Q.An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.In a class consisting of 40 boys and 30 girls, 30% of the boys and 40% of the girls are good at Mathematics. If a student selected at random from that class is found to be a girl, then the probability that she is not good at Mathematics is (A) $\dfrac{3}{5}$ (B) $\dfrac{2}{5}$ (C) $\dfrac{3}{10}$ (D) $\dfrac{7}{10}$

   ›Reveal solutionSolution

   Conditioning on "the student is a girl" restricts the sample space to the 30 girls only; the answer is $\frac35$.

   Concept and Intuition

   This is conditional probability with a twist: the condition ("selected student is a girl") is given as a fact, not something to be computed via Bayes' theorem. Once we know the student is a girl, the boys' statistics become irrelevant — we simply work within the group of girls.

   Step-by-Step Solution

   1. Total girls $=30$.
   2. Girls good at Mathematics $= 40\%$ of $30 = 12$.
   3. Girls not good at Mathematics $= 30-12=18$.
   4. Since we are told the selected student is a girl, the relevant sample space is just these 30 girls.
   5. $P(\text{not good at maths} \mid \text{girl}) = \frac{18}{30} = \frac{3}{5}.$ …

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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 $E$: $P(E)=P(E|A)P(A)+P(E|B)P(B)+P(E|C)P(C)$ $=0.6(0.30)+0.3(0.50)+0.1(0.20)=0.18+0.15+0.02=0.35$ …

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3. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.In a school there are 3 sections A, B and C. Section A contains 20 girls and 30 boys, section B contains 40 girls and 20 boys and section C contains 10 girls and 30 boys. The probabilities of selecting the section A, B and C are 0.2, 0.3 and 0.5 respectively. If a student selected at random from the school is a girl, then the probability that she belongs to section A is (A) $\dfrac{121}{200}$ (B) $\dfrac{16}{121}$ (C) $\dfrac{14}{81}$ (D) $\dfrac{16}{81}$

   ›Reveal solutionSolution

   A three-section Bayes'-theorem problem; the answer is $\dfrac{16}{81}$, option (D).

   Concept and Intuition

   Given a randomly selected student is a girl, we want to reverse-condition on which section she came from. This requires the total probability of "selecting a girl" (summed over all three sections weighted by section-selection probability), then applying Bayes' theorem to isolate section A's contribution.

   Step-by-Step Solution

   1. Conditional probabilities of picking a girl within each section: $P(\text{girl}\mid A)=\dfrac{20}{50}=\dfrac{2}{5}$; $P(\text{girl}\mid B)=\dfrac{40}{60}=\dfrac{2}{3}$; $P(\text{girl}\mid C)=\dfrac{10}{40}=\dfrac14$.
   2. Section-selection probabilities: $P(A)=\dfrac15,\ P(B)=\dfrac{3}{10},\ P(C)=\dfrac12$.
   3. Joint terms: $P(A)P(\text{girl}\mid A)=\dfrac15\cdot\dfrac25=\dfrac{2}{25}$; $P(B)P(\text{girl}\mid B)=\dfrac{3}{10}\cdot\dfrac23=\dfrac15$; $P(C)P(\text{girl}\mid C)=\dfrac12\cdot\dfrac14=\dfrac18$. …

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4. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.In a test a student either guesses or copies or knows the answer to answer a multiple choice question with four choices. The probability that he makes a guess is $1/3$ and the probability that he copies the answer is $1/6$. The probability that his answer is correct, given that he copied it is $1/8$. The probability that he knew the answer to the question, given that he answered it correctly is (A) $\dfrac{29}{24}$ (B) $\dfrac{22}{29}$ (C) $\dfrac{24}{29}$ (D) $\dfrac{23}{29}$

   ›Reveal solutionSolution

   A classic Bayes'-theorem problem with three mutually exclusive causes (guess/copy/know) for a correct answer. Answer: $24/29$.

