Q.If $P(A) = \dfrac{2}{5}$, $P(B) = \dfrac{3}{10}$ and $P(A \cap B) = \dfrac{1}{5}$, then $P(A' \mid B') \cdot P(B' \mid A')$ is equal to
(A) $\dfrac{5}{6}$
(B) $\dfrac{5}{7}$
(C) $\dfrac{25}{42}$
(D) $1$

Showing the 12 most recent of 69 on this concept.

1. 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 $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$. …

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2. 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. $P(B')=1-P(B)=1-0.4=0.6$. …

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3. 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 $P(\bar A\cap\bar B)=1-P(A\cup B)$. …

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4. 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. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$. …

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5. 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, $P(A\cap B)=\dfrac3{36}$. …

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6. 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 $P(A\cup\bar B)=0.7+0.6-0.5=0.8$. …

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7. 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). …

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8. 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. $P(B^c)=1-P(B)$. …

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9. 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 $=1+1=2$. …

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10. 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 …

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11. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.A bag 'A' contains 2 black and 3 white balls. Another bag 'B' contains 3 black and 2 white balls. Two balls are drawn randomly from 'A' and placed in 'B'. Later, if two balls are drawn randomly from 'B', then the probability of getting a black ball and a white ball from it is (A) $\dfrac{59}{105}$ (B) $\dfrac{46}{105}$ (C) $\dfrac{59}{210}$ (D) $\dfrac{67}{210}$

    ›Reveal solutionSolution

    A two-stage random-transfer problem — condition on what got transferred from A to B, then apply the law of total probability.

    Concept and Intuition

    Since the composition of bag B after the transfer depends on which 2 balls were drawn from A, we must split into the three possible transfer outcomes (BB, BW, WW), compute the probability of each, then the conditional probability of drawing one black and one white ball from the resulting bag B, and combine via the law of total probability.

    Step-by-Step Solution

    1. Bag A has 5 balls (2B, 3W); ways to choose 2: $\binom{5}{2}=10$.
       • $P(\text{BB from A}) = \binom{2}{2}/\binom{5}{2} = 1/10$.
       • $P(\text{BW from A}) = \binom{2}{1}\binom{3}{1}/\binom{5}{2} = 6/10 = 3/5$.
       • $P(\text{WW from A}) = \binom{3}{2}/\binom{5}{2} = 3/10$.
    2. Bag B originally has 3B, 2W (5 balls); after adding 2 balls it has 7 balls, and we want $P(1B,1W)$ drawn from it: $\binom{7}{2}=21$ ways.
       • If BB added: B = 5B, 2W. $P(1B,1W) = \binom{5}{1}\binom{2}{1}/21 = 10/21$.
       • If BW added: B = 4B, 3W. $P(1B,1W) = \binom{4}{1}\binom{3}{1}/21 = 12/21$.
       • If WW added: B = 3B, 4W. $P(1B,1W) = \binom{3}{1}\binom{4}{1}/21 = 12/21$.
    3. Total probability: …

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12. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. If one ball is drawn at random from one of the bags and is found to be red, then the probability that it was drawn from the bag B is (A) $\dfrac{23}{54}$ (B) $\dfrac{25}{51}$ (C) $\dfrac{25}{52}$ (D) $\dfrac{27}{55}$

    ›Reveal solutionSolution

    A classic Bayes'-theorem problem: given the ball drawn is red, the probability it came from bag B is $\dfrac{25}{52}$.

    Concept and Intuition

    This is Bayes' theorem: we're given the outcome (red ball drawn) and want to find the probability of which "cause" (bag A or B) produced it, using the prior probability of choosing each bag (each $\frac12$) and each bag's own probability of yielding red.

    Step-by-Step Solution

    1. Bag A: 2 white, 3 red (5 total) $\Rightarrow P(\text{red}|A)=\dfrac35$.
    2. Bag B: 4 white, 5 red (9 total) $\Rightarrow P(\text{red}|B)=\dfrac59$.
    3. $P(A)=P(B)=\dfrac12$ (bag chosen at random).
    4. Total probability of red: $P(\text{red})=P(A)P(\text{red}|A)+P(B)P(\text{red}|B)=\dfrac12\cdot\dfrac35+\dfrac12\cdot\dfrac59=\dfrac3{10}+\dfrac5{18}$.
    5. Common denominator 90: $\dfrac3{10}=\dfrac{27}{90}$, $\dfrac5{18}=\dfrac{25}{90}$, sum $=\dfrac{52}{90}=\dfrac{26}{45}$.
    6. Bayes: $P(B|\text{red})=\dfrac{P(B)P(\text{red}|B)}{P(\text{red})}=\dfrac{(1/2)(5/9)}{26/45}=\dfrac{5/18}{26/45}$. …

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