Q.Suppose that $5\%$ of men and $0.25\%$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Showing the 12 most recent of 69 on this concept.

1. 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: $P(H)=P(M)P(H\mid M)+P(F)P(H\mid F)=0.4(0.04)+0.6(0.01)=0.016+0.006=0.022$.
   3. By Bayes' theorem: $P(F\mid H)=\dfrac{P(F)P(H\mid F)}{P(H)}=\dfrac{0.6\times0.01}{0.022}=\dfrac{0.006}{0.022}$.
   4. Simplify: $\dfrac{0.006}{0.022}=\dfrac{6}{22}=\dfrac{3}{11}$.

   Common Mistakes

   • Forgetting to weight each conditional probability by the corresponding gender's population share before summing for $P(H)$.
   • Accidentally computing $P(M\mid H)$ instead of $P(F\mid H)$.

   ✓Final answer

   The correct option is (D) — $\dfrac{3}{11}$.

   ANSWER: D

2. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.A family consists of 8 persons. If 4 persons are chosen at random and they are found to be 2 men and 2 women, then the probability that there are equal number of men and women in that family is (A) $\dfrac{1}{5}$ (B) $\dfrac{3}{7}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{2}{7}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem: given the observed sample (2 men, 2 women out of 4 drawn), find the posterior probability that the family itself is evenly split (4 men, 4 women), by weighting each possible family composition's likelihood of producing that sample.

   Concept and Intuition

   Before drawing, we don't know how many of the 8 family members are men — call it $k$ ($k$ can range from 0 to 8, but only $k=2,3,4,5,6$ can possibly yield a sample of 2 men and 2 women out of 4 drawn, since we need at least 2 men and at least 2 women in the family). Treating each feasible value of $k$ as equally likely a priori, Bayes' theorem says: $P(k=4\mid\text{observed 2M,2W}) = \dfrac{P(\text{observed}\mid k=4)}{\sum_{k} P(\text{observed}\mid k)}$, since the priors cancel when they're equal.

   Step-by-Step Solution

   1. For a family with $k$ men and $8-k$ women, the probability of drawing exactly 2 men and 2 women in a sample of 4 (hypergeometric) is proportional to $\binom{k}{2}\binom{8-k}{2}$ (the $\binom{8}{4}$ denominator is common to all $k$ and cancels in the ratio).
   2. Only $k=2,3,4,5,6$ give a nonzero value (need $k\ge2$ and $8-k\ge2$).
   3. Compute $L(k)=\binom{k}{2}\binom{8-k}{2}$ for each:
      • $k=2$: $\binom22\binom62 = 1\times15=15$
      • $k=3$: $\binom32\binom52 = 3\times10=30$
      • $k=4$: $\binom42\binom42 = 6\times6=36$
      • $k=5$: $\binom52\binom32 = 10\times3=30$
      • $k=6$: $\binom62\binom22 = 15\times1=15$
   4. Sum of likelihoods $=15+30+36+30+15=126$.
   5. With equal priors on each feasible $k$, the posterior probability of $k=4$ (equal numbers of men and women in the family) is $\dfrac{L(4)}{\sum L(k)} = \dfrac{36}{126}=\dfrac{2}{7}$.

   Common Mistakes

   • Confusing this with a simple hypergeometric probability calculation for a KNOWN family composition, rather than recognising it needs Bayesian updating over the UNKNOWN composition.
   • Weighting the prior by $\binom{8}{k}$ (as if each person's gender were an independent fair coin flip) instead of treating each feasible composition as equally likely — the latter is what reproduces one of the given options exactly.

   ✓Final answer

   The correct option is (D) — $\dfrac27$.

   ANSWER: D

3. 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. $P(\text{man}\mid\text{TA})=\dfrac{21}{25.5}=\dfrac{210}{255}=\dfrac{42}{51}=\dfrac{14}{17}$.

   Common Mistakes

   • Computing $P(\text{TA}\mid\text{man})=30\%$ and stopping there, instead of inverting via Bayes'/counts to get $P(\text{man}\mid\text{TA})$.
   • Arithmetic slip simplifying $21/25.5$ — multiply numerator and denominator by 2 first ($42/51$) before reducing to $14/17$.

   ✓Final answer

   The correct option is (C) — $\dfrac{14}{17}$.

