Q.While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards to be of different colours
(A) $\frac{29}{52}$
(B) $\frac{1}{2}$
(C) $\frac{26}{51}$
(D) $\frac{27}{51}$

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If six students, including two particular students A and B, stand in a row randomly, then the probability that they stand in such a way that A and B are separated by one student in between them, is (A) $\dfrac{2}{15}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{4}{15}$ (D) $\dfrac{8}{15}$

   ›Reveal solutionSolution

   Reduce a seating-arrangement probability to just the two special people's positions, since the other four students fill the remaining seats interchangeably regardless of the outcome.

   Concept and Intuition

   When a probability question only concerns the relative arrangement of two particular objects among $n$ objects placed in a row, we don't need to consider all $n!$ full arrangements — the other objects are symmetric and cancel out. It suffices to count ordered position-pairs for A and B among the 6 slots.

   Step-by-Step Solution

   1. Total ways to assign ordered positions to A and B among 6 slots: $6\times5=30$.
   2. "Separated by exactly one student" means the position numbers of A and B differ by exactly $2$ (one seat in between).
   3. Pairs of positions (unordered) with difference $2$ among $\{1,\dots,6\}$: $(1,3),(2,4),(3,5),(4,6)$ — that's $4$ pairs. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If a number $x$ is chosen at random from the set $\{1, 2, 3, \ldots, 100\}$, then the probability of getting that $x$ which satisfies $x + \dfrac{100}{x} > 29$ is (A) $0.76$ (B) $0.77$ (C) $0.78$ (D) $0.75$

   ›Reveal solutionSolution

   Convert the inequality into a quadratic, find its roots, and count how many integers from 1 to 100 lie outside the root interval.

   Concept and Intuition

   Multiplying through by the positive $x$ turns a rational inequality into a quadratic one, whose sign pattern (positive outside the roots, since the leading coefficient is positive) tells us exactly which integers satisfy the original condition.

   Step-by-Step Solution

   1. For $x\in\{1,\dots,100\}$, $x>0$, so multiplying $x+\dfrac{100}{x}>29$ by $x$ preserves the inequality direction: $x^2+100>29x \Rightarrow x^2-29x+100>0$.
   2. Solve $x^2-29x+100=0$: discriminant $=841-400=441=21^2$. Roots $=\dfrac{29\pm21}{2}=25$ or $4$.
   3. Since the parabola opens upward, $x^2-29x+100>0$ outside the roots: $x<4$ or $x>25$.
   4. Integers in $[1,100]$ with $x<4$: $1,2,3$ — 3 values. …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If b and c are numbers chosen at random from the set $\{1, 2, 3, \ldots, 10\}$ with replacement, then the probability that $x^2 + bx + c = 0$ has real roots is (A) $0.52$ (B) $0.54$ (C) $0.58$ (D) $0.62$

   ›Reveal solutionSolution

   Real roots of $x^2+bx+c=0$ require discriminant $b^2-4c\ge0$. Count favorable $(b,c)$ pairs out of the $100$ equally-likely pairs from $\{1,\dots,10\}^2$; the count is $62$, giving probability 0.62 — option (D).

   Concept and Intuition

   A quadratic $x^2+bx+c=0$ has real roots exactly when its discriminant is non-negative: $b^2-4c\ge0$, i.e. $c\le \dfrac{b^2}{4}$. Since $b,c$ are drawn independently and uniformly (with replacement) from $\{1,2,\dots,10\}$, there are $10\times10=100$ equally likely ordered pairs, and we just need to count how many satisfy $c\le b^2/4$ (also capped at $c\le10$ since $c$ can't exceed 10).

   Step-by-Step Solution

   1. Total outcomes: $10\times10=100$.
   2. For each value of $b$ from 1 to 10, count how many $c\in\{1,\dots,10\}$ satisfy $c\le \lfloor b^2/4\rfloor$ (and $c\ge1$):
      • $b=1$: $b^2/4=0.25\Rightarrow$ no valid $c\ge1$ → count $0$
      • $b=2$: $1\Rightarrow c=1$ → count $1$
      • $b=3$: $2.25\Rightarrow c\le2$ → count $2$
      • $b=4$: $4\Rightarrow c\le4$ → count $4$
      • $b=5$: $6.25\Rightarrow c\le6$ → count $6$
      • $b=6$: $9\Rightarrow c\le9$ → count $9$
      • $b=7$: $12.25$, capped at $10\Rightarrow$ count $10$ …

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Two persons A and B are alternately throwing two dice indefinitely. If A starts the game and the person who gets a prime number on one die and a composite number on the other for the first time wins the game, then the probability that B wins the game is (A) $\dfrac{3}{5}$ (B) $\dfrac{2}{5}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   This is a "race to succeed first" problem in alternating independent trials; find the single-trial winning probability, then sum the geometric series for the second player. Answer: $\dfrac25$.