   Concept and Intuition

   This is a textbook application of Bayes' theorem: we're given the prior probabilities of three exhaustive, mutually exclusive ways of answering (guessing, copying, knowing) along with the conditional probability of being correct under each, and asked to reverse the conditioning — find the probability the student knew the answer, GIVEN that they got it right.

   Step-by-Step Solution

   1. The three ways of answering are exhaustive: $P(\text{know})=1-P(\text{guess})-P(\text{copy})=1-\tfrac13-\tfrac16=1-\tfrac12=\tfrac12$.
   2. Conditional correctness probabilities: $P(C\mid\text{guess})=\tfrac14$ (one of four choices), $P(C\mid\text{copy})=\tfrac18$ (given), $P(C\mid\text{know})=1$.
   3. Total probability of being correct: $P(C)=\tfrac13\cdot\tfrac14+\tfrac16\cdot\tfrac18+\tfrac12\cdot1=\tfrac1{12}+\tfrac1{48}+\tfrac12$. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.In a bolt factory, machines A, B, C manufacture 25%, 35%, 40% of the total output respectively. There is a chance of having 5%, 4%, 2% defective bolts manufactured by A, B, C respectively. If a bolt is drawn at random from the output, then the probability that it is defective is (A) $\dfrac{69}{2000}$ (B) $\dfrac{59}{2000}$ (C) $\dfrac{79}{2000}$ (D) $\dfrac{89}{2000}$

   ›Reveal solutionSolution

   Classic total-probability ("law of total probability") setup: weight each machine's defect rate by its share of output and sum. The answer is (A).

   Concept and Intuition

   When an item can come from several mutually exclusive, exhaustive sources (here, machines A, B, C), and each source has its own conditional probability of producing a defect, the overall probability of a defect is the weighted average:

   $P(D) = \sum_i P(\text{source}_i)\,P(D\mid \text{source}_i).$

   This is the law of total probability — it's the natural way to combine "how much each machine contributes" with "how likely each machine's own output is defective."

   Step-by-Step Solution

   1. Let $A, B, C$ denote drawing a bolt from each machine: $P(A)=0.25$, $P(B)=0.35$, $P(C)=0.40$ (these sum to 1, as they must).
   2. Conditional defect probabilities: $P(D\mid A)=0.05$, $P(D\mid B)=0.04$, $P(D\mid C)=0.02$.
   3. By the law of total probability:

   $P(D) = P(A)P(D\mid A) + P(B)P(D\mid B) + P(C)P(D\mid C).$

   4. Compute each term: $0.25\times0.05=0.0125$; $0.35\times0.04=0.014$; $0.40\times0.02=0.008$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.70% of the total employees of a factory are men. Among the employees of that factory, 30% of men and 15% of women are technical assistants. If an employee chosen at random is found to be a technical assistant, then the probability that this employee is a man is (A) $\dfrac{9}{23}$ (B) $\dfrac{3}{17}$ (C) $\dfrac{14}{17}$ (D) $\dfrac{14}{23}$

   ›Reveal solutionSolution

   A direct Bayes'/total-probability computation using a convenient total of 100 employees. The answer is $\dfrac{14}{17}$.

   Concept and Intuition

   We want $P(\text{man}\mid\text{technical assistant})$. Since we're given the proportion of men and women, and what fraction of each group are technical assistants, the cleanest approach is to work with actual counts (out of a convenient total like 100) rather than abstract probabilities — it avoids fraction juggling.

   Step-by-Step Solution

   1. Assume 100 employees: Men $=70$, Women $=30$.
   2. Technical assistants among men: $30\%$ of $70=21$.
   3. Technical assistants among women: $15\%$ of $30=4.5$.
   4. Total technical assistants $=21+4.5=25.5$.
   5. $P(\text{man}\mid\text{TA})=\dfrac{21}{25.5}=\dfrac{210}{255}=\dfrac{42}{51}=\dfrac{14}{17}$. …

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7. 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. …

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8. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If it is known that a woman has two children and she has at least one girl child, the probability that the woman has both girl children is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   This is a classic conditional-probability trap: conditioning on "at least one girl" (not "the elder/first child is a girl") leaves 3 equally likely outcomes, of which 1 is both-girls, giving $\dfrac13$.

