Q.In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

Showing the 12 most recent of 69 on this concept.

1. 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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2. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A manufacturing company of bulbs has 3 units A, B and C which produce 25%, 35% and 40% of the bulbs respectively. Out of the bulbs produced by A, B, C units, 5%, 4% and 2% are defective respectively. If a bulb is chosen at random and found to be defective, then the probability that it is produced by unit B is (A) $\dfrac{28}{69}$ (B) $\dfrac{28}{71}$ (C) $\dfrac{29}{67}$ (D) $\dfrac{25}{69}$

   ›Reveal solutionSolution

   This tests Bayes' theorem: given the bulb is defective, find the (reversed) probability it came from unit B. Answer: $\dfrac{28}{69}$.

   Concept and Intuition

   We know the forward probabilities (which unit makes what fraction, and each unit's defect rate), but we're asked a reverse question: given the bulb turned out defective, which unit is it likely from? Bayes' theorem reweights the prior production shares by how likely each unit was to have produced this particular (defective) outcome.

   Step-by-Step Solution

   1. Let $A,B,C$ be the events "bulb from unit A/B/C", with $P(A)=0.25,\ P(B)=0.35,\ P(C)=0.40$.
   2. Conditional defect rates: $P(D\mid A)=0.05,\ P(D\mid B)=0.04,\ P(D\mid C)=0.02$.
   3. Total probability of a defective bulb: $P(D)=0.25(0.05)+0.35(0.04)+0.40(0.02)=0.0125+0.014+0.008=0.0345$ …

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3. 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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4. 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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5. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.There are 2 bags each containing 3 white and 5 black balls and 4 bags each containing 6 white and 4 black balls. If a ball drawn randomly from a bag is found to be black, then the probability that this ball is from the first set of bags is (A) $\dfrac{25}{57}$ (B) $\dfrac{25}{41}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   A classic Bayes'-theorem (inverse-probability) question: given the ball drawn is black, find the probability it came from the first group of bags. Answer: $\dfrac{25}{57}$.

   Concept and Intuition

   When an experiment happens in two stages — first a bag is picked (from one of two groups of bags), then a ball is drawn from it — and we're told the result of the second stage (a black ball came out), Bayes' theorem lets us reverse the direction of reasoning and find the probability about the first stage (which group the bag was from). The prior probability of picking from a group is proportional to how many bags are in that group (each individual bag is equally likely to be picked), and then we weight by how likely that group is to produce the observed outcome.

   Step-by-Step Solution

   1. Groups and priors. Group 1 (call it $S_1$) has 2 bags (each 3 white, 5 black — 8 balls). Group 2 ($S_2$) has 4 bags (each 6 white, 4 black — 10 balls). Total bags $=6$, each equally likely to be chosen, so

   $P(S_1)=\frac{2}{6}=\frac13,\qquad P(S_2)=\frac{4}{6}=\frac23.$

   2. Likelihoods of drawing black. From a group-1 bag: $P(B\mid S_1)=\dfrac{5}{8}$. From a group-2 bag: $P(B\mid S_2)=\dfrac{4}{10}=\dfrac25$.
   3. Total probability of drawing a black ball (law of total probability): $P(B) = P(S_1)P(B\mid S_1)+P(S_2)P(B\mid S_2) = \frac13\cdot\frac58+\frac23\cdot\frac25 = \frac{5}{24}+\frac{4}{15}.$ …

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6. 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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7. 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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8. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Bag A contains 3 white and 4 black balls. Bag B contains 4 white and 3 black balls. Bag C contains 2 white and 5 black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag C is (A) $\dfrac{1}{6}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{2}{13}$

   ›Reveal solutionSolution

   This is a direct Bayes'-theorem (inverse probability) question. Answer: $\dfrac{2}{9}$.

   Concept and Intuition

   We're given the outcome (a white ball was drawn) and asked for the probability of a particular cause (it came from bag C). This is exactly Bayes' theorem: $P(C\mid W) = \dfrac{P(C)P(W\mid C)}{\sum_i P(\text{bag}_i)P(W\mid \text{bag}_i)}$.

   Step-by-Step Solution

   1. Each bag is chosen with probability $\frac13$.
   2. $P(W\mid A) = \dfrac{3}{7}$ (3 white out of 7 total in bag A).
   3. $P(W\mid B) = \dfrac{4}{7}$ (4 white out of 7 in bag B).
   4. $P(W\mid C) = \dfrac{2}{7}$ (2 white out of 7 in bag C).
   5. Total probability of white: $P(W) = \frac13\left(\dfrac37+\dfrac47+\dfrac27\right) = \frac13\cdot\dfrac{9}{7} = \dfrac{9}{21} = \dfrac{3}{7}$. …

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9. 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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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.An item is tested on a device for its defectiveness. The probability that such an item is defective is 0.3. The device gives accurate result in 8 out of 10 such tests. If the device reports that an item tested is not defective, then the probability that it is actually defective is (A) $\dfrac{2}{15}$ (B) $\dfrac{3}{29}$ (C) $\dfrac{3}{31}$ (D) $\dfrac{4}{51}$

    ›Reveal solutionSolution

    A direct Bayes'-theorem inversion problem; the item is actually defective with probability $\dfrac{3}{31}$ given a "not defective" report — option (C).

    Concept and Intuition

    The device's report can be wrong. To find the true state given an observed (possibly wrong) report, we must weigh both ways the report could have arisen: a genuinely non-defective item correctly reported as such, or a genuinely defective item incorrectly reported as non-defective. Bayes' theorem combines these.

    Step-by-Step Solution

    1. Let $D$: item defective, $P(D)=0.3$; $D'$: item not defective, $P(D')=0.7$.
    2. Device accuracy $=0.8$, so it errs with probability $0.2$.
    3. "Reports not defective" while the item is defective means the device made an error: $P(\text{report ND}\mid D)=0.2$.
    4. "Reports not defective" while the item is not defective means the device was accurate: $P(\text{report ND}\mid D')=0.8$. …

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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-2023-05-15-FN1 markMCQ
    Q.Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is chosen at random and a ball is drawn from it at random, then the probability that the ball drawn is white, is (A) $\dfrac{1}{42}$ (B) $\dfrac{42}{32}$ (C) $\dfrac{33}{42}$ (D) $\dfrac{23}{42}$

    ›Reveal solutionSolution

    Total-probability rule over the two equally likely bags gives $P(\text{white})=\frac12\cdot\frac46+\frac12\cdot\frac37=\frac{23}{42}$.

    Concept and Intuition

    The ball drawn depends on which bag was chosen first. Since the bag choice is random with $P(B_1)=P(B_2)=\frac12$, and the draw is conditionally independent given the bag, the Law of Total Probability adds the two conditional probabilities weighted by how likely each bag is.

    Step-by-Step Solution

    1. $B_1$: 4 white, 2 black out of 6 $\Rightarrow P(\text{white}\mid B_1)=\frac{4}{6}=\frac23$.
    2. $B_2$: 3 white, 4 black out of 7 $\Rightarrow P(\text{white}\mid B_2)=\frac37$.
    3. $P(\text{white})=\frac12\cdot\frac23+\frac12\cdot\frac37=\frac13+\frac{3}{14}$. …

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