   Concept and Intuition

   On each of the two dice, the outcomes 1–6 split into prime $\{2,3,5\}$, composite $\{4,6\}$, and neither (just $1$). The winning event per throw of two dice is "one die prime, the other composite" — its probability $p$ is fixed and the same for every throw. Since A and B alternate and A goes first, B only gets a chance to win on throw 2 (if A failed on throw 1), throw 4 (if both failed on throws 1–3), and so on — a geometric pattern where the common ratio is $q^2$ (both players failing one full round).

   Step-by-Step Solution

   1. On a single die, $P(\text{prime})=P(\{2,3,5\})=\frac36=\frac12$ and $P(\text{composite})=P(\{4,6\})=\frac26=\frac13$ (note: 1 is neither prime nor composite).
   2. Winning event per two-dice throw: (die1 prime AND die2 composite) OR (die1 composite AND die2 prime). $p=\frac12\cdot\frac13+\frac13\cdot\frac12=\frac16+\frac16=\frac13$.
   3. Failure probability per throw: $q=1-p=\frac23$.
   4. A throws first (throw 1), B throws second (throw 2), etc. B wins if: A fails throw 1 (prob $q$) and B succeeds throw 2 (prob $p$) — probability $qp$; or both A and B fail their first throws and A fails again (prob $q^3$) and B succeeds — probability $q^3p$; and so on. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If a number $x$ is to be chosen randomly from the set of numbers $\{1, 2, 3, \ldots 30\}$, then the probability of getting an $x$ that is a multiple of 3 such that $\left(x - \dfrac{26}{x}\right) > 25$, is (A) $\dfrac{1}{15}$ (B) $\dfrac{2}{15}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{5}$

   ›Reveal solutionSolution

   Combine "multiple of 3" with the algebraic condition $x-\frac{26}{x}>25$ (which reduces to $x>26$) to count the favourable outcomes out of 30. Answer: $\dfrac1{15}$.

   Concept and Intuition

   This is a classical-probability problem: with all 30 numbers equally likely, we need the count satisfying BOTH stated conditions divided by 30. The inequality condition simplifies to a clean threshold once cleared of the fraction, so the real work is finding which integers exceed that threshold and are also multiples of 3.

   Step-by-Step Solution

   1. Multiples of 3 in $\{1,\ldots,30\}$: $3,6,9,12,15,18,21,24,27,30$ — 10 numbers total, so $P(\text{multiple of }3)$ alone would be $\frac{10}{30}=\frac13$, but we need the further restriction below.
   2. Solve $x-\dfrac{26}{x}>25$ for $x>0$: multiply both sides by $x$ (positive, so inequality direction is preserved): $x^2-26>25x \Rightarrow x^2-25x-26>0$.
   3. Factor: $x^2-25x-26=(x-26)(x+1)$. So the inequality is $(x-26)(x+1)>0$.
   4. Since $x>0$ means $x+1>0$ always, the inequality holds exactly when $x-26>0$, i.e. $x>26$. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If 5 boys and 4 girls are arranged in a row randomly, then the probability of occurrence of the arrangement in which either all boys sit together or no two boys sit together in a row is (A) $\dfrac{5}{16}$ (B) $\dfrac{11}{16}$ (C) $\dfrac{1}{21}$ (D) $\dfrac{20}{21}$

   ›Reveal solutionSolution

   Count "all boys together" and "no two boys together" separately (they can't overlap), divide by $9!$; the answer simplifies to $1/21$.

   Concept and Intuition

   With 5 boys and 4 girls, "all 5 boys together" and "no two boys adjacent" are opposite extremes and can never happen simultaneously (since there are more than 1 boy). So the two favourable counts simply add.

   Step-by-Step Solution

   1. Total ways to arrange 9 distinct people in a row $=9!=362880$.
   2. All boys together: glue the 5 boys into one block. Now there are $4+1=5$ units to arrange: $5!$ ways. The boys inside the block can be permuted in $5!$ ways. Total $=5!\times5!=120\times120=14400$.
   3. No two boys together: first seat the 4 girls: $4!=24$ ways. This creates 5 slots: $\_G\_G\_G\_G\_$. Since there are exactly 5 boys and 5 slots, every slot must take exactly one boy (no slot can be empty or hold two), giving $5!=120$ arrangements of the boys into the slots. Total $=24\times120=2880$.
   4. Since a single boy cannot be simultaneously "in a block of 5" and "isolated from all other boys" (there are 4 other boys), these two cases never overlap, so we add them: $14400+2880=17280$. …

