Q.Probability of solving specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that

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1. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.A and B are two independent events of a random experiment and $P(A) > P(B)$. If the probability that both A and B occur is $\dfrac{1}{6}$ and neither of them occurs is $\dfrac{1}{3}$, then the probability of the occurance of B is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{8}$

   ›Reveal solutionSolution

   Translate the two given probabilities into a sum and a product of $P(A), P(B)$ using independence, then solve the resulting quadratic and pick the smaller root.

   Concept and Intuition

   For independent events, $P(A\cap B)=P(A)P(B)$ and $P(\overline A\cap\overline B)=P(\overline A)P(\overline B)=(1-P(A))(1-P(B))$. These two equations in $p=P(A)$ and $q=P(B)$ can be converted into their sum and product, which is exactly what's needed to solve a quadratic for $p,q$.

   Step-by-Step Solution

   1. Let $p=P(A)$, $q=P(B)$, with $p>q$ (given $P(A)>P(B)$).
   2. $P(A\cap B)=pq=\dfrac16$ (independence).
   3. $P(\overline A\cap\overline B)=(1-p)(1-q)=\dfrac13$.
   4. Expand: $1-p-q+pq=\dfrac13$. Substitute $pq=\dfrac16$: $1-(p+q)+\dfrac16=\dfrac13 \Rightarrow p+q = 1+\dfrac16-\dfrac13=\dfrac56$.
   5. So $p,q$ are roots of $t^2-\dfrac56t+\dfrac16=0$, i.e. $6t^2-5t+1=0$. …

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2. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Let A and B be two independent events of a random experiment. If the probability that both A and B occur is $\dfrac{1}{6}$ and the probability that neither of them occur is $\dfrac{1}{3}$, then the probability of occurrence of A is (A) $0$ or $1$ (B) $\dfrac{1}{2}$ or $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ or $\dfrac{1}{3}$ (D) $\dfrac{1}{2}$ or $\dfrac{1}{7}$

   ›Reveal solutionSolution

   From $P(A)P(B)=\frac16$ and $(1-P(A))(1-P(B))=\frac13$, we get $P(A)+P(B)=\frac56$, and solving the resulting quadratic gives $P(A)=\frac12$ or $\frac13$.

   Concept and Intuition

   Independence lets us write both given probabilities as simple products: $P(A\cap B)=P(A)P(B)$ and $P(\bar A\cap\bar B)=P(\bar A)P(\bar B)$ (since complements of independent events are also independent). This turns the problem into finding two numbers whose product and a related "complement product" are both known — a job for Vieta's formulas.

   Step-by-Step Solution

   1. Let $p=P(A)$, $q=P(B)$. Given: $pq=\dfrac16$ (product rule for independent events).
   2. Also given: $P(\text{neither})=P(A'\cap B')=(1-p)(1-q)=\dfrac13$.
   3. Expand: $1-p-q+pq=\dfrac13\Rightarrow1-(p+q)+\dfrac16=\dfrac13\Rightarrow p+q=1+\dfrac16-\dfrac13=\dfrac56$.
   4. So $p,q$ are roots of $t^2-\dfrac56t+\dfrac16=0$. Multiply by 6: $6t^2-5t+1=0$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $A, B, C$ are independent events of a random experiment such that $P(A) = 3/4$, $P(B) = 5/6$ and $P(C) = 2/3$, then the probability that exactly one of the events occur is (A) $\dfrac{1}{4}$ (B) $\dfrac{5}{36}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{71}{72}$

   ›Reveal solutionSolution

   Sum the three mutually exclusive ways exactly one of three independent events can occur. Answer: $\dfrac5{36}$.

   Concept and Intuition

   For independent events, "exactly one occurs" means one event happens and the other two fail, and there are three such disjoint scenarios (which one occurs). Because the events are independent, each scenario's probability is simply the product of the individual (or complementary) probabilities, and we add the three scenarios since they can't happen simultaneously.

   Step-by-Step Solution

   1. Complements: $P(A')=1-\frac34=\frac14$, $P(B')=1-\frac56=\frac16$, $P(C')=1-\frac23=\frac13$.
   2. Scenario "only $A$": $P(A)P(B')P(C')=\frac34\cdot\frac16\cdot\frac13=\frac{3}{72}$.
   3. Scenario "only $B$": $P(A')P(B)P(C')=\frac14\cdot\frac56\cdot\frac13=\frac{5}{72}$. …

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4. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$E_1$ and $E_2$ are two independent events of a random experiment such that $P(E_1)=\dfrac{1}{2}$ and $P(E_1\cup E_2)=\dfrac{2}{3}$. Then match the items of List-I with the items of List-II. List-I: (A) $P(E_2)$ (B) $P(E_1/E_2)$ (C) $P(\overline{E_2}/E_1)$ (D) $P(\overline{E_1}\cup\overline{E_2})$ List-II:

   (i) $\dfrac{1}{2}$

   (ii) $\dfrac{5}{6}$

   (iii) $\dfrac{1}{3}$

   (iv) $\dfrac{1}{6}$

   (v) $\dfrac{2}{3}$ The correct match is (A) A-iii; B-iv; C-i; D-v (B) A-iii; B-i; C-v; D-ii (C) A-i; B-v; C-ii; D-iv (D) A-v; B-i; C-iii; D-ii

   ›Reveal solutionSolution

   Using independence, $P(E_2)=\frac13$, and all four quantities reduce cleanly, matching A-iii, B-i, C-v, D-ii.

