Showing the 12 most recent of 69 on this concept.

1. 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: $P(\text{Red}) = \dfrac13\left(\dfrac25+\dfrac49+\dfrac29\right) = \dfrac13\left(\dfrac25+\dfrac23\right) = \dfrac13\cdot\dfrac{16}{15}=\dfrac{16}{45}$.
   5. Bayes: $P(B_2\mid \text{Red}) = \dfrac{P(B_2)P(\text{Red}\mid B_2)}{P(\text{Red})} = \dfrac{\frac13\cdot\frac49}{\frac{16}{45}} = \dfrac{\frac{4}{27}}{\frac{16}{45}} = \dfrac{4}{27}\times\dfrac{45}{16}=\dfrac{5}{12}$.

   Common Mistakes

   • Forgetting to weight each box's conditional probability by its prior $1/3$.
   • Arithmetic slip when adding the three fractions with different denominators (5 and 9).

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{12}$.

   ANSWER: B

2. 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

3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Two balls are drawn at random from a box containing 4 white, 6 black balls one after the other without replacement. If it is known that second ball drawn is black, then the probability that the first ball drawn is also black is (A) $\dfrac{5}{11}$ (B) $\dfrac{5}{9}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{5}{13}$

   ›Reveal solutionSolution

   Conditional probability with sampling without replacement, solved with Bayes' theorem; answer is $\frac{5}{9}$.

   Concept and Intuition

   When balls are drawn one after another without replacement, the marginal probability that any particular draw (say the 2nd) is black equals the overall proportion of black balls, $\frac{6}{10}$ — position doesn't matter for the marginal event by symmetry. To find the conditional probability of the first draw given information about the second, we use Bayes' theorem: we need the joint probability of both events and divide by the marginal probability of the conditioning event.

   Step-by-Step Solution

   1. Total balls: 4 white (W) + 6 black (B) = 10.
   2. $P(\text{2nd is black}) = P(\text{1st B, 2nd B}) + P(\text{1st W, 2nd B})$ $= \frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{6}{9} = \frac{30}{90}+\frac{24}{90} = \frac{54}{90} = \frac{3}{5}$.
   3. $P(\text{1st black and 2nd black}) = \frac{6}{10}\cdot\frac{5}{9} = \frac{30}{90} = \frac{1}{3}$.
   4. By Bayes' theorem: $P(\text{1st black}\mid \text{2nd black}) = \dfrac{P(\text{1st B, 2nd B})}{P(\text{2nd B})} = \dfrac{1/3}{3/5} = \dfrac{5}{9}$.

   Common Mistakes

   • Assuming $P(\text{2nd black}) = \frac{5}{9}$ (the conditional probability given 1st was black) instead of computing the correct marginal $\frac{3}{5}$.
   • Forgetting to include both cases (1st W/2nd B and 1st B/2nd B) when finding the joint denominator event.

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{9}$.

   ANSWER: B

4. 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. $P(\text{white})=\frac12\cdot\frac23+\frac12\cdot\frac37=\frac13+\frac{3}{14}$.
   4. Common denominator 42: $\frac13=\frac{14}{42}$, $\frac{3}{14}=\frac{9}{42}$, sum $=\frac{23}{42}$.

   Common Mistakes

   • Simplifying $\frac46$ incorrectly or forgetting to weight by $\frac12$ for each bag.
   • Adding numerators/denominators directly instead of finding a common denominator (a classic fraction-addition slip).

   ✓Final answer

   The correct option is (D) — $\dfrac{23}{42}$.

   ANSWER: D

5. 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: $P(W) = \frac13\left(\dfrac37+\dfrac47+\dfrac27\right) = \frac13\cdot\dfrac{9}{7} = \dfrac{9}{21} = \dfrac{3}{7}$.
   6. By Bayes' theorem: $P(C\mid W) = \dfrac{\frac13\cdot\frac27}{\frac37} = \dfrac{\frac{2}{21}}{\frac37} = \dfrac{2}{21}\times\dfrac{7}{3} = \dfrac{14}{63} = \dfrac{2}{9}$.

   Common Mistakes

   • Forgetting to divide by the total probability $P(W)$ and stopping at the numerator $\frac{2}{21}$.
   • Mixing up which bag's white-ball fraction goes with which term.

   ✓Final answer

   The correct option is (B) — $\dfrac{2}{9}$.

   ANSWER: B

6. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red is (A) $\frac{39}{70}$ (B) $\frac{41}{70}$ (C) $\frac{29}{70}$ (D) $\frac{17}{35}$

   ›Reveal solutionSolution

   Total probability theorem over the two equally-likely bags gives $\frac{17}{35}$.

