Q.Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Two cards are drawn at random from a pack of 52 playing cards. If both the cards drawn are found to be black in colour, then the probability that atleast one of them is a face card is (A) $\dfrac{3}{13}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{9}{65}$ (D) $\dfrac{27}{65}$

   ›Reveal solutionSolution

   This is a conditional probability restricted to the 26 black cards. Using the complement (no face card among the two) is the fastest route, giving $27/65$.

   Concept and Intuition

   Once we're told both drawn cards are black, the sample space shrinks to just the 26 black cards (13 spades + 13 clubs). Among these, 6 are face cards (J, Q, K of spades and clubs) and 20 are non-face cards. "At least one face card" is easiest via the complement: $1-P(\text{no face card})$.

   Step-by-Step Solution

   1. Black cards $=26$; black face cards $=6$ (J,Q,K $\times$ 2 suits); black non-face cards $=20$.
   2. Total ways to pick 2 from the 26 black cards: $\binom{26}{2}=325$.
   3. Ways with no face card (both from the 20 non-face black cards): $\binom{20}{2}=190$.
   4. $P(\text{no face card})=\dfrac{190}{325}=\dfrac{38}{65}$.
   5. $P(\text{at least one face card})=1-\dfrac{38}{65}=\dfrac{27}{65}$.

   Common Mistakes

   • Forgetting to restrict the sample space to black cards only (using all 52 cards instead of 26) — the conditioning on "both black" changes the base.
   • Miscounting black face cards as 12 (all face cards) instead of 6 (only the black-suited ones).

   ✓Final answer

   The correct option is (D) — $\dfrac{27}{65}$.

   ANSWER: D

2. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Two cards are drawn from a pack of 52 playing cards one after the other without replacement. If the first card drawn is a queen, then the probability of getting a face card from a black suit in the second draw is (A) $\dfrac{11}{663}$ (B) $\dfrac{11}{1326}$ (C) $\dfrac{11}{312}$ (D) $\dfrac{11}{156}$

   ›Reveal solutionSolution

   This is a sequential (without-replacement) draw problem: count ordered pairs where the first card is a queen and the second is a black-suit face card, being careful that a black queen is itself both a queen and a black face card.

   Concept and Intuition

   When two cards are drawn one after another without replacement, we can count favourable ordered outcomes out of all $52\times51$ equally-likely ordered outcomes. The subtlety here is that black-suit face cards (J, Q, K of spades and clubs — six cards total) overlap with queens: the queen of spades and queen of clubs are both "a queen" and "a black-suit face card." So the count of eligible second-draw cards depends on whether the queen drawn first was itself black or red.

   Step-by-Step Solution

   1. There are $4$ queens total and $6$ black-suit face cards (spades J, Q, K and clubs J, Q, K).
   2. If the first card drawn is a black queen (spade Q or club Q — $2$ choices), that card is removed from the black-face-card pool too, leaving $6-1=5$ black face cards for the second draw. Contribution: $2\times5=10$ ordered pairs.
   3. If the first card drawn is a red queen (heart Q or diamond Q — $2$ choices), the black-face-card pool is untouched at $6$. Contribution: $2\times6=12$ ordered pairs.
   4. Total favourable ordered pairs: $10+12=22$.
   5. Total ordered pairs of two distinct cards from $52$: $52\times51=2652$.
   6. Probability $=\dfrac{22}{2652}=\dfrac{11}{1326}$ (dividing numerator and denominator by $2$).

   Common Mistakes

   • Treating the six black face cards as unaffected by which queen is drawn first, ignoring that two of them (the black queens) coincide with the "queen" event.
   • Forgetting to reduce the fraction, or mis-simplifying $22/2652$.

   ✓Final answer

   The correct option is (B) — $\dfrac{11}{1326}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If two cards are drawn at a time at random from a well shuffled pack of 52 playing cards and they are found to be a king card and a card with prime number, then the probability that they are a black king card and a card with an odd prime number is (A) $\dfrac{32}{663}$ (B) $\dfrac{12}{663}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{5}{8}$

   ›Reveal solutionSolution

   A conditional-probability counting problem: restrict the sample space to (King, prime-numbered-card) pairs, then count how many of those pairs are (black King, odd-prime card).

   Concept and Intuition

   A standard deck has 4 suits (2 black: spades, clubs; 2 red: hearts, diamonds), each with 13 ranks: A, 2–10, J, Q, K. "Cards with a prime number" means cards whose rank value is a prime, i.e. rank 2, 3, 5, or 7 — four ranks × 4 suits = 16 cards. Kings are a separate rank (not a "number" card), 4 total, 2 of them black (spade, club). Since we are told the two drawn cards are exactly one King and one prime-numbered card, we treat every (King, prime-card) pairing as equally likely and count favourable outcomes among them.

   Step-by-Step Solution

   1. Total King cards = 4; total prime-numbered cards (ranks 2,3,5,7) = $4\times4=16$.
   2. Sample space size (ways to have one King and one prime card) = $4\times16=64$.
   3. "Odd prime number" cards are ranks 3, 5, 7 (2 is the only even prime, excluded): $3\times4=12$ cards.
   4. Black Kings = 2 (spade King, club King).
   5. Favourable outcomes = (black King) × (odd-prime card) = $2\times12=24$.
   6. Required probability $=\dfrac{24}{64}=\dfrac{3}{8}$.