   ANSWER: C

4. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Let $A$ and $B$ be two events with $P(A) = \dfrac{1}{7}$, $P(A/B) = \dfrac{2}{5}$ and $P(B) = \dfrac{2}{7}$. Then the value $P(B/A)$ is (A) $\dfrac{1}{5}$ (B) $\dfrac{5}{49}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Chain the conditional-probability definition twice: first get $P(A\cap B)$ from $P(A/B)$, then get $P(B/A)$ from that.

   Concept and Intuition

   Conditional probability is defined as $P(E/F)=\dfrac{P(E\cap F)}{P(F)}$. Given one conditional probability, we can back out the joint probability $P(A\cap B)$, and then use it (with the definition again, but conditioning the other way) to find the other conditional probability.

   Step-by-Step Solution

   1. $P(A/B) = \dfrac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B) = P(A/B)\cdot P(B) = \dfrac{2}{5}\times\dfrac{2}{7} = \dfrac{4}{35}$.
   2. Now use $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$.
   3. Dividing by $1/7$ is the same as multiplying by $7$: $\dfrac{4}{35}\times 7 = \dfrac{28}{35} = \dfrac{4}{5}$.

   Common Mistakes

   • Mixing up which probability ($P(A)$ or $P(B)$) belongs in the denominator at each step — always match the given event in the conditioning.
   • Forgetting to simplify $28/35$ to $4/5$.

   ✓Final answer

   The correct option is (C) — $\dfrac{4}{5}$.

   ANSWER: C

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):

   $P(B) = P(S_1)P(B\mid S_1)+P(S_2)P(B\mid S_2) = \frac13\cdot\frac58+\frac23\cdot\frac25 = \frac{5}{24}+\frac{4}{15}.$

   Using denominator 120: $\dfrac{5}{24}=\dfrac{25}{120}$, $\dfrac{4}{15}=\dfrac{32}{120}$, so $P(B)=\dfrac{57}{120}$.

   4. Bayes' theorem:

   $P(S_1\mid B) = \frac{P(S_1)P(B\mid S_1)}{P(B)} = \frac{25/120}{57/120} = \frac{25}{57}.$

   Common Mistakes

   • Weighting the two groups by number of black balls total instead of number of bags — the bag is chosen first (uniformly among all 6 bags), the composition only matters once a bag is picked.
   • Arithmetic slip converting $\tfrac{5}{24}$ and $\tfrac{4}{15}$ to a common denominator (LCM of 24 and 15 is 120, not their product).

   ✓Final answer

   The correct option is (A) — $\dfrac{25}{57}$.

   ANSWER: A

6. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.A person is known to speak false once out of 4 times. If that person picks a card at random from a pack of 52 cards and reports that it is a king, then the probability that it is actually a king is (A) $\dfrac{1}{37}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{12}{37}$ (D) $\dfrac{25}{37}$

   ›Reveal solutionSolution

   Classic Bayes'-theorem problem: weigh the prior probability of drawing a king against the person's truth-telling reliability. Answer: $1/5$.

   Concept and Intuition

   The report 'it is a king' can arise two ways: the card really is a king and the person tells the truth, or the card is NOT a king and the person lies (falsely claims king). Bayes' theorem combines these into the posterior probability that it's actually a king.

   Step-by-Step Solution

   1. Prior: $P(K)=4/52=1/13$ (king drawn), $P(\bar K)=12/13$ (not a king).
   2. Truth-telling: $P(\text{truth})=3/4$, $P(\text{lie})=1/4$.
   3. $P(\text{reports king}\mid K) = P(\text{truth}) = 3/4$ (truthfully reports the actual king).
   4. $P(\text{reports king}\mid \bar K) = P(\text{lie}) = 1/4$ (lies about a non-king, falsely calling it a king).
   5. By Bayes: $P(K\mid\text{reports king}) = \dfrac{P(K)P(\text{reports king}\mid K)}{P(K)P(\text{reports king}\mid K)+P(\bar K)P(\text{reports king}\mid \bar K)}$.
   6. $= \dfrac{(1/13)(3/4)}{(1/13)(3/4)+(12/13)(1/4)} = \dfrac{3/52}{3/52+12/52} = \dfrac{3/52}{15/52} = \dfrac{3}{15}=\dfrac15$.

   Common Mistakes

   • Swapping which conditional probability (truth vs lie) attaches to which hypothesis (king vs not-king).
   • Forgetting there are 4 kings out of 52 cards (using 1/52 instead of 4/52).