   Concept and Intuition

   Since the bag is chosen at random (each with probability $\frac12$), the overall probability of drawing red is the weighted average of the conditional probabilities of drawing red from each bag.

   Step-by-Step Solution

   1. Bag 1: 4 red, 3 black (7 total) → $P(\text{red}|\text{Bag1})=\dfrac{4}{7}$.
   2. Bag 2: 2 red, 3 black (5 total) → $P(\text{red}|\text{Bag2})=\dfrac{2}{5}$.
   3. $P(\text{red}) = \dfrac12\cdot\dfrac47+\dfrac12\cdot\dfrac25 = \dfrac{4}{14}+\dfrac{2}{10}=\dfrac{2}{7}+\dfrac15$.
   4. Common denominator 35: $\dfrac{10}{35}+\dfrac{7}{35}=\dfrac{17}{35}$.

   Common Mistakes

   • Averaging the counts of red balls across bags instead of averaging the conditional probabilities.

   ✓Final answer

   The correct option is (D) — $\dfrac{17}{35}$.

   ANSWER: D

7. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Two urns identical in appearance contain respectively 3 green and 2 black balls and 2 green and 5 black balls. One urn is selected at random and a ball is drawn from it. The probability that it is black is ________ (A) $\dfrac{39}{70}$ (B) $\dfrac{37}{70}$ (C) $\dfrac{41}{70}$ (D) $\dfrac{33}{70}$

   ›Reveal solutionSolution

   Apply the law of total probability over the two equally-likely urns to get $P(\text{black}) = \dfrac{39}{70}$.

   Concept and Intuition

   When an experiment first randomly selects between two scenarios (here, two urns, each equally likely) and then performs a further random step (drawing a ball), the overall probability of an outcome is the weighted average of the outcome's probability under each scenario, weighted by the scenario's own probability.

   Step-by-Step Solution

   1. Urn 1 has $3$ green $+ 2$ black $=5$ balls, so $P(\text{black}\mid\text{Urn 1}) = \dfrac{2}{5}$.
   2. Urn 2 has $2$ green $+ 5$ black $=7$ balls, so $P(\text{black}\mid\text{Urn 2}) = \dfrac{5}{7}$.
   3. Each urn is chosen with probability $\dfrac12$.
   4. Total probability: $P(\text{black}) = \dfrac12\cdot\dfrac{2}{5} + \dfrac12\cdot\dfrac{5}{7} = \dfrac12\left(\dfrac{14}{35}+\dfrac{25}{35}\right) = \dfrac12\cdot\dfrac{39}{35} = \dfrac{39}{70}$.

   Common Mistakes

   • Adding the balls across urns as if drawing from one combined urn of 12 balls (that ignores the two-stage random selection of urns).
   • Arithmetic slip finding a common denominator for $2/5$ and $5/7$ (it's $35$).

   ✓Final answer

   The correct option is (A) — $\dfrac{39}{70}$.

   ANSWER: A

8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.An urn A contains 4 white and 1 black ball; urn B contains 3 white and 2 black balls and urn C contains 2 white and 3 black balls. One ball is transferred randomly from A to B; later one ball is transferred randomly from B to C. Finally, if a ball is drawn randomly from C, then the probability that it is a black ball is (A) $\dfrac{7}{12}$ (B) $\dfrac{89}{180}$ (C) $\dfrac{101}{180}$ (D) $\dfrac{17}{36}$

   ›Reveal solutionSolution

   A two-stage transfer-then-draw problem, solved by branching over all four possible transfer outcomes: probability of black $=\dfrac{101}{180}$.

   Concept and Intuition

   This is a sequential conditional-probability problem: each transfer changes the composition of the receiving urn, so we must branch over every possible outcome of each transfer and weight the final draw accordingly (total probability theorem, applied twice).