   Concept and Intuition

   For independent events, conditional probabilities collapse to unconditional ones: $P(E_1|E_2)=P(E_1)$ and $P(\overline{E_2}|E_1)=P(\overline{E_2})$. Also $P(\overline{E_1}\cup\overline{E_2}) = 1-P(E_1\cap E_2) = 1-P(E_1)P(E_2)$ by De Morgan's law.

   Step-by-Step Solution

   1. $P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1)P(E_2)$: $\frac23 = \frac12+P(E_2)-\frac12P(E_2) \Rightarrow \frac23-\frac12=\frac12P(E_2)\Rightarrow \frac16=\frac12P(E_2)\Rightarrow P(E_2)=\frac13$.
   2. (A) $P(E_2)=\frac13$ → matches (iii).
   3. (B) $P(E_1/E_2)=P(E_1)=\frac12$ (independence) → matches (i).
   4. (C) $P(\overline{E_2}/E_1)=P(\overline{E_2})=1-\frac13=\frac23$ (independence) → matches (v). …

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5. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If A,B and C are three independent events of a random experiment such that $P(A\cap B^c\cap C^c)=\dfrac{1}{4}$, $P(A^c\cap B\cap C^c)=\dfrac{1}{8}$ and $P(A^c\cap B^c\cap C^c)=\dfrac{1}{4}$, then P(A), P(B) and P(C) are respectively (A) $\dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{5}$ (B) $1, \dfrac{1}{2}, \dfrac{1}{3}$ (C) $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}$ (D) $\dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}$

   ›Reveal solutionSolution

   Comparing the equation for $P(A\cap B^c\cap C^c)$ with the one for $P(A^c\cap B^c\cap C^c)$ immediately forces $P(A)=\frac12$; back-substitution then gives $P(B)=\frac13, P(C)=\frac14$.

   Concept and Intuition

   For independent events, $P(A\cap B^c\cap C^c)=p(1-q)(1-r)$ and similar products for the other combinations. Since two of the given equations share the identical factor $(1-q)(1-r)$, equating them directly isolates $p$ without solving a full system.

   Step-by-Step Solution

   1. $P(A\cap B^c\cap C^c)=p(1-q)(1-r)=\dfrac14$ ... (i)
   2. $P(A^c\cap B\cap C^c)=(1-p)q(1-r)=\dfrac18$ ... (ii)
   3. $P(A^c\cap B^c\cap C^c)=(1-p)(1-q)(1-r)=\dfrac14$ ... (iii)
   4. From (i) and (iii): $p(1-q)(1-r)=(1-p)(1-q)(1-r)$; since $(1-q)(1-r)\ne0$, this gives $p=1-p\Rightarrow p=\dfrac12$.
   5. Substitute into (i): $\dfrac12(1-q)(1-r)=\dfrac14\Rightarrow(1-q)(1-r)=\dfrac12$.
   6. Substitute into (ii): $\dfrac12\,q(1-r)=\dfrac18\Rightarrow q(1-r)=\dfrac14$. …

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6. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $A$ and $B$ be independent events with $P(A) = \frac{1}{3}$ and $P(B) = \frac{2}{7}$, then the value of $P(A/B^C)$ is (A) $\frac{1}{3}$ (B) $\frac{2}{7}$ (C) $\frac{2}{21}$ (D) $\frac{5}{21}$

   ›Reveal solutionSolution

   Independence of $A$ and $B$ implies independence of $A$ and $B^c$ too, so conditioning on $B^c$ doesn't change $P(A)$.

   Concept and Intuition

   If two events are independent, an event is also independent of the complement of the other — knowing $B$ didn't happen carries no information about $A$ either.

