Q.In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?

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-2025-05-24-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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3. 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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4. 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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5. 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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6. 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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7. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.Three boxes $B_1$, $B_2$ and $B_3$ contain balls with different colors as follows:

   	White	Black	Red
   $B_1$	2	1	2
   $B_2$	3	2	4
   $B_3$	4	3	2

   A die is thrown. Box $B_1$ is chosen if either 1 or 2 turns up. Box $B_2$ is chosen if 3 or 4 turns up and box $B_3$ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is drawn at random from that box. If the ball drawn is found to be Red, then the probability that it is drawn from box $B_2$ is (A) $\dfrac{7}{12}$ (B) $\dfrac{5}{12}$ (C) $\dfrac{1}{12}$ (D) $\dfrac{3}{26}$

   ›Reveal solutionSolution

   A Bayes'-theorem problem: reverse the conditional probability "ball is Red, which box?" using the total-probability formula.

   Concept and Intuition

   Bayes' theorem lets us "invert" a conditional probability. We know $P(\text{Red}\mid B_i)$ for each box and the prior $P(B_i)$ (each $1/3$ since the die is fair and each box corresponds to two faces); we want the posterior $P(B_2 \mid \text{Red})$.

   Step-by-Step Solution

   1. Totals in each box: $B_1$: $2+1+2=5$; $B_2$: $3+2+4=9$; $B_3$: $4+3+2=9$.
   2. $P(\text{Red}\mid B_1)=\dfrac{2}{5}$, $P(\text{Red}\mid B_2)=\dfrac{4}{9}$, $P(\text{Red}\mid B_3)=\dfrac{2}{9}$.
   3. Each box has prior probability $\dfrac13$ (each box corresponds to two die faces out of six).
   4. Total probability: $P(\text{Red}) = \dfrac13\left(\dfrac25+\dfrac49+\dfrac29\right) = \dfrac13\left(\dfrac25+\dfrac23\right) = \dfrac13\cdot\dfrac{16}{15}=\dfrac{16}{45}$. …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.On every evening, a student either watches TV or reads a book. The probability of watching TV is $\dfrac{4}{5}$. If he watches TV, the probability that he will fall asleep is $\dfrac{3}{4}$ and it is $\dfrac{1}{4}$ when he reads a book. If the student is found to be asleep on an evening, the probability that he watched the TV is (A) $\dfrac{11}{13}$ (B) $\dfrac{12}{13}$ (C) $\dfrac{2}{13}$ (D) $\dfrac{4}{13}$

   ›Reveal solutionSolution

   This tests Bayes' theorem: finding the probability of a cause (watched TV) given an observed effect (fell asleep). The answer is $\dfrac{12}{13}$.

   Concept and Intuition

   When asked "given the observed outcome, what's the probability of a particular cause," that's a Bayes'-theorem setup: compute the joint probability of that cause with the outcome, and divide by the total probability of the outcome (summed over all causes).

   Step-by-Step Solution

   1. $P(\text{TV})=\tfrac45$, $P(\text{book})=\tfrac15$.
   2. $P(\text{asleep}\mid\text{TV})=\tfrac34$, $P(\text{asleep}\mid\text{book})=\tfrac14$.
   3. Total probability of falling asleep: $P(\text{asleep})=P(\text{TV})P(\text{asleep}\mid\text{TV})+P(\text{book})P(\text{asleep}\mid\text{book})=\tfrac45\cdot\tfrac34+\tfrac15\cdot\tfrac14=\tfrac35+\tfrac1{20}=\tfrac{12}{20}+\tfrac1{20}=\tfrac{13}{20}$. …

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10. 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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11. 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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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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