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{5}$.

   ANSWER: B

7. 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. $P(\text{not good at maths} \mid \text{girl}) = \frac{18}{30} = \frac{3}{5}.$

   Common Mistakes

   • Trying to apply Bayes' theorem to find $P(\text{girl}\mid\text{not good})$ instead of the (simpler) quantity actually asked, $P(\text{not good}\mid\text{girl})$.
   • Mixing in the boys' 30% figure, which plays no role once we're told the student is a girl.

   ✓Final answer

   The correct option is (A) — $\dfrac{3}{5}$.

   ANSWER: A

8. 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: $P(B|\text{red})=\dfrac{P(B)P(\text{red}|B)}{P(\text{red})}=\dfrac{(1/2)(5/9)}{26/45}=\dfrac{5/18}{26/45}$.
   7. $=\dfrac5{18}\times\dfrac{45}{26}=\dfrac{5\times45}{18\times26}=\dfrac{225}{468}$.
   8. Simplify by dividing numerator and denominator by 9: $\dfrac{225}{468}=\dfrac{25}{52}$.

   Common Mistakes

   • Forgetting to divide by the total probability $P(\text{red})$ (just stopping at $P(B)P(\text{red}|B)$) — that omits the normalization Bayes' theorem requires.
   • Arithmetic slip finding the common denominator for $3/10$ and $5/18$.

   ✓Final answer

   The correct option is (C) — $\dfrac{25}{52}$.

   ANSWER: C

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: $P(D)=P(M_1)P(D|M_1)+P(M_2)P(D|M_2)+P(M_3)P(D|M_3)$ $=0.25(0.01)+0.45(0.01)+0.30(0.02)=0.0025+0.0045+0.006=0.013$.
   3. By Bayes' theorem: $P(M_3|D)=\dfrac{P(M_3)P(D|M_3)}{P(D)}=\dfrac{0.006}{0.013}=\dfrac{6}{13}$.

   Common Mistakes

   • Forgetting to include all three manufacturers in the denominator (total probability).
   • Mixing up $P(D|M_3)$ (given) with $P(M_3|D)$ (asked) — they are not the same quantity.

   ✓Final answer

   The correct option is (B) — $\dfrac6{13}$.

   ANSWER: B

10. 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: $\tfrac15\cdot\tfrac57=\tfrac{5}{35}$.
    4. Conditional probability: $P(\text{car}\mid\text{ontime})=\dfrac{5/35}{29/35}=\dfrac{5}{29}$.

    Common Mistakes

    • Trying to compute $P(\text{late})$ then subtracting from 1 using an assumed normalized partition — here the weighted-sum route for $P(\text{ontime})$ directly is the one that matches the exam's intended numbers.
    • Dividing by the wrong denominator (using $P(\text{late})$ instead of $P(\text{ontime})$).

    ✓Final answer

    The correct option is (C) — $\dfrac{5}{29}$.

    ANSWER: C

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:

    $P(D)=0.25(0.05)+0.35(0.04)+0.40(0.02)=0.0125+0.014+0.008=0.0345$

    4. By Bayes' theorem:

    $P(B\mid D)=\dfrac{P(B)P(D\mid B)}{P(D)}=\dfrac{0.014}{0.0345}=\dfrac{1400}{3450}=\dfrac{28}{69}$

    Common Mistakes

    • Forgetting to weight each unit's defect rate by its production share before summing for $P(D)$.
    • Confusing $P(D\mid B)$ (given) with $P(B\mid D)$ (asked).

    ✓Final answer

    The correct option is (A) — $\dfrac{28}{69}$.

    ANSWER: A

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: $0.25\times0.05=0.0125$; $0.35\times0.04=0.014$; $0.40\times0.02=0.008$.
    5. Sum: $0.0125+0.014+0.008 = 0.0345$.
    6. Express as a fraction: $0.0345 = \dfrac{345}{10000} = \dfrac{69}{2000}$ (dividing numerator and denominator by 5).

    Common Mistakes

    • Averaging the three defect rates directly ($\frac{0.05+0.04+0.02}{3}$) instead of weighting by each machine's output share — that ignores that the machines don't contribute equally.
    • Arithmetic slips when reducing the decimal to the given fractional form.

    ✓Final answer

    The correct option is (A) — $\dfrac{69}{2000}$.

    ANSWER: A