   Step-by-Step Solution

   1. $A,B$ independent $\Rightarrow P(A\cap B)=P(A)P(B)$.
   2. $P(A\cap B^c) = P(A)-P(A\cap B) = P(A)-P(A)P(B) = P(A)(1-P(B)) = P(A)P(B^c)$. …

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7. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A, B, C are three independent events of a random experiment such that $P(C) = 2/3$. If probabilities of occurrence of only A, only B and only C are respectively $\dfrac{4}{60}, \dfrac{3}{60}$ and $\dfrac{2}{60}$, then the ratio of the probability of the occurrence of A to the occurrence of B is (A) $16:15$ (B) $4:3$ (C) $12:25$ (D) $2:3$

   ›Reveal solutionSolution

   This tests setting up "only-one-event" probabilities for independent events as products with complements, then solving the resulting system for $P(A)$ and $P(B)$.

   Concept and Intuition

   For independent events, "only $A$ occurs" means $A$ happens and both $B,C$ fail: $P(A)(1-P(B))(1-P(C))$. Writing this out for all three "only" events gives three equations in the two unknowns $p=P(A)$ and $q=P(B)$ (since $P(C)$ is already known) — enough to solve completely, with the third equation serving as a consistency check.

   Step-by-Step Solution

   1. Let $p=P(A)$, $q=P(B)$, and $P(C)=\frac23$ so $1-P(C)=\frac13$.
   2. Only $A$: $p(1-q)\cdot\frac13=\frac4{60}=\frac1{15}\Rightarrow p(1-q)=\frac15$.
   3. Only $B$: $(1-p)q\cdot\frac13=\frac3{60}=\frac1{20}\Rightarrow(1-p)q=\frac3{20}$.
   4. Only $C$: $(1-p)(1-q)\cdot\frac23=\frac2{60}=\frac1{30}\Rightarrow(1-p)(1-q)=\frac1{20}$.
   5. From steps 2 and 3: $p-pq=\frac15$ and $q-pq=\frac3{20}$; subtracting, $p-q=\frac15-\frac3{20}=\frac1{20}$, so $p=q+\frac1{20}$.
   6. Expand step 4: $1-p-q+pq=\frac1{20}$. Using $pq=p-\frac15$ from step 2: $1-p-q+p-\frac15=\frac1{20}\Rightarrow\frac45-q=\frac1{20}\Rightarrow q=\frac45-\frac1{20}=\frac{15}{20}=\frac34$. …

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8. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Two brothers X and Y appeared for an exam. Let A be the event that X has passed the exam and B is the event that Y has passed. The probability of A is $\frac{1}{7}$ and of B is $\frac{2}{9}$. Then, the probability that both of them pass the exam is (A) $\frac{1}{63}$ (B) $\frac{2}{35}$ (C) $\frac{2}{63}$ (D) $\frac{9}{14}$

   ›Reveal solutionSolution

   Two brothers' exam outcomes are independent events, so their joint probability is the product; the answer is $\frac{2}{63}$.

   Concept and Intuition

   When two events are independent (one brother's pass/fail doesn't affect the other's), the probability that both occur is the product of their individual probabilities: $P(A\cap B)=P(A)P(B)$.

   Step-by-Step Solution

   1. $P(A)=\dfrac{1}{7}$, $P(B)=\dfrac{2}{9}$.
   2. Both pass (independent events): $P(A\cap B)=P(A)\times P(B)=\dfrac{1}{7}\times\dfrac{2}{9}=\dfrac{2}{63}$. …

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9. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If A and B are two independent events such that $P(A) = 0.3$, $P(B) = x$ and $P(A\cup B) = 0.44$, then $x=$ (A) $0.1$ (B) $0.4$ (C) $0.3$ (D) $0.2$

   ›Reveal solutionSolution

   This tests the union formula for independent events; solving $0.44=0.3+x-0.3x$ gives $x=0.2$.

   Concept and Intuition

   For any two events, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. When $A$ and $B$ are independent,

   $P(A\cap B)=P(A)P(B)$, which turns the union formula into a single equation in the unknown

   probability.

   Step-by-Step Solution

   1. $P(A)=0.3,\ P(B)=x,\ P(A\cup B)=0.44$, with $A,B$ independent.
   2. $P(A\cup B)=P(A)+P(B)-P(A)P(B) = 0.3+x-0.3x$.
   3. Set equal to $0.44$: $0.3+0.7x=0.44$.
   4. $0.7x=0.14 \Rightarrow x=0.2$.

   Common Mistakes …

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10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.Let $E_1, E_2$ and $E_3$ be mutually independent events Statement I: $E_1$ and $E_2 \cup E_3$ are independent Statement II: $E_1$ and $E_2 \cap E_3$ are independent Which one of the following options is correct? (A) Both I and II are true (B) Only I is true (C) Only II is true (D) Both I and II are false

    ›Reveal solutionSolution

    Mutual independence of $E_1,E_2,E_3$ guarantees $E_1$ is independent of every Boolean combination of $E_2,E_3$ (unions, intersections, complements). Both statements are true — option (A).

