Q.A box contains $3$ orange balls, $3$ green balls and $2$ blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing $2$ green balls and one blue ball is
(A) $\dfrac{3}{28}$
(B) $\dfrac{2}{21}$
(C) $\dfrac{1}{28}$
(D) $\dfrac{167}{168}$

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A student has gone to a book fair and found 6 different mathematics books, 5 different physics books, 4 different chemistry books meant for TGEAPCET examination. If he likes to buy at least one book of each subject, then the total number of ways in which he can buy the books is (A) 29295 (B) 32768 (C) 4210 (D) 5120

   ›Reveal solutionSolution

   To find the total number of ways to buy at least one book of each subject, we first calculate the number of ways to choose at least one book for each subject independently, and then multiply these results. The total number of ways is $\boxed{29295}$.

   When we need to select items from a group, and the condition is "at least one", it's often easier to think about the total possibilities and subtract the cases where no items are selected.

   Consider a set of $n$ distinct items. For each item, we have two choices: either we select it or we don't.

   So, for $n$ items, the total number of ways to select any number of items (including selecting none) is $2 \times 2 \times \dots \times 2$ ($n$ times), which is $2^n$.

   If we want to select "at least one" item, we simply exclude the single case where no items are selected.

   The number of ways to select at least one item from $n$ distinct items is $2^n - 1$.

   Since the choice of books for one subject does not affect the choice of books for another subject, these are independent events. When independent events occur, the total number of ways is found by multiplying the number of ways for each event. This is known as the multiplication principle.

   Let's apply this concept step-by-step to the given problem.

   1. Mathematics Books:

      There are 6 different mathematics books. The student wants to buy at least one mathematics book.

      Using the formula, the number of ways to choose at least one mathematics book is $2^6 - 1$.

      $2^6 = 64$.

      So, the number of ways to choose mathematics books is $64 - 1 = 63$.

   2. Physics Books:

      There are 5 different physics books. The student wants to buy at least one physics book.

      Using the formula, the number of ways to choose at least one physics book is $2^5 - 1$.

      $2^5 = 32$.

      So, the number of ways to choose physics books is $32 - 1 = 31$. …

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2. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Urn A contains 6 white and 2 black balls; urn B contains 5 white and 3 black balls and urn C contains 4 white and 4 black balls. If an urn is chosen at random and a ball is drawn at random from it, then the probability that the ball drawn is white is (A) $\dfrac{3}{8}$ (B) $\dfrac{5}{8}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   The problem is a law of total probability scenario: pick an urn uniformly at random, then draw a ball. The overall probability of white is the average of the three urn-specific white probabilities, which works out to $\frac{15}{24} = \frac{5}{8}$.

   We have three urns, each equally likely to be chosen. Within each urn, the chance of drawing a white ball is simply the fraction of white balls in that urn. Because the urn is chosen first (at random) and then the ball is drawn, the total probability of white is the weighted average of the individual white probabilities, with weights equal to the probability of picking each urn.

   Step-by-step reasoning

   1. Identify the probabilities of choosing each urn.

      Since an urn is chosen at random from three urns, each has probability $\frac{1}{3}$.

   2. Find the probability of drawing a white ball from each urn.

      • Urn A: 6 white, 2 black → total 8 balls. So $P(\text{white} \mid A) = \frac{6}{8} = \frac{3}{4}$.
      • Urn B: 5 white, 3 black → total 8 balls. So $P(\text{white} \mid B) = \frac{5}{8}$.
      • Urn C: 4 white, 4 black → total 8 balls. So $P(\text{white} \mid C) = \frac{4}{8} = \frac{1}{2}$.

   3. Apply the law of total probability.

      The overall probability of white is:

   $P(\text{white}) = P(A)\cdot P(\text{white}\mid A) + P(B)\cdot P(\text{white}\mid B) + P(C)\cdot P(\text{white}\mid C)$

   Substitute the values:

   $P(\text{white}) = \frac{1}{3}\cdot\frac{3}{4} + \frac{1}{3}\cdot\frac{5}{8} + \frac{1}{3}\cdot\frac{1}{2}$

   4. Simplify each term.

      • First term: $\frac{1}{3}\cdot\frac{3}{4} = \frac{3}{12} = \frac{1}{4}$
      • Second term: $\frac{1}{3}\cdot\frac{5}{8} = \frac{5}{24}$
      • Third term: $\frac{1}{3}\cdot\frac{1}{2} = \frac{1}{6}$

   5. Add them with a common denominator.

      The least common multiple of 4, 24, and 6 is 24.

      $$$$

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3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If all the letters of the word ACADEMICIAN are permuted in all possible ways then the number of permutations in which no two A’s are together and all the consonants are together is (A) 7200 (B) 14400 (C) 3600 (D) 1800

   ›Reveal solutionSolution

   The number of permutations is $7200$, so the answer is (A).