   Concept and Intuition

   With two children, birth order matters for counting equally likely outcomes: $BB, BG, GB, GG$, each with probability $1/4$. The event "at least one girl" is a set of outcomes, not a statement about a specific (e.g. first) child, so it correctly removes only $BB$ and keeps three outcomes, not two. This distinguishes it from the simpler (and different) question "given the elder child is a girl," which would leave only $\{GB,GG\}$ and give probability $1/2$.

   Step-by-Step Solution

   1. Sample space (ordered by birth, say elder-younger): $\{BB,BG,GB,GG\}$, each with probability $\dfrac14$.
   2. Event $E$ = "at least one girl" = $\{BG,GB,GG\}$, so $P(E)=\dfrac34$. …

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9. 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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10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.There are three families $F_1, F_2, F_3$. $F_1$ has 2 boys and 1 girl; $F_2$ has 1 boy and 2 girls; $F_3$ has 1 boy and 1 girl. A family is randomly chosen and a child is chosen from that family randomly. If it is known that the child thus selected is a girl, then the probability that she is from $F_2$ is (A) $\dfrac{4}{9}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{3}{7}$ (D) $\dfrac{5}{7}$

    ›Reveal solutionSolution

    Applying Bayes' theorem across the three equally-likely families gives $P(F_2\mid\text{girl})=\dfrac49$.

    Concept and Intuition

    This is a textbook application of Bayes' theorem: we're given the reverse conditional probabilities (family $\to$ probability of picking a girl) and asked for the forward one (girl picked $\to$ probability she's from a specific family).

    Step-by-Step Solution

    1. Prior: $P(F_1)=P(F_2)=P(F_3)=\dfrac13$.
    2. $P(\text{girl}\mid F_1)=\dfrac13$ (1 girl out of 3 children), $P(\text{girl}\mid F_2)=\dfrac23$, $P(\text{girl}\mid F_3)=\dfrac12$.
    3. Total probability: $P(\text{girl})=\frac13\cdot\frac13+\frac13\cdot\frac23+\frac13\cdot\frac12=\frac13\left(\frac13+\frac23+\frac12\right)=\frac13\cdot\frac{2+4+3}{6}=\frac13\cdot\frac96=\frac12$
    4. $P(F_2\cap\text{girl})=\dfrac13\cdot\dfrac23=\dfrac29$. …

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11. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.In a toy factory, the machines A, B and C are used to manufacture 30%, 40% and 30% of the output, respectively. The probabilities of toys made by machines A, B, and C to be defective are respectively 2%, 3% and 1%. A toy is taken from the factory and is found to be defective. The probability that it was manufactured by the machine B is (A) $4/5$ (B) $2/9$ (C) $3/4$ (D) $4/7$

    ›Reveal solutionSolution

    This tests Bayes' theorem for finding the probability of a specific cause given an observed

    effect (a defective toy); the answer is $4/7$.

    Concept and Intuition

    When an outcome (a defective toy) could have come from several sources, each with its own prior

    probability and its own conditional probability of producing that outcome, Bayes' theorem lets us

    "invert" the conditioning: given that the outcome occurred, what's the probability it came from a

    particular source? The key is to first find the total probability of the outcome by summing over

    all sources (the law of total probability), then take the one source's contribution as a fraction of

    that total.

    Step-by-Step Solution

    1. Prior probabilities: $P(A)=0.3,\ P(B)=0.4,\ P(C)=0.3$.
    2. Conditional defect rates: $P(D\mid A)=0.02,\ P(D\mid B)=0.03,\ P(D\mid C)=0.01$.
    3. Total probability of a defective toy (law of total probability):

    $P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C) = 0.3(0.02)+0.4(0.03)+0.3(0.01)$

    $= 0.006+0.012+0.003 = 0.021.$

    4. By Bayes' theorem: …

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12. 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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