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7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Two symmetric cubical dice are rolled once. Match the items of column-I with the items of column-II Column-I: A. Probability that the numbers appearing on them are equal B. Probability that the numbers shown on them are all distinct C. Probability that the sum of the numbers on their faces is 10 D. Probability that the sum of the numbers on their faces is 6 Column-II: I. $\dfrac{1}{12}$ II. $\dfrac{5}{36}$ III. $\dfrac{1}{6}$ IV. $\dfrac{4}{36}$ V. $\dfrac{5}{6}$ The correct match is (A) A-III, B-I, C-V, D-II (B) A-III, B-V, C-I, D-II (C) A-V, B-IV, C-I, D-II (D) A-V, B-III, C-IV, D-I

   ›Reveal solutionSolution

   Direct counting over the 36 equally likely outcomes of two dice gives each of the four probabilities, which match column-II items III, V, I, II respectively.

   Concept and Intuition

   With two fair dice there are $36$ equally likely ordered outcomes. Each event here is a simple counting problem: count favourable ordered pairs and divide by $36$.

   Step-by-Step Solution

   1. A. Equal numbers: $(1,1),(2,2),\dots,(6,6)$ — 6 outcomes. $P=6/36=1/6$ — matches III.
   2. B. All distinct (numbers not equal): complement of A, so $P=1-1/6=5/6$ — matches V.
   3. C. Sum $=10$: $(4,6),(5,5),(6,4)$ — 3 outcomes. $P=3/36=1/12$ — matches I.
   4. D. Sum $=6$: $(1,5),(2,4),(3,3),(4,2),(5,1)$ — 5 outcomes. $P=5/36$ — matches II. …

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8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Let $S$ be the set of all words formed by arranging all the letters of the word HOMOGENEOUS. If a word is randomly chosen from the set $S$, then the probability that the word selected has all the consonants together is (A) $\dfrac{6!\,6!}{11!}$ (B) $\dfrac{7!\,5!}{11!}$ (C) $\dfrac{6!\,5!}{11!}$ (D) $\dfrac{8!\,5!}{11!}$

   ›Reveal solutionSolution

   Arrangement probability with repeated letters; consonants-together condition gives $\frac{7!\,5!}{11!}$.

   Concept and Intuition

   When letters repeat, the total number of distinct arrangements is $\frac{n!}{\text{(product of repeat factorials)}}$. For a "group together" condition, glue the required letters into a single block, arrange the block internally, then arrange the block along with the rest as a smaller set of units (again dividing out any remaining repeats).

   Step-by-Step Solution

   1. HOMOGENEOUS = H,O,M,O,G,E,N,E,O,U,S (11 letters). Counts: O×3, E×2, and H,M,G,N,S,U each ×1.
   2. Vowels: O,O,O,E,E,U → 6 vowels (with repeats O×3, E×2). Consonants: H,M,G,N,S → 5 distinct consonants.
   3. Total distinct arrangements of all 11 letters: $\dfrac{11!}{3!\,2!}$ (dividing by repeats of O and E; U is unique so no extra division needed). …

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9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.Two dice are thrown and the sum of the numbers appeared on the dice is noted. If A is the event of getting a prime number as their sum and B is the event of getting a number greater than 8 as their sum, then $P(A \cap \bar B) =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{13}{36}$ (C) $\dfrac{2}{9}$ (D) $\dfrac{5}{18}$

   ›Reveal solutionSolution

   $A\cap\overline B$ restricts to prime sums that are also $\le 8$; count the favourable dice outcomes out of the 36 equally likely ones.

   Concept and Intuition

   $\overline B$ is the complement of "sum $>8$", i.e. "sum $\le 8$". So $A\cap\overline B$ is exactly "sum is prime and sum $\le 8$" — this narrows the prime sums $\{2,3,5,7,11\}$ down to $\{2,3,5,7\}$ since $11>8$ is excluded.

   Step-by-Step Solution

   1. Total outcomes for two dice: $36$, all equally likely.
   2. Prime sums possible with two dice (sums range 2–12): $2,3,5,7,11$.
   3. $\overline B$ = sum $\le 8$, so from the prime list, only $2,3,5,7$ qualify ($11$ is excluded).
   4. Count ways for each: sum$=2$: $(1,1)$ → 1 way. sum$=3$: $(1,2),(2,1)$ → 2 ways. sum$=5$: $(1,4),(2,3),(3,2),(4,1)$ → 4 ways. sum$=7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ → 6 ways. …

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10. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.If five digit numbers are formed from the digits 0, 1, 2, 3, 4 using every digit exactly only once, then the probability that a randomly chosen number from those numbers is divisible by 4, is (A) $\dfrac{5}{16}$ (B) $\dfrac{3}{16}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{7}{16}$

    ›Reveal solutionSolution

    Count 5-digit permutations of $\{0,1,2,3,4\}$ divisible by 4 (last two digits divisible by 4, first digit nonzero) against the total valid 5-digit arrangements. Probability is $5/16$.

