    Q.In a school there are 3 sections A, B and C. Section A contains 20 girls and 30 boys, section B contains 40 girls and 20 boys and section C contains 10 girls and 30 boys. The probabilities of selecting the section A, B and C are 0.2, 0.3 and 0.5 respectively. If a student selected at random from the school is a girl, then the probability that she belongs to section A is (A) $\dfrac{121}{200}$ (B) $\dfrac{16}{121}$ (C) $\dfrac{14}{81}$ (D) $\dfrac{16}{81}$

    ›Reveal solutionSolution

    A three-section Bayes'-theorem problem; the answer is $\dfrac{16}{81}$, option (D).

    Concept and Intuition

    Given a randomly selected student is a girl, we want to reverse-condition on which section she came from. This requires the total probability of "selecting a girl" (summed over all three sections weighted by section-selection probability), then applying Bayes' theorem to isolate section A's contribution.

    Step-by-Step Solution

    1. Conditional probabilities of picking a girl within each section: $P(\text{girl}\mid A)=\dfrac{20}{50}=\dfrac{2}{5}$; $P(\text{girl}\mid B)=\dfrac{40}{60}=\dfrac{2}{3}$; $P(\text{girl}\mid C)=\dfrac{10}{40}=\dfrac14$.
    2. Section-selection probabilities: $P(A)=\dfrac15,\ P(B)=\dfrac{3}{10},\ P(C)=\dfrac12$.
    3. Joint terms: $P(A)P(\text{girl}\mid A)=\dfrac15\cdot\dfrac25=\dfrac{2}{25}$; $P(B)P(\text{girl}\mid B)=\dfrac{3}{10}\cdot\dfrac23=\dfrac15$; $P(C)P(\text{girl}\mid C)=\dfrac12\cdot\dfrac14=\dfrac18$.
    4. Common denominator 200: $\dfrac{2}{25}=\dfrac{16}{200}$, $\dfrac15=\dfrac{40}{200}$, $\dfrac18=\dfrac{25}{200}$. Sum $=\dfrac{81}{200}=P(\text{girl})$.
    5. Bayes: $P(A\mid\text{girl})=\dfrac{16/200}{81/200}=\dfrac{16}{81}$.

    Common Mistakes

    • Using the section sizes directly as weights instead of the given selection probabilities $0.2,0.3,0.5$.
    • Arithmetic slips when combining fractions with different denominators — using a common denominator (200) avoids this.

    ✓Final answer

    The correct option is (D) — $\dfrac{16}{81}$.

    ANSWER: D

12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.A, B, C are mutually exclusive and exhaustive events of a random experiment and E is an event that occurs in conjunction with one of the events A, B, C. The conditional Probabilities of E given the happening of A, B, C are respectively 0.6, 0.3 and 0.1. If $P(A) = 0.30$ and $P(B) = 0.50$, then $P(C|E) =$ (A) $\dfrac{2}{35}$ (B) $\dfrac{15}{35}$ (C) $\dfrac{18}{35}$ (D) $\dfrac{17}{35}$

    ›Reveal solutionSolution

    This is a direct Bayes' theorem application over three mutually exclusive, exhaustive causes; $P(C|E)=2/35$.

    Concept and Intuition

    When an event $E$ can occur alongside any of several mutually exclusive, exhaustive causes $A,B,C$, Bayes' theorem lets us find the probability of a particular cause given that $E$ has occurred, by weighing each cause's prior probability by how likely $E$ is under it, then normalizing.

    Step-by-Step Solution

    1. Since $A,B,C$ are mutually exclusive and exhaustive, $P(C)=1-P(A)-P(B)=1-0.30-0.50=0.20$.
    2. Total probability of $E$: $P(E)=P(E|A)P(A)+P(E|B)P(B)+P(E|C)P(C)$

    $=0.6(0.30)+0.3(0.50)+0.1(0.20)=0.18+0.15+0.02=0.35$

    3. By Bayes' theorem: $P(C|E)=\dfrac{P(E|C)\,P(C)}{P(E)}=\dfrac{0.1\times0.2}{0.35}=\dfrac{0.02}{0.35}$.
    4. Simplify: $\dfrac{0.02}{0.35}=\dfrac{2}{35}$.

    Common Mistakes

    • Forgetting to first compute $P(C)$ from the exhaustiveness condition.
    • Using $P(E)$ instead of $P(E|C)P(C)$ in the numerator (or vice versa).

    ✓Final answer

    The correct option is (A) — $\dfrac2{35}$.

    ANSWER: A