    Concept and Intuition

    Mutual independence of three events means every subset of them satisfies the product rule: $P(E_i\cap E_j)=P(E_i)P(E_j)$ for all pairs, and $P(E_1\cap E_2\cap E_3)=P(E_1)P(E_2)P(E_3)$. A standard consequence is that any event generated from $E_2$ and $E_3$ alone (via $\cup,\cap,{}^c$) remains independent of $E_1$, because independence of $E_1$ from $E_2$ and from $E_3$ combines nicely through inclusion-exclusion and De Morgan's laws.

    Step-by-Step Solution

    1. Statement I: Check $P(E_1\cap(E_2\cup E_3)) = P(E_1)P(E_2\cup E_3)$.

    $P(E_1\cap(E_2\cup E_3)) = P((E_1\cap E_2)\cup(E_1\cap E_3)) = P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3)$

    $= P(E_1)P(E_2)+P(E_1)P(E_3)-P(E_1)P(E_2)P(E_3) \quad(\text{using mutual independence})$

    $= P(E_1)\left[P(E_2)+P(E_3)-P(E_2)P(E_3)\right] = P(E_1)\,P(E_2\cup E_3).$

    So Statement I is true.

    2. Statement II: Check $P(E_1\cap(E_2\cap E_3)) = P(E_1)P(E_2\cap E_3)$.

    $P(E_1\cap E_2\cap E_3) = P(E_1)P(E_2)P(E_3) \quad(\text{mutual independence, directly})$ …

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11. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.A coin is tossed three times. Let A be the event of "getting three heads" and B be the event of "getting a head on the first toss". Then A and B are (A) Dependent events (B) Independent events (C) Impossible events (D) Certain events

    ›Reveal solutionSolution

    $P(A\cap B)=\tfrac18$ but $P(A)P(B)=\tfrac1{16}$, so $A$ and $B$ are dependent.

    Concept and Intuition

    Two events are independent exactly when knowing one has occurred does not change the probability of the other — formally $P(A\cap B)=P(A)\,P(B)$, equivalently $P(A\mid B)=P(A)$.

    Here $B$ ("first toss is a head") is part of the requirement for $A$ ("all three heads"). Learning that $B$ happened is genuine good news for $A$: it eliminates half the sample space, all of it unfavourable to $A$. So $A$ must depend on $B$. Note carefully: the three tosses are independent of one another, but these two events are not — they are defined on overlapping information.

    Step-by-Step Solution

    1. Sample space of three tosses: $\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$ — $8$ equally likely outcomes.
    2. $A=\{HHH\}\Rightarrow P(A)=\dfrac18$.
    3. $B=\{HHH,HHT,HTH,HTT\}\Rightarrow P(B)=\dfrac48=\dfrac12$.
    4. $A\cap B=\{HHH\}\Rightarrow P(A\cap B)=\dfrac18$.
    5. Test independence: $P(A)\cdot P(B)=\dfrac18\times\dfrac12=\dfrac1{16}$, while $P(A\cap B)=\dfrac18$. Since $\dfrac1{16}\neq\dfrac18$, the events are dependent. …

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12. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.A and B are independent events of a random experiment if and only if (A) $P(A|B) \ne P(A \cap B)$ (B) $P(A|B) = P(B|A)$ (C) $P(A|B) \ne P(A|B^c)$ (D) $P(A|B) = P(A|B^c)$

    ›Reveal solutionSolution

    Independence means knowing whether B occurred (or didn't) doesn't affect A's probability — precisely captured by $P(A|B)=P(A|B^c)$.

    Concept and Intuition

    Two events A and B are independent exactly when $P(A\cap B) = P(A)P(B)$, equivalently $P(A|B) = P(A)$ (given $P(B)>0$). A useful equivalent characterization avoids referencing the unconditional $P(A)$ directly: independence holds iff $P(A|B) = P(A|B^c)$. Intuitively, if learning "B happened" gives exactly the same probability for A as learning "B didn't happen," then knowledge of B carries no information about A — which is the essence of independence.

    Step-by-Step Solution

    1. Suppose A, B independent: $P(A|B)=P(A)$. Also $P(A|B^c) = \dfrac{P(A\cap B^c)}{P(B^c)} = \dfrac{P(A)-P(A\cap B)}{1-P(B)} = \dfrac{P(A)-P(A)P(B)}{1-P(B)} = \dfrac{P(A)(1-P(B))}{1-P(B)} = P(A)$.
    2. So independence implies $P(A|B) = P(A|B^c) (=P(A))$.
    3. Conversely, if $P(A|B)=P(A|B^c) = k$ for some constant $k$, then using the law of total probability, $P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = k\,P(B)+k\,P(B^c) = k$. So $P(A|B)=P(A)=k$, which is exactly the definition of independence.
    4. Hence $P(A|B)=P(A|B^c)$ is a necessary and sufficient condition for independence. …

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