Q.If $A$ and $B$ are such events that $P(A) > 0$ and $P(B) \neq 1$, then $P(A' \mid B')$ equals
(A) $1 - P(A \mid B)$
(B) $1 - P(A' \mid B)$
(C) $\dfrac{1 - P(A \cup B)}{P(B')}$
(D) $\dfrac{P(A')}{P(B')}$

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.For two events A and B, a true statement among the following is (A) $P(\bar{A} \cup \bar{B}) = 1 - P(A)P\left(\dfrac{B}{A}\right)$ (B) $P(\bar{A} \cup \bar{B}) = 1 - P(A \cup B)$ (C) $P(\bar{A} \cup \bar{B}) = P(A \cup B)$ (D) $P(\bar{A} \cup \bar{B}) = P(\bar{A}) + P(\bar{B})$

   ›Reveal solutionSolution

   De Morgan's law gives $P(\bar A\cup\bar B)=1-P(A\cap B)$, and since $P(A\cap B)=P(A)P(B|A)$ always (definition of conditional probability), option (A) is the universally true statement.

   Concept and Intuition

   $\bar A\cup\bar B$ is the complement of $A\cap B$ (De Morgan's law: $\overline{A\cap B}=\bar A\cup\bar B$). So $P(\bar A\cup\bar B)=1-P(A\cap B)$ always holds, regardless of independence. The multiplication rule $P(A\cap B)=P(A)P(B|A)$ is also always true by definition of conditional probability.

   Step-by-Step Solution

   1. By De Morgan's law: $\bar A\cup\bar B = \overline{A\cap B}$, so $P(\bar A\cup\bar B) = 1-P(A\cap B)$.
   2. By definition of conditional probability, $P(B|A)=\frac{P(A\cap B)}{P(A)}$, so $P(A\cap B)=P(A)P(B|A)$.
   3. Substituting: $P(\bar A\cup\bar B)=1-P(A)P(B/A)$ — this is option (A), and it holds for ANY events A, B. …

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2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The probability for a contractor to get a road contract is $\dfrac{2}{9}$ and to get a building contract is $\dfrac{5}{9}$. if the probability to get both the contract is $\dfrac{1}{6}$ then what is the probability to get neither of these two contracts? (A) $\dfrac{7}{9}$ (B) $\dfrac{4}{9}$ (C) $\dfrac{7}{18}$ (D) $\dfrac{4}{18}$

   ›Reveal solutionSolution

   This tests the addition rule for probability and the complement rule. The answer is $\dfrac{7}{18}$.

   Concept and Intuition

   "Neither" is the complement of "at least one", i.e. $1-P(A\cup B)$. First find $P(A\cup B)$ using the inclusion-exclusion formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, then subtract from 1.

   Step-by-Step Solution

   1. $P(A)=\dfrac29$ (road), $P(B)=\dfrac59$ (building), $P(A\cap B)=\dfrac16$ (both).
   2. $P(A\cup B) = \dfrac29+\dfrac59-\dfrac16 = \dfrac79-\dfrac16$.
   3. Common denominator 18: $\dfrac79=\dfrac{14}{18}$, $\dfrac16=\dfrac{3}{18}$, so $P(A\cup B)=\dfrac{14-3}{18}=\dfrac{11}{18}$. …

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3. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Two natural numbers are chosen at random from 1 to 100 and are multiplied. If A is the event that the product is an even number and B is the event that the product is divisible by 4, then $P(A \cap \bar{B}) =$ (A) $\dfrac{25}{198}$ (B) $\dfrac{49}{198}$ (C) $\dfrac{25}{99}$ (D) $\dfrac{50}{99}$

   ›Reveal solutionSolution

   Split 1–100 by 2-adic valuation (odd / $\equiv2\bmod4$ / divisible by 4); the event 'even but not divisible by 4' happens only for an odd–$(\equiv2\bmod4)$ pair. Probability is $25/99$.

   Concept and Intuition

   Whether a product of two numbers is divisible by 4 depends on the total power of 2 across both factors. Splitting the range into three classes by how many factors of 2 each number contributes (0, exactly 1, or $\ge2$) turns this into simple counting.

   Step-by-Step Solution

   1. Among 1–100: odd numbers (0 factors of 2) — 50 of them. Numbers $\equiv2\pmod4$ (exactly 1 factor of 2, e.g. 2,6,10,...,98) — 25 of them. Numbers divisible by 4 (at least 2 factors of 2) — 25 of them.
   2. Two distinct numbers are chosen (without replacement) from 1–100; total ways $=\binom{100}{2}=4950$.
   3. Event $A\cap\bar B$ = product is even AND not divisible by 4, i.e. the total power of 2 across the two numbers is exactly 1.
   4. This happens only when one number is odd (0 power) and the other is $\equiv2\pmod4$ (exactly 1 power) — any other combination gives total power 0 (both odd, not in $A$) or $\ge2$ (one or both contribute $\ge1$ in a way that sums to $\ge2$, landing in $B$). …

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4. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Two candidates A and B have attended an interview conducted by a recruitment board for two jobs. If the probability that candidate A will get the job is 0.8 and the probability that candidate B will get the job is 0.7, then the probability that atleast one of them will get the job is (A) 0.96 (B) 0.94 (C) 0.92 (D) 0.9

   ›Reveal solutionSolution

   This tests the complement rule for "at least one" events with two independent trials. Answer: 0.94.