   Common Mistakes

   • Including the Ace as a "number 1" and mistakenly treating 1 as prime — 1 is not prime and Ace is not a numbered card in this context.
   • Including 2 among the "odd primes" — 2 is prime but even, so it is excluded from "odd prime."
   • Forgetting only 2 of the 4 Kings are black.

   ✓Final answer

   The correct option is (C) — $\dfrac{3}{8}$.

   ANSWER: C

4. 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

5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.From a pack of 52 playing cards, one card was found missing. From the remaining cards, two cards are drawn at random and found to be spade cards. The probability that the missing card is a spade card is (A) $\dfrac{39}{50}$ (B) $\dfrac{27}{51}$ (C) $\dfrac{11}{50}$ (D) $\dfrac{11}{100}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem: update the prior probability that the missing card is a spade (1/4) using the evidence that two randomly drawn cards from the remaining 51 both turned out to be spades. The posterior is $\dfrac{11}{50}$.

   Concept and Intuition

   Before drawing, the missing card is spade with prior probability $13/52=1/4$ and not-spade with probability $3/4$. Observing two spades drawn is more likely if the missing card is NOT a spade (since more spades remain in the deck in that case), so we must weight by how likely the observed evidence is under each hypothesis — this is exactly Bayes' theorem.

   Step-by-Step Solution

   1. Let $H_1$: missing card is a spade ($P(H_1)=13/52=1/4$); $H_2$: missing card is not a spade ($P(H_2)=39/52=3/4$).
   2. If $H_1$ holds, 12 spades remain among the 51 cards: $P(E\mid H_1)=\dfrac{\binom{12}{2}}{\binom{51}{2}}=\dfrac{66}{1275}$.
   3. If $H_2$ holds, 13 spades remain among the 51 cards: $P(E\mid H_2)=\dfrac{\binom{13}{2}}{\binom{51}{2}}=\dfrac{78}{1275}$.
   4. By Bayes' theorem: $P(H_1\mid E)=\dfrac{P(H_1)P(E\mid H_1)}{P(H_1)P(E\mid H_1)+P(H_2)P(E\mid H_2)}=\dfrac{\tfrac14\cdot66}{\tfrac14\cdot66+\tfrac34\cdot78}$.
   5. $=\dfrac{66}{66+234}=\dfrac{66}{300}=\dfrac{11}{50}$.

   Common Mistakes

   • Ignoring the prior and just answering $1/4$, or ignoring the evidence's differential likelihood.
   • Forgetting the common factor $\binom{51}{2}$ cancels, so it doesn't even need to be computed.

   ✓Final answer

   The correct option is (C) — $\dfrac{11}{50}$.

   ANSWER: C

6. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.One card is missing in a pack of 52 playing cards. If two cards are drawn randomly from the remaining cards at a time and are found to be spades, then the probability that the missing card is not a spade is (A) $\dfrac{3}{50}$ (B) $\dfrac{39}{50}$ (C) $\dfrac{39}{52}$ (D) $\dfrac{38}{52}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem: use the prior probability the missing card is/isn't a spade together with the likelihood of drawing two spades in each case.

   Concept and Intuition

   Before any draw, $P(\text{missing is spade})=\frac14$ and $P(\text{missing is not spade})=\frac34$. After observing "two cards drawn are both spades," Bayes' theorem updates these priors using how likely that observation is under each scenario (fewer spades left if the missing card was a spade).

   Step-by-Step Solution

   1. Let $M$: missing card is a spade ($P(M)=13/52=1/4$); $M'$: missing card is not a spade ($P(M')=39/52=3/4$).
   2. If $M$: 12 spades remain among 51 cards, so $P(\text{2 spades drawn}\mid M)=\dfrac{\binom{12}{2}}{\binom{51}{2}}=\dfrac{66}{1275}$.
   3. If $M'$: 13 spades remain among 51 cards, so $P(\text{2 spades drawn}\mid M')=\dfrac{\binom{13}{2}}{\binom{51}{2}}=\dfrac{78}{1275}$.
   4. By Bayes' theorem: $P(M'\mid E)=\dfrac{P(E|M')P(M')}{P(E|M')P(M')+P(E|M)P(M)}=\dfrac{78\cdot\frac34}{78\cdot\frac34+66\cdot\frac14}=\dfrac{58.5}{58.5+16.5}=\dfrac{58.5}{75}$.
   5. Simplify: $\dfrac{58.5}{75}=\dfrac{234}{300}=\dfrac{39}{50}$.

   Common Mistakes

   • Forgetting to weight the likelihoods by the prior probabilities $P(M),P(M')$ (i.e. just comparing $\binom{12}{2}$ vs $\binom{13}{2}$ directly).

   ✓Final answer

   The correct option is (B) — $\dfrac{39}{50}$.

   ANSWER: B

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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