Q.Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

Showing the 12 most recent of 69 on this concept.

1. 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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2. 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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3. 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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4. 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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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: $P(E\mid H_2)=\dfrac{\binom{13}{2}}{\binom{51}{2}}=\dfrac{78}{1275}$. …

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

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7. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A die is thrown three times. If the sum of the numbers thrown is 15, then the probability that the first throw was a Four, is (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{5}{108}$ (D) $\dfrac{1}{108}$

   ›Reveal solutionSolution

   A conditional probability found by directly counting favourable die-triples against all triples summing to the given total.

   Concept and Intuition

   $P(\text{first}=4\mid\text{sum}=15)=\dfrac{\#\{\text{triples with first}=4,\ \text{sum}=15\}}{\#\{\text{triples with sum}=15\}}$ — a straightforward application of conditional probability by counting, since all $6^3$ triples are equally likely.

   Step-by-Step Solution

   1. If the first throw is $4$, the other two throws (each from $1$ to $6$) must sum to $15-4=11$.
   2. Pairs of dice summing to $11$: $(5,6)$ and $(6,5)$ — 2 ways.
   3. Now count all triples $(a,b,c)$, each in $1$–$6$, with $a+b+c=15$. Substitute $a'=6-a,\,b'=6-b,\,c'=6-c$ (each in $0$–$5$): then $a'+b'+c'=18-15=3$.
   4. Number of non-negative integer solutions to $a'+b'+c'=3$ is $\binom{3+2}{2}=10$; since $3<5$ none violate the upper bound of $5$, so all $10$ are valid.
   5. So there are $10$ triples with sum $15$ in total. …

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8. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.A bag contains 2 white, 3 green and 5 red balls. If three balls are drawn one after the other without replacement, then the probability that the last ball drawn was red is (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{5}{9}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   By the symmetry of random sampling without replacement, the probability that the last ball drawn is red equals the overall fraction of red balls in the bag: $\dfrac{5}{10}=\dfrac12$.

   Concept and Intuition

   When balls are drawn one after another without replacement, every ball is equally likely to occupy any given position in the drawing order (all $10!$ orderings of the bag's balls are equally likely). Hence the marginal probability that the ball in any fixed position (first, second, ..., last) is red is just (number of red balls)/(total balls), independent of which position we pick.

   Step-by-Step Solution

   1. Bag: $2$ white $+3$ green $+5$ red $=10$ balls total.
   2. Consider a full random permutation of all 10 balls (drawing three is just looking at the first three positions of such a permutation, but the argument works for any position).
   3. By symmetry, $P(\text{ball in position }k\text{ is red})=\dfrac{5}{10}$ for every position $k$, including the third (last) position drawn here. …

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9. 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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10. 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. $P(\text{white})=\frac12\cdot\frac23+\frac12\cdot\frac37=\frac13+\frac{3}{14}$. …

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11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.For a biased die, the probabilities for different faces to turn up are given below

    Face	1	2	3	4	5	6
    Probability	0.1	0.32	0.21	0.15	0.05	0.17

    The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is (A) $\dfrac{10}{33}$ (B) $\dfrac{5}{21}$ (C) $\dfrac{8}{21}$ (D) $\dfrac{1}{42}$

    ›Reveal solutionSolution

    Conditioning on "face 1 or face 2" just means renormalizing the two individual probabilities so they add to 1.

    Concept and Intuition

    $P(\text{face 1}\mid\text{face 1 or 2})=\dfrac{P(\text{face 1})}{P(\text{face 1})+P(\text{face 2})}$, since these two events are mutually exclusive and their union is the conditioning event.

    Step-by-Step Solution

    1. $P(1)=0.1$, $P(2)=0.32$.
    2. $P(1\text{ or }2)=0.1+0.32=0.42$.
    3. $P(1\mid1\text{ or }2)=\dfrac{0.1}{0.42}=\dfrac{10}{42}=\dfrac{5}{21}$.

    Common Mistakes …

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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.2 aero planes I and II bond a target in succession. The probabilities of I and II scoring a hit correctly is 0.3 and 0.2 respectively. The second plane will bomb only if first misses the target. The probability that the target is hit by the $2^{nd}$ plane is (A) 0.06 (B) 0.14 (C) 0.32 (D) 0.7

    ›Reveal solutionSolution

    The 2nd plane gets a chance only after the 1st fails, so $P = 0.7\times 0.2 = 0.14$. Answer: (B).

    Concept and Intuition

    This is a sequential (conditional) experiment: the second trial happens only when the first fails. The event 'the target is hit by the 2nd plane' is therefore a compound event —

    $\{\text{I misses}\}\cap\{\text{II hits}\}$

    and because the two planes' performances are independent, the probability of the intersection is the product of the probabilities.

    A useful picture is a probability tree:

             ┌── I hits (0.3)  ────────────────► target hit by plane I  (0.3)
    Start ───┤
             └── I misses (0.7) ─┬── II hits (0.2) ──► hit by plane II  (0.7 × 0.2 = 0.14)
                                 └── II misses (0.8) ► target not hit   (0.7 × 0.8 = 0.56)
    

    The three leaves sum to $0.3+0.14+0.56=1$ ✓ — a good check that the model is complete.

    Step-by-Step Solution

    1. Let $H_1$ = plane I hits, with $P(H_1)=0.3$, so $P(\overline{H_1})=1-0.3=0.7$.
    2. Let $H_2$ = plane II hits (given it bombs), with $P(H_2)=0.2$.
    3. Plane II bombs only if plane I missed. Hence

    $P(\text{target hit by 2nd plane}) = P(\overline{H_1}\cap H_2) = P(\overline{H_1})\,P(H_2)$

    4. Substitute: $= 0.7\times 0.2 = 0.14$ …

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