   Concept and Intuition

   When two events are independent, the easiest way to find $P(\text{at least one occurs})$ is to compute the probability that neither occurs (multiply the individual "failure" probabilities) and subtract from 1.

   Step-by-Step Solution

   1. $P(A\text{ gets job})=0.8\Rightarrow P(A\text{ doesn't})=0.2$.
   2. $P(B\text{ gets job})=0.7\Rightarrow P(B\text{ doesn't})=0.3$.
   3. Assuming independence, $P(\text{neither gets the job})=0.2\times0.3=0.06$.
   4. $P(\text{at least one gets the job})=1-0.06=0.94$.

   Common Mistakes …

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5. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.The probability that A speaks truth is 75% and the probability that B speaks truth is 80%. The probability that they contradict each other when asked to speak on a fact is (A) $\dfrac{3}{20}$ (B) $\dfrac{4}{20}$ (C) $\dfrac{7}{20}$ (D) $\dfrac{5}{20}$

   ›Reveal solutionSolution

   "Contradiction" means exactly one of A, B tells the truth; summing the two mutually exclusive cases gives $\dfrac7{20}$.

   Concept and Intuition

   A and B "contradict" each other exactly when one speaks truth and the other lies (if both tell the truth or both lie, they'd agree, not contradict). These are two independent, mutually exclusive scenarios whose probabilities add.

   Step-by-Step Solution

   1. $P(A\text{ true})=0.75$, so $P(A\text{ false})=0.25$.
   2. $P(B\text{ true})=0.80$, so $P(B\text{ false})=0.20$.
   3. Contradiction case 1: A true, B false: $0.75\times0.20=0.15$.
   4. Contradiction case 2: A false, B true: $0.25\times0.80=0.20$. …

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6. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked randomly. The probability that it is neither red nor green is (A) $1/3$ (B) $3/4$ (C) $7/19$ (D) $8/21$

   ›Reveal solutionSolution

   "Neither red nor green" simply means "blue"; the probability is just the blue count over the total count.

   Concept and Intuition

   For a simple random draw from a finite set with equally likely outcomes, the probability of an event equals the count of favourable outcomes divided by the total count.

   Step-by-Step Solution

   1. Total balls $= 8\ (\text{red}) + 7\ (\text{blue}) + 6\ (\text{green}) = 21$.
   2. The complement of "red or green" among these three categories is simply "blue" (since every ball is one of the three colours).
   3. Number of blue balls $=7$.
   4. $P(\text{blue}) = \dfrac{7}{21} = \dfrac13$.

   Common Mistakes …

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7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.A bag contains 3 white, 2 blue and 5 red balls. One ball is drawn at random from this bag. Then, the probability that the ball drawn is not red is (A) 3/10 (B) 1/5 (C) 1/2 (D) 4/5

   ›Reveal solutionSolution

   "Not red" simply means white or blue; count those and divide by the total.

   Concept and Intuition

   For equally likely outcomes, probability is favourable outcomes over total outcomes — here "not red" is the complement of "red" within the same sample space.

   Step-by-Step Solution

   1. Total balls $=3\text{ white}+2\text{ blue}+5\text{ red}=10$.
   2. Not-red balls $=3+2=5$.
   3. $P(\text{not red})=\dfrac{5}{10}=\dfrac12$.

   Common Mistakes …

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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$U_1, U_2, U_3$ are three urns. $U_1$ contains 5 red, 3 white, 2 black balls; $U_2$ contains 4 red, 4 white, 2 black balls and $U_3$ contains 3 red, 4 white, 3 black balls. If a ball is chosen at random from an urn chosen at random, then the probability of not getting a black ball is (A) $\dfrac{7}{30}$ (B) $\dfrac{23}{30}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{11}{30}$

   ›Reveal solutionSolution

   Use the Law of Total Probability, averaging the "not black" probability over the three equally likely urns. Answer: $\dfrac{23}{30}$.

   Concept and Intuition

   This is a two-stage random experiment: first an urn is picked (uniformly at random among the three), then a ball is drawn from that urn. The overall probability of an event is the weighted average of its probability conditional on each urn, weighted by the probability of picking that urn — this is exactly the Law of Total Probability.

