Q.A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Showing the 12 most recent of 69 on this concept.

1. 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: $P(E\mid H_2)=\dfrac{\binom{13}{2}}{\binom{51}{2}}=\dfrac{78}{1275}$. …

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2. 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 $P(\text{2 spades drawn}\mid M')=\dfrac{\binom{13}{2}}{\binom{51}{2}}=\dfrac{78}{1275}$. …

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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). …

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4. 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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5. 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. …

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6. 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. $P(\text{no face card})=\dfrac{190}{325}=\dfrac{38}{65}$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.A bag contains 5 balls of unknown colours. There are equal chances that out of these five balls, there may be 0 or 1 or 2 or 3 or 4 or 5 red balls. A ball is taken out from the bag at random and is found to be red. The probability that it is the only red ball in the bag is (A) $\dfrac{1}{5}$ (B) $\dfrac{1}{6}$ (C) $\dfrac{1}{15}$ (D) $\dfrac{1}{30}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem over the six equally likely compositions of red balls; the posterior probability that exactly one ball is red, given a red ball was drawn, is $1/15$.

   Concept and Intuition

   Bayes' theorem updates the prior (uniform belief over how many red balls there are) using the evidence (a red ball was drawn) — compositions with more red balls make drawing red more likely, so they get more posterior weight, but we want specifically the $R=1$ case.

   Step-by-Step Solution

   1. Prior: $P(R=r)=\dfrac16$ for $r=0,1,2,3,4,5$.
   2. Likelihood of drawing red given $R=r$ red balls among 5: $P(\text{red}\mid R=r)=\dfrac{r}{5}$.
   3. Total probability of drawing red: $P(\text{red})=\sum_{r=0}^{5}\dfrac16\cdot\dfrac{r}{5}=\dfrac{1}{30}(0+1+2+3+4+5)=\dfrac{15}{30}=\dfrac12$. …

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8. 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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9. 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. $P(\text{1st black and 2nd black}) = \frac{6}{10}\cdot\frac{5}{9} = \frac{30}{90} = \frac{1}{3}$. …

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10. 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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11. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.A bag contains 6 balls. If three balls are drawn at a time and all of them are found to be green, then the probability that exactly 5 of the balls in the bag are green is: (A) $\dfrac{4}{35}$ (B) $\dfrac{5}{35}$ (C) $\dfrac{2}{7}$ (D) $\dfrac{1}{7}$

    ›Reveal solutionSolution

    This is a Bayes'-theorem problem: given a uniform prior on the unknown number of green balls in the bag, update the belief after observing 3 green balls drawn. Answer: $2/7$.

    Concept and Intuition

    Before drawing, any number of green balls from $0$ to $6$ is equally likely (7 equally likely hypotheses). Observing "all 3 drawn balls are green" is much more likely under hypotheses with more green balls, so Bayes' theorem reweights the prior toward higher $k$. We want the posterior probability that $k=5$.

    Step-by-Step Solution

    1. Prior: $P(k \text{ green balls})=\dfrac17$ for $k=0,1,\dots,6$.
    2. Likelihood of drawing 3 balls, all green, given $k$ green balls out of 6: $P(\text{all green}\mid k)=\dfrac{\binom{k}{3}}{\binom{6}{3}}=\dfrac{\binom{k}{3}}{20}$ (zero for $k<3$).
    3. Compute: $k=3:\binom33=1\Rightarrow1/20$; $k=4:\binom43=4\Rightarrow4/20$; $k=5:\binom53=10\Rightarrow10/20$; $k=6:\binom63=20\Rightarrow20/20=1$. …

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12. 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: …

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