Q.Assume that the chances of a patient having a heart attack is $40\%$. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30\%$ and prescription of certain drug reduces its chances by $25\%$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

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 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 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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4. 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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5. 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). …

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6. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.Two persons P and Q are considering to apply for a job. The probability that P applies for the job is 1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for the job is (A) $4/5$ (B) $5/6$ (C) $7/8$ (D) $11/12$

   ›Reveal solutionSolution

   Chain the given conditional probabilities to find $P(Q)$ and $P(P\cap Q)$, then use the complement rule — the answer is (A) $4/5$.

   Concept and Intuition

   Conditional probability definitions let us cross-multiply to recover the joint probability $P(P\cap Q)$ from either conditional. Once $P(P), P(Q), P(P\cap Q)$ are all known, De Morgan's law converts "neither event" into the complement of the union.

   Step-by-Step Solution

   1. Given: $P(P)=\tfrac14$, $P(P|Q)=\tfrac12$, $P(Q|P)=\tfrac13$.
   2. $P(P\cap Q) = P(Q|P)\cdot P(P) = \tfrac13\times\tfrac14 = \tfrac1{12}$.
   3. Also $P(P\cap Q) = P(P|Q)\cdot P(Q) \Rightarrow \tfrac1{12} = \tfrac12\cdot P(Q) \Rightarrow P(Q)=\tfrac16$.
   4. $P(P\cup Q) = P(P)+P(Q)-P(P\cap Q) = \tfrac14+\tfrac16-\tfrac1{12} = \tfrac{3}{12}+\tfrac{2}{12}-\tfrac{1}{12}=\tfrac{4}{12}=\tfrac13$.
   5. By De Morgan's law, "neither P nor Q applies" is the complement of $P\cup Q$: $P(\overline P\cap\overline Q) = 1-\tfrac13=\tfrac23$.
   6. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$. …

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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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