   Step-by-Step Solution

   1. Each urn is equally likely to be chosen: $P(U_1)=P(U_2)=P(U_3)=\dfrac13$.
   2. Totals per urn: $U_1$ has $5+3+2=10$ balls, $U_2$ has $4+4+2=10$ balls, $U_3$ has $3+4+3=10$ balls.
   3. Probability of NOT drawing black from each urn:

   $P(\text{not black}\mid U_1)=\frac{5+3}{10}=\frac{8}{10}$

   $P(\text{not black}\mid U_2)=\frac{4+4}{10}=\frac{8}{10}$

   $P(\text{not black}\mid U_3)=\frac{3+4}{10}=\frac{7}{10}$

   4. By the Law of Total Probability: …

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9. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.In a shoe rack there are 4 pairs of shoes and 4 shoes are drawn one after the other at random without replacement. Then the probability of getting atleast one correct pair of shoes among the four shoes drawn is (A) $\dfrac{8}{35}$ (B) $\dfrac{27}{35}$ (C) $\dfrac{1679}{1680}$ (D) $\dfrac{1}{1680}$

   ›Reveal solutionSolution

   This tests complementary counting for "at least one" events. Answer: $27/35$.

   Concept and Intuition

   "At least one correct pair" is hard to count directly (could be exactly 1 pair, or 2 pairs), but its complement — "no correct pair at all" — is very restrictive and easy to count. With exactly 4 pairs and 4 shoes drawn, avoiding any complete pair forces one shoe from every single pair (there's no pair left over to skip).

   Step-by-Step Solution

   1. Total ways to choose 4 shoes out of 8: $\binom{8}{4} = 70$.
   2. Count the complement: no pair among the 4 drawn is a matched (left+right) pair. Since there are exactly 4 pairs and we draw exactly 4 shoes, "no complete pair" means each of the 4 pairs contributes exactly one shoe (if any pair contributed 0 shoes, another would have to contribute 2, forming a complete pair — contradiction).
   3. For each of the 4 pairs, there are 2 choices (which shoe of that pair to include), giving $2^4 = 16$ ways with no complete pair. …

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10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.The probability that a person chosen at random is left handed (in hand writing) is 0.1. Then the probability that in a group of 10 people there is a left handed person is (A) $(0.9)^9$ (B) $(0.9)^8$ (C) $(0.9)^6$ (D) $0.9$

    ›Reveal solutionSolution

    This is a binomial "exactly one success" computation; the factor of $10\times0.1$ conveniently equals $1$, leaving a clean power of $0.9$.

    Concept and Intuition

    With $X\sim B(10,0.1)$ counting left-handed people in the group, $P(X=1)=\binom{10}{1}(0.1)(0.9)^9$. The combinatorial factor $10$ exactly cancels the $0.1$, collapsing the whole expression to a pure power of $0.9$ — which is exactly the pattern the answer choices are built around.

    Step-by-Step Solution

    1. $X\sim B(n=10,p=0.1)$.
    2. $P(X=1)=\binom{10}{1}\,(0.1)^1\,(0.9)^{9}$. …

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11. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.Two students appeared simultaneously for an entrance exam. If the probability that the first student gets qualified in the exam is $\dfrac{1}{4}$ and the probability that the second student gets qualified in the same exam is $\dfrac{2}{5}$, then the probability that atleast one of them gets qualified in that exam is (A) $\dfrac{1}{10}$ (B) $\dfrac{7}{20}$ (C) $\dfrac{6}{10}$ (D) $\dfrac{11}{20}$

    ›Reveal solutionSolution

    This tests the complement rule for "at least one" probability with two independent events. The probability at least one student qualifies is $\dfrac{11}{20}$.

    Concept and Intuition

    When asked for $P(\text{at least one of several independent events occurs})$, it is almost always easiest to compute the complement — the probability that none occur — and subtract from 1, since "none occur" for independent events is just the product of each event's complement probability.

    Step-by-Step Solution

    1. $P(\text{student 1 qualifies})=\tfrac14\Rightarrow P(\text{student 1 fails})=\tfrac34$.
    2. $P(\text{student 2 qualifies})=\tfrac25\Rightarrow P(\text{student 2 fails})=\tfrac35$.
    3. Assuming independence, $P(\text{neither qualifies})=\tfrac34\times\tfrac35=\tfrac{9}{20}$. …

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12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.In a lottery, containing 35 tickets, exactly 10 tickets bear a prize. If a ticket is drawn at random, then the probability of not getting a prize is. (A) $1/10$ (B) $2/5$ (C) $2/7$ (D) $5/7$

    ›Reveal solutionSolution

    This tests basic classical probability: favourable outcomes over total outcomes for a single random draw. The probability of not winning is $\dfrac57$.

    Concept and Intuition

    For a single random draw from a finite set with equally likely outcomes, the probability of an event is simply the count of outcomes satisfying that event divided by the total count. Here the event is "drawing a non-prize ticket".

    Step-by-Step Solution

    1. Total tickets $=35$; prize-bearing tickets $=10$.
    2. Non-prize tickets $=35-10=25$.
    3. $P(\text{not getting a prize}) = \dfrac{25}{35}$. …

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