   Step-by-Step Solution

   1. Urn A: 4W,1B. Transfer to B: $P(\text{W})=\frac45$, $P(\text{B})=\frac15$.
   2. If white moved to B: B becomes 4W,2B (6 balls). If black moved to B: B becomes 3W,3B (6 balls).
   3. From B (4W,2B): transfer white to C with $P=\frac{4}{6}=\frac23$ (C becomes 3W,3B), or black with $P=\frac26=\frac13$ (C becomes 2W,4B).
   4. From B (3W,3B): transfer white to C with $P=\frac12$ (C becomes 3W,3B), or black with $P=\frac12$ (C becomes 2W,4B).
   5. Final draw from C: $P(\text{black}\mid 3W3B)=\frac12$; $P(\text{black}\mid 2W4B)=\frac{4}{6}=\frac23$.
   6. Combine all four branches:
      • A-white, B-white: $\frac45\cdot\frac23\cdot\frac12=\frac{8}{30}=\frac{4}{15}$
      • A-white, B-black: $\frac45\cdot\frac13\cdot\frac23=\frac{8}{45}$
      • A-black, B-white: $\frac15\cdot\frac12\cdot\frac12=\frac1{20}$
      • A-black, B-black: $\frac15\cdot\frac12\cdot\frac23=\frac1{15}$
   7. Convert to a common denominator (180): $\frac{48}{180}+\frac{32}{180}+\frac{9}{180}+\frac{12}{180}=\frac{101}{180}$.

   Common Mistakes

   • Forgetting that after each transfer the receiving urn has one MORE ball (6, not 5), which changes every subsequent probability.
   • Missing one of the four branches or mismatching which composition of C follows from which transfer outcome.

   ✓Final answer

   The correct option is (C) — $\dfrac{101}{180}$.

   ANSWER: C

9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A box P contains 3 white and 7 red balls. A bag Q contains 4 green and 5 blue balls. Two balls are randomly drawn from box P. If both are of same color, one ball is drawn from bag Q and if the two balls are of different color, 2 balls are drawn from the bag Q. If it is known that there is exactly one green ball among the ball or balls drawn from bag Q, then the probability that the two balls drawn from box P are of different colors is (A) $\dfrac{35}{67}$ (B) $\dfrac{21}{62}$ (C) $\dfrac{20}{43}$ (D) $\dfrac{32}{67}$

   ›Reveal solutionSolution

   A two-stage Bayes' theorem problem; conditioning on "exactly one green ball from Q" gives $P(\text{different colors from P}) = \frac{35}{67}$.

   Concept and Intuition

   This is a compound experiment: the outcome in box P (same-color vs different-color) determines how many balls are drawn from bag Q, and hence changes the probability model for "exactly one green." We must compute, for each P-outcome, the probability of observing exactly one green from Q, then combine via Bayes' theorem using the P-outcome's prior probability.

   Step-by-Step Solution

   1. Box P has 3 white + 7 red = 10 balls. $P(\text{same color}) = \dfrac{{}^3C_2+{}^7C_2}{{}^{10}C_2} = \dfrac{3+21}{45}=\dfrac{24}{45}=\dfrac{8}{15}$. $P(\text{different colors}) = 1-\dfrac{8}{15}=\dfrac{7}{15}$ (check: $\dfrac{{}^3C_1{}^7C_1}{{}^{10}C_2}=\dfrac{21}{45}=\dfrac{7}{15}$ ✓).
   2. If same color (S): draw 1 ball from Q (4 green, 5 blue, 9 total). "Exactly one green" among 1 ball drawn just means that ball is green: $P(\text{1 green}\mid S)=\dfrac{4}{9}$.
   3. If different colors (D): draw 2 balls from Q. $P(\text{exactly one green}\mid D) = \dfrac{{}^4C_1\cdot{}^5C_1}{{}^9C_2} = \dfrac{20}{36}=\dfrac{5}{9}$.
   4. By Bayes' theorem:

   $P(D\mid \text{one green}) = \dfrac{P(D)P(\text{one green}\mid D)}{P(S)P(\text{one green}\mid S)+P(D)P(\text{one green}\mid D)} = \dfrac{\frac{7}{15}\cdot\frac{5}{9}}{\frac{8}{15}\cdot\frac{4}{9}+\frac{7}{15}\cdot\frac{5}{9}}$

   5. Numerator $=\dfrac{35}{135}$; denominator $=\dfrac{32}{135}+\dfrac{35}{135}=\dfrac{67}{135}$. So $P(D\mid\text{one green}) = \dfrac{35}{67}$.

   Common Mistakes

   • Using the same probability model for "one green" regardless of whether 1 or 2 balls were drawn from Q — the two cases have genuinely different sample spaces.
   • Forgetting to weight each conditional probability by the correct prior $P(S)$ or $P(D)$ before combining.

   ✓Final answer

   The correct option is (A) — $\dfrac{35}{67}$.

   ANSWER: A

10. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.A bag 'A' contains 2 black and 3 white balls. Another bag 'B' contains 3 black and 2 white balls. Two balls are drawn randomly from 'A' and placed in 'B'. Later, if two balls are drawn randomly from 'B', then the probability of getting a black ball and a white ball from it is (A) $\dfrac{59}{105}$ (B) $\dfrac{46}{105}$ (C) $\dfrac{59}{210}$ (D) $\dfrac{67}{210}$

    ›Reveal solutionSolution

    A two-stage random-transfer problem — condition on what got transferred from A to B, then apply the law of total probability.

