Q.Evaluate $P(A \cup B)$, if $2P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$.

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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 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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3. 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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4. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.On every evening, a student either watches TV or reads a book. The probability of watching TV is $\dfrac{4}{5}$. If he watches TV, the probability that he will fall asleep is $\dfrac{3}{4}$ and it is $\dfrac{1}{4}$ when he reads a book. If the student is found to be asleep on an evening, the probability that he watched the TV is (A) $\dfrac{11}{13}$ (B) $\dfrac{12}{13}$ (C) $\dfrac{2}{13}$ (D) $\dfrac{4}{13}$

   ›Reveal solutionSolution

   This tests Bayes' theorem: finding the probability of a cause (watched TV) given an observed effect (fell asleep). The answer is $\dfrac{12}{13}$.

   Concept and Intuition

   When asked "given the observed outcome, what's the probability of a particular cause," that's a Bayes'-theorem setup: compute the joint probability of that cause with the outcome, and divide by the total probability of the outcome (summed over all causes).

   Step-by-Step Solution

   1. $P(\text{TV})=\tfrac45$, $P(\text{book})=\tfrac15$.
   2. $P(\text{asleep}\mid\text{TV})=\tfrac34$, $P(\text{asleep}\mid\text{book})=\tfrac14$.
   3. Total probability of falling asleep: $P(\text{asleep})=P(\text{TV})P(\text{asleep}\mid\text{TV})+P(\text{book})P(\text{asleep}\mid\text{book})=\tfrac45\cdot\tfrac34+\tfrac15\cdot\tfrac14=\tfrac35+\tfrac1{20}=\tfrac{12}{20}+\tfrac1{20}=\tfrac{13}{20}$. …

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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 $=1+1=2$. …

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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. $P(B')=1-P(B)=1-0.4=0.6$. …

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7. 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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8. 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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9. 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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10. 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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11. 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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12. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.A shopkeeper buys a particular type of electric bulbs from three manufacturers $M_1, M_2$ and $M_3$. He buys 25% of his requirement from $M_1$, 45% from $M_2$ and 30% from $M_3$. Based on past experience he found that 2% of type $M_3$ bulbs are defective, where as only 1% of type $M_1$ and type $M_2$ are defective. If a bulb chosen by him at random is defective, then the probability that it was of type $M_3$ is (A) $\dfrac{5}{13}$ (B) $\dfrac{6}{13}$ (C) $\dfrac{7}{13}$ (D) $\dfrac{8}{13}$

    ›Reveal solutionSolution

    A direct application of Bayes' theorem with the law of total probability for the defective-bulb probability. Answer: (B).

    Concept and Intuition

    This is a classic "reverse conditional probability" problem: we know how likely a defect is given the source, and want the reverse — the probability the source was $M_3$ given a defect was observed. Bayes' theorem converts one into the other via the law of total probability.

    Step-by-Step Solution

    1. Given: $P(M_1)=0.25,\ P(M_2)=0.45,\ P(M_3)=0.30$; $P(D|M_1)=0.01,\ P(D|M_2)=0.01,\ P(D|M_3)=0.02$.
    2. Total probability of a defective bulb: $P(D)=P(M_1)P(D|M_1)+P(M_2)P(D|M_2)+P(M_3)P(D|M_3)$ $=0.25(0.01)+0.45(0.01)+0.30(0.02)=0.0025+0.0045+0.006=0.013$. …

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