    Concept and Intuition

    Divisibility by 4 depends only on the last two digits of the number. So fix which two digits occupy the last two positions (in the order that makes that 2-digit number divisible by 4), then freely arrange the remaining three digits in the first three positions — except we must subtract cases where the leading digit is 0.

    Step-by-Step Solution

    1. Total 5-digit numbers using all of 0,1,2,3,4 exactly once, first digit $\ne0$: total permutations $5!=120$, minus those starting with 0 ($4!=24$), giving $96$.
    2. For the last two digits (positions 4,5) to form a number divisible by 4, since divisibility by 4 requires the number even, the units digit must be 0, 2, or 4.
    3. Check all valid (tens,units) pairs from distinct digits of $\{0,1,2,3,4\}$: using $2a+b\equiv0\pmod4$ (since $10a+b\equiv 2a+b\pmod 4$):
       • units=0: tens even, $\ne0$: (2,0),(4,0) — both valid (20,40 divisible by 4).
       • units=2: tens odd: (1,2),(3,2) — valid (12,32 divisible by 4).
       • units=4: tens even, $\ne4$: (0,4),(2,4) — valid (04→4, 24 divisible by 4).
    4. For each pair, the remaining 3 digits fill the first 3 positions; count arrangements excluding leading zero:
       • (2,0): remaining {1,3,4}, no zero, $3!=6$ valid. …

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11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.A box P contains one white ball, three red balls and two black balls. Another box Q contains two white balls, three red balls and four black balls. If one ball is drawn at random from each one of the two boxes, then the probability that the balls drawn are of different colour is (A) $\dfrac{29}{54}$ (B) $\dfrac{25}{42}$ (C) $\dfrac{35}{54}$ (D) $\dfrac{39}{52}$

    ›Reveal solutionSolution

    It's easier to compute $P(\text{same colour})$ and subtract from 1 than to directly sum the different-colour cross terms. Answer: $35/54$.

    Concept and Intuition

    When picking one item from each of two independent boxes, 'different colour' is the complement of 'same colour', and 'same colour' splits neatly into the sum of matching-colour probabilities (white–white, red–red, black–black), each a simple product since the draws are independent.

    Step-by-Step Solution

    1. Box P (total 6): $P(\text{white})=1/6,\ P(\text{red})=3/6,\ P(\text{black})=2/6$.
    2. Box Q (total 9): $P(\text{white})=2/9,\ P(\text{red})=3/9,\ P(\text{black})=4/9$.
    3. $P(\text{both white}) = \frac16\cdot\frac29=\frac{2}{54}$.
    4. $P(\text{both red}) = \frac36\cdot\frac39=\frac{9}{54}$.
    5. $P(\text{both black}) = \frac26\cdot\frac49=\frac{8}{54}$.
    6. $P(\text{same colour}) = \frac{2+9+8}{54}=\frac{19}{54}$. …

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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If each of the coefficients a, b, c in the equation $ax^2+bx+c=0$ is determined by throwing a die, then the probability that the equation will have equal roots, is (A) $\dfrac{1}{36}$ (B) $\dfrac{1}{72}$ (C) $\dfrac{7}{216}$ (D) $\dfrac{5}{216}$

    ›Reveal solutionSolution

    Equal roots of $ax^2+bx+c=0$ need $b^2=4ac$; counting die-outcome triples satisfying this out of $216$ total gives probability $\dfrac{5}{216}$.

    Concept and Intuition

    The quadratic has equal (repeated) roots exactly when its discriminant $b^2-4ac=0$. Since $a,b,c$ are each independent outcomes of a fair die (1 to 6), we just need to count triples satisfying $b^2=4ac$.

    Step-by-Step Solution

    1. For $b^2=4ac$ to have an integer $ac$, $b$ must be even (so $b^2/4$ is an integer). Possible $b\in\{2,4,6\}$.
    2. $b=2$: need $ac=1$. Only $(a,c)=(1,1)$. → 1 triple.
    3. $b=4$: need $ac=4$. Pairs in $\{1,\dots,6\}$: $(1,4),(2,2),(4,1)$. → 3 triples.
    4. $b=6$: need $ac=9$. Pairs in $\{1,\dots,6\}$: only $(3,3)$ (since $9=1\times9$ needs a factor $>6$). → 1 triple. …

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