    Concept and Intuition

    Since the composition of bag B after the transfer depends on which 2 balls were drawn from A, we must split into the three possible transfer outcomes (BB, BW, WW), compute the probability of each, then the conditional probability of drawing one black and one white ball from the resulting bag B, and combine via the law of total probability.

    Step-by-Step Solution

    1. Bag A has 5 balls (2B, 3W); ways to choose 2: $\binom{5}{2}=10$.
       • $P(\text{BB from A}) = \binom{2}{2}/\binom{5}{2} = 1/10$.
       • $P(\text{BW from A}) = \binom{2}{1}\binom{3}{1}/\binom{5}{2} = 6/10 = 3/5$.
       • $P(\text{WW from A}) = \binom{3}{2}/\binom{5}{2} = 3/10$.
    2. Bag B originally has 3B, 2W (5 balls); after adding 2 balls it has 7 balls, and we want $P(1B,1W)$ drawn from it: $\binom{7}{2}=21$ ways.
       • If BB added: B = 5B, 2W. $P(1B,1W) = \binom{5}{1}\binom{2}{1}/21 = 10/21$.
       • If BW added: B = 4B, 3W. $P(1B,1W) = \binom{4}{1}\binom{3}{1}/21 = 12/21$.
       • If WW added: B = 3B, 4W. $P(1B,1W) = \binom{3}{1}\binom{4}{1}/21 = 12/21$.
    3. Total probability:

    $P = \frac{1}{10}\cdot\frac{10}{21} + \frac{3}{5}\cdot\frac{12}{21} + \frac{3}{10}\cdot\frac{12}{21}$

    4. Convert each term to a denominator of 210: $\frac{10}{210} + \frac{72}{210} + \frac{36}{210} = \frac{118}{210} = \frac{59}{105}$.

    Common Mistakes

    • Forgetting that bag B's composition changes with the transfer outcome, and just using the original bag B composition.
    • Arithmetic slip when combining fractions with different denominators — always reduce to a common denominator before adding.

    ✓Final answer

    The correct option is (A) — $\dfrac{59}{105}$.

    ANSWER: A

11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it randomly. If the number on the card is found to be a non-prime number, the probability that the card was drawn from Box I is (A) $\dfrac{4}{17}$ (B) $\dfrac{8}{17}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{2}{3}$

    ›Reveal solutionSolution

    Bayes' theorem with the counts of non-prime numbers in each box gives $P(\text{Box I}\mid\text{non-prime})=8/17$.

    Concept and Intuition

    This is a classic "which urn/box did it come from" Bayes' problem: we're given the outcome (a non-prime card) and asked to find the probability of the cause (which box). We need $P(\text{non-prime}\mid\text{box})$ for each box, weighted by the prior $P(\text{box})=1/2$.

    Step-by-Step Solution

    1. Primes from 1 to 30: $2,3,5,7,11,13,17,19,23,29$ — that's $10$ primes, so $30-10=20$ non-primes in Box I.
    2. Primes from 31 to 50: $31,37,41,43,47$ — that's $5$ primes, so $20-5=15$ non-primes in Box II.
    3. $P(\text{non-prime}\mid\text{Box I})=\dfrac{20}{30}=\dfrac23$, $P(\text{non-prime}\mid\text{Box II})=\dfrac{15}{20}=\dfrac34$.
    4. $P(\text{Box I})=P(\text{Box II})=\dfrac12$.
    5. $P(\text{Box I and non-prime})=\dfrac12\cdot\dfrac23=\dfrac13$; $P(\text{Box II and non-prime})=\dfrac12\cdot\dfrac34=\dfrac38$.
    6. $P(\text{non-prime})=\dfrac13+\dfrac38=\dfrac{8}{24}+\dfrac{9}{24}=\dfrac{17}{24}$.
    7. $P(\text{Box I}\mid\text{non-prime})=\dfrac{1/3}{17/24}=\dfrac13\cdot\dfrac{24}{17}=\dfrac{8}{17}$.

    Common Mistakes

    • Forgetting that $1$ is not prime (it's non-prime), which would miscount Box I's non-primes.
    • Skipping the proper Bayes' normalization and just comparing raw counts across boxes of different sizes.

    ✓Final answer

    The correct option is (B) — $\dfrac{8}{17}$.

    ANSWER: B

12. 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