Q.In a college, $30\%$ students fail in physics, $25\%$ fail in mathematics and $10\%$ fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is
(A) $\dfrac{1}{10}$
(B) $\dfrac{2}{5}$
(C) $\dfrac{9}{20}$
(D) $\dfrac{1}{3}$

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 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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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 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In a college, the heights of 4% of male and 1% of female students are more than 1.8 meters. If 60% of the total students are female and a student selected at random has height more than 1.8 meters, then the probability that this student is a female, is (A) $\dfrac{8}{11}$ (B) $\dfrac{6}{11}$ (C) $\dfrac{5}{11}$ (D) $\dfrac{3}{11}$

   ›Reveal solutionSolution

   A direct application of Bayes' theorem: given the height is above 1.8m, find the probability the student is female.

   Concept and Intuition

   This is the classic "reverse conditional probability" setup — we know $P(\text{tall}\mid\text{gender})$ for each gender and the gender proportions, and want $P(\text{gender}\mid\text{tall})$. Bayes' theorem converts one direction of conditioning into the other by weighting each gender's tall-probability by its population share and normalizing.

   Step-by-Step Solution

   1. Let $M$ = male, $F$ = female, $H$ = height $>1.8$m. Given: $P(F)=0.6$, $P(M)=0.4$, $P(H\mid M)=0.04$, $P(H\mid F)=0.01$.
   2. Total probability of height $>1.8$m: $P(H)=P(M)P(H\mid M)+P(F)P(H\mid F)=0.4(0.04)+0.6(0.01)=0.016+0.006=0.022$. …

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5. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The probability that a person goes to college by car is $\dfrac{1}{5}$; by bus $\dfrac{2}{5}$ and by train is $\dfrac{3}{5}$ respectively. The probabilities that he reaches the college late if he takes car, bus, train are $\dfrac{2}{7}, \dfrac{4}{7}$ and $\dfrac{1}{7}$ respectively. If he reaches the college in time, the probability that he travelled by car is (A) $\dfrac{6}{29}$ (B) $\dfrac{24}{29}$ (C) $\dfrac{5}{29}$ (D) $\dfrac{23}{29}$

   ›Reveal solutionSolution

   Bayes' theorem: weight each mode's "on-time" probability by its usage probability, then take the car-share of the total.

   Concept and Intuition

   This is a direct application of Bayes' theorem / total probability: to find $P(\text{cause}\mid\text{effect})$, build the denominator as the weighted sum over every possible cause of reaching the observed effect (here, arriving on time), then take the numerator's share of it.

   Step-by-Step Solution

   1. On-time probabilities per mode: car $1-\tfrac27=\tfrac57$; bus $1-\tfrac47=\tfrac37$; train $1-\tfrac17=\tfrac67$.
   2. Total on-time probability: $P(\text{ontime})=\tfrac15\cdot\tfrac57+\tfrac25\cdot\tfrac37+\tfrac35\cdot\tfrac67=\tfrac{5}{35}+\tfrac{6}{35}+\tfrac{18}{35}=\tfrac{29}{35}$.
   3. Joint probability of taking the car AND being on time: $\tfrac15\cdot\tfrac57=\tfrac{5}{35}$. …

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6. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.One ticket is selected at random from 50 tickets numbered $00, 01, 02, \ldots 49$. The probability that sum of the digits is 10, given that product of the digits is 9 is (A) $\dfrac{9}{10}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{2}{25}$

   ›Reveal solutionSolution

   Only two tickets (19 and 33) have digit-product 9, and only one of them (19) also has digit-sum 10, giving conditional probability $1/2$.

   Concept and Intuition

   Conditional probability $P(\text{sum}=10\mid\text{product}=9)$ restricts attention entirely to the tickets satisfying the "given" condition (product $=9$), then asks what fraction of those also satisfy the target condition (sum $=10$).

   Step-by-Step Solution

   1. List all two-digit combinations $(d_1,d_2)$ with $d_1\in\{0,1,2,3,4\}$ (tens digit, since tickets run 00–49) and $d_2\in\{0,\dots,9\}$ (units digit) whose product is 9:
      • $d_1=1,d_2=9$: ticket $19$.
      • $d_1=3,d_2=3$: ticket $33$.
      • No other integer pairs with $d_1\le4$ give product 9.
   2. So the "product = 9" event has exactly 2 tickets: $\{19, 33\}$.
   3. Check digit sums: $19\to1+9=10$ ✓; $33\to3+3=6$ ✗.
   4. Only 1 out of these 2 tickets also has digit-sum 10. …

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7. 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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8. 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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9. 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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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 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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12. 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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