Q.Given that E and F are events such that $P(E) = 0.6$, $P(F) = 0.3$ and $P(E \cap F) = 0.2$, find $P(E|F)$ and $P(F|E)$.

Showing the 12 most recent of 69 on this concept.

1. 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

2. 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

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 2021Set eng-2021-08-23-AN1 markMCQ
   Q.2 aero planes I and II bond a target in succession. The probabilities of I and II scoring a hit correctly is 0.3 and 0.2 respectively. The second plane will bomb only if first misses the target. The probability that the target is hit by the $2^{nd}$ plane is (A) 0.06 (B) 0.14 (C) 0.32 (D) 0.7

   ›Reveal solutionSolution

   The 2nd plane gets a chance only after the 1st fails, so $P = 0.7\times 0.2 = 0.14$. Answer: (B).

   Concept and Intuition

   This is a sequential (conditional) experiment: the second trial happens only when the first fails. The event 'the target is hit by the 2nd plane' is therefore a compound event —

   $\{\text{I misses}\}\cap\{\text{II hits}\}$

   and because the two planes' performances are independent, the probability of the intersection is the product of the probabilities.

   A useful picture is a probability tree:

            ┌── I hits (0.3)  ────────────────► target hit by plane I  (0.3)
   Start ───┤
            └── I misses (0.7) ─┬── II hits (0.2) ──► hit by plane II  (0.7 × 0.2 = 0.14)
                                └── II misses (0.8) ► target not hit   (0.7 × 0.8 = 0.56)
   

   The three leaves sum to $0.3+0.14+0.56=1$ ✓ — a good check that the model is complete.

   Step-by-Step Solution

   1. Let $H_1$ = plane I hits, with $P(H_1)=0.3$, so $P(\overline{H_1})=1-0.3=0.7$.
   2. Let $H_2$ = plane II hits (given it bombs), with $P(H_2)=0.2$.
   3. Plane II bombs only if plane I missed. Hence

   $P(\text{target hit by 2nd plane}) = P(\overline{H_1}\cap H_2) = P(\overline{H_1})\,P(H_2)$

   4. Substitute:

   $= 0.7\times 0.2 = 0.14$

   5. ⇒ option (B).

   Common Mistakes

   • Answering $0.2$ — that is the conditional probability $P(H_2\mid \overline{H_1})$, i.e. the chance the 2nd plane hits given it actually bombs, not the unconditional probability asked for.
   • Answering $0.3\times0.2=0.06$ (distractor A) — that would be 'both hit', but plane II never bombs if plane I has already hit.
   • Answering $0.3+0.2-0.06=0.44$ or the 'at least one hit' value — the question asks specifically for a hit by the second plane.

   ✓Final answer

   The correct option is (B) — 0.14.

   ANSWER: B

6. 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

7. 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

8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.For a biased die, the probabilities for different faces to turn up are given below

   Face	1	2	3	4	5	6
   Probability	0.1	0.32	0.21	0.15	0.05	0.17

   The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is (A) $\dfrac{10}{33}$ (B) $\dfrac{5}{21}$ (C) $\dfrac{8}{21}$ (D) $\dfrac{1}{42}$

   ›Reveal solutionSolution

   Conditioning on "face 1 or face 2" just means renormalizing the two individual probabilities so they add to 1.

   Concept and Intuition

   $P(\text{face 1}\mid\text{face 1 or 2})=\dfrac{P(\text{face 1})}{P(\text{face 1})+P(\text{face 2})}$, since these two events are mutually exclusive and their union is the conditioning event.

   Step-by-Step Solution

   1. $P(1)=0.1$, $P(2)=0.32$.
   2. $P(1\text{ or }2)=0.1+0.32=0.42$.
   3. $P(1\mid1\text{ or }2)=\dfrac{0.1}{0.42}=\dfrac{10}{42}=\dfrac{5}{21}$.

   Common Mistakes

   • Dividing by the total probability $1$ instead of the restricted event's probability $0.42$.
   • Arithmetic slip simplifying $10/42$ (should reduce to $5/21$).

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{21}$.

   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 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

11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.Two persons P and Q are considering to apply for a job. The probability that P applies for the job is 1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for the job is (A) $4/5$ (B) $5/6$ (C) $7/8$ (D) $11/12$

    ›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

    1. Given: $P(P)=\tfrac14$, $P(P|Q)=\tfrac12$, $P(Q|P)=\tfrac13$.
    2. $P(P\cap Q) = P(Q|P)\cdot P(P) = \tfrac13\times\tfrac14 = \tfrac1{12}$.
    3. Also $P(P\cap Q) = P(P|Q)\cdot P(Q) \Rightarrow \tfrac1{12} = \tfrac12\cdot P(Q) \Rightarrow P(Q)=\tfrac16$.
    4. $P(P\cup Q) = P(P)+P(Q)-P(P\cap Q) = \tfrac14+\tfrac16-\tfrac1{12} = \tfrac{3}{12}+\tfrac{2}{12}-\tfrac{1}{12}=\tfrac{4}{12}=\tfrac13$.
    5. By De Morgan's law, "neither P nor Q applies" is the complement of $P\cup Q$: $P(\overline P\cap\overline Q) = 1-\tfrac13=\tfrac23$.
    6. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$.
    7. $P(\overline P|\overline Q) = \dfrac{P(\overline P\cap\overline Q)}{P(\overline Q)} = \dfrac{2/3}{5/6} = \dfrac23\times\dfrac65 = \dfrac{12}{15}=\dfrac45$.

    Common Mistakes

    • Mixing up which conditional probability to use for computing the joint probability first (must use $P(Q|P)P(P)$, not $P(P|Q)P(P)$, to directly match given data).
    • Forgetting to convert to complements at the final step (computing $P(P|Q)$-type instead of $P(\overline P|\overline Q)$).

    ✓Final answer

    The correct option is (A) — $4/5$.

    ANSWER: A

12. 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