   Letters of ACADEMICIAN (11 in all)

   • Vowels: $A,A,A,E,I,I$
   • Consonants: $C,C,D,M,N$

   Step 1 — keep all consonants together. Treat the five consonants as one block. Its internal arrangements: $\dfrac{5!}{2!}=60$ (two C's).

   Step 2 — arrange the block with the non-A vowels. The items that are not A's are: the block, $E, I, I$ — $4$ items, giving $\dfrac{4!}{2!}=12$ arrangements. …

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4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If all the letters of the word ACADEMICIAN are permuted in all possible ways then the number of permutations in which no two A’s are together and all the consonants are together is (A) 1800 (B) 14400 (C) 7200 (D) 3600

   ›Reveal solutionSolution

   $7200$ permutations — option (C).

   The word ACADEMICIAN has $11$ letters: vowels $A,A,A,E,I,I$ (6) and consonants $C,C,D,M,N$ (5).

   Consonants together: treat the five consonants as one block. Internal arrangements $= \dfrac{5!}{2!} = 60$ (two C's).

   No two A's together: first arrange the block together with the non-A vowels $E, I, I$. These four entities (block, $E, I, I$) arrange in …

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5. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.There are two boxes each containing 10 balls. In each box, few of them are black balls and rest are white. A ball is drawn at random from one of the boxes and found that it is black. If the probability that the black ball drawn is from the second box is $\dfrac{1}{5}$, then number of black balls in the first box is (A) 5 or 10 (B) 2 or 7 (C) 4 or 8 (D) 3 or 6 or 9

   ›Reveal solutionSolution

   We use Bayes’ theorem to relate the conditional probability that the black ball came from the second box to the unknown numbers of black balls. Solving the resulting equation gives two possible values for the number of black balls in the first box, which match one of the given options.

   Concept and intuition

   This is a classic Bayes’ theorem problem: we have two boxes, each with a certain number of black and white balls. A black ball is drawn, and we are told the probability it came from the second box is $\frac{1}{5}$. That means the black ball is more likely to have come from the first box. The key is to set up the conditional probability using Bayes’ rule, letting the unknown number of black balls in the first box be $x$ and in the second box be $y$. Since each box has 10 balls total, the number of white balls is $10-x$ and $10-y$ respectively. The probability of choosing either box is $\frac{1}{2}$. Then we solve for possible integer values of $x$ and $y$ that satisfy the given condition.

   Step-by-step solution

   1. Define variables

      Let the first box have $x$ black balls (and $10-x$ white).

      Let the second box have $y$ black balls (and $10-y$ white).

      Both $x$ and $y$ are integers from 0 to 10.

   2. Write the probability of drawing a black ball from each box

      • $P(\text{black} \mid \text{Box 1}) = \frac{x}{10}$
      • $P(\text{black} \mid \text{Box 2}) = \frac{y}{10}$
      • The probability of picking either box is $P(\text{Box 1}) = P(\text{Box 2}) = \frac{1}{2}$.

   3. Apply Bayes’ theorem

      We want $P(\text{Box 2} \mid \text{black}) = \frac{1}{5}$.

      Bayes’ theorem says:

   $P(\text{Box 2} \mid \text{black}) = \frac{P(\text{black} \mid \text{Box 2}) \cdot P(\text{Box 2})}{P(\text{black})}$

   The total probability of drawing a black ball is:

   $P(\text{black}) = P(\text{Box 1}) \cdot \frac{x}{10} + P(\text{Box 2}) \cdot \frac{y}{10} = \frac{1}{2} \cdot \frac{x}{10} + \frac{1}{2} \cdot \frac{y}{10} = \frac{x+y}{20}$

   4. Plug into Bayes’ formula

   $\frac{1}{5} = \frac{\frac{y}{10} \cdot \frac{1}{2}}{\frac{x+y}{20}} = \frac{\frac{y}{20}}{\frac{x+y}{20}} = \frac{y}{x+y}$

   So we have:

   $\frac{y}{x+y} = \frac{1}{5}$

   5. Solve the equation Cross-multiply:

   $5y = x + y \quad \Rightarrow \quad 4y = x \quad \Rightarrow \quad x = 4y$

   Since $x$ and $y$ are integers between 0 and 10, and $x = 4y$, the possible pairs are:

   • $y = 0$ → $x = 0$
   • $y = 1$ → $x = 4$
   • $y = 2$ → $x = 8$
   • $y = 3$ → $x = 12$ (invalid, since $x \le 10$) …

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6. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.There are two boxes each containing 10 balls. In each box, few of them are black balls and rest are white. A ball is drawn at random from one of the boxes and found that it is black. If the probability that the black ball drawn is from the second box is $\dfrac{1}{5}$, then number of black balls in the first box is (A) 3 or 6 or 9 (B) 5 or 10 (C) 4 or 8 (D) 2 or 7

   ›Reveal solutionSolution

   Use Bayes’ theorem to relate the conditional probability that the black ball came from the second box to the unknown numbers of black balls in each box. The condition leads to a Diophantine equation whose integer solutions give the possible black-ball counts in the first box.

   The problem gives two boxes, each with 10 balls. Let the first box contain $x$ black balls (and $10-x$ white), and the second box contain $y$ black balls (and $10-y$ white). A box is chosen at random (each with probability $\frac12$), then a ball is drawn from that box. The drawn ball is black. We are told that the probability that this black ball came from the second box is $\frac15$.

   This is a classic Bayes’ theorem situation: we have a prior probability of choosing each box, a likelihood of drawing a black ball from each box, and we want the posterior probability that the ball came from the second box given that it is black.

   1. Set up the probabilities. The probability of drawing a black ball from the first box is $\frac{x}{10}$, and from the second box is $\frac{y}{10}$. The prior probability of choosing either box is $\frac12$. By the law of total probability, the overall probability of drawing a black ball is

   $P(\text{black}) = \frac12 \cdot \frac{x}{10} + \frac12 \cdot \frac{y}{10} = \frac{x+y}{20}.$

   2. Apply Bayes’ theorem. The probability that the black ball came from the second box is

   $P(\text{box 2} \mid \text{black}) = \frac{P(\text{box 2}) \cdot P(\text{black} \mid \text{box 2})}{P(\text{black})} = \frac{\frac12 \cdot \frac{y}{10}}{\frac{x+y}{20}} = \frac{y/20}{(x+y)/20} = \frac{y}{x+y}.$

   The problem states this equals $\frac15$. So

   $\frac{y}{x+y} = \frac15.$

   3. Solve the equation. Cross-multiplying:

   $5y = x + y \quad\Rightarrow\quad 4y = x \quad\Rightarrow\quad x = 4y.$

   Since $x$ and $y$ are integers between 0 and 10 (inclusive), and each box has exactly 10 balls, $y$ can be 0, 1, 2, or 3 (because $x = 4y \le 10$). …

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7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Among the 5 married couples, if the names of 5 men are matched with the names of their wives randomly, then the probability that no man is matched with name of his wife is (A) $\dfrac{9}{20}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{11}{30}$ (D) $\dfrac{17}{60}$

   ›Reveal solutionSolution

   This is a classic derangement problem for 5 items. The probability that no man is matched with his own wife is the number of derangements of 5 items divided by 5! = 120, which equals 44/120 = 11/30. The correct option is (C).

   Concept & Intuition

   We have 5 men and their 5 wives. A random matching is a permutation of the wives' names assigned to the men. We want the probability that no man gets his own wife — i.e., the permutation has no fixed points. Such permutations are called derangements. The key idea: count all permutations (5! = 120), then count those with zero fixed points using inclusion–exclusion. The formula for derangements of $n$ items is

   $!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}.$

   For $n=5$, this gives the exact count.

   Step-by-step solution

   1. Total number of matchings There are 5 men and 5 wives. A matching is a bijection from men to wives — a permutation of the 5 wives. Total permutations:

   $5! = 120.$

   2. Define the event we want

      Let $A_i$ be the event that the $i$-th man is matched with his own wife. We want the probability that none of $A_1, A_2, \dots, A_5$ occur. That is, the number of permutations with no fixed points (derangements) divided by 120.

   3. Use inclusion–exclusion to count derangements

      Count permutations where at least one man gets his own wife, then subtract from total.

      • Number of permutations where a specific set of $k$ men get their own wives: the remaining $5-k$ wives can be arranged arbitrarily among the remaining $5-k$ men, so $(5-k)!$ ways.
      • There are $\binom{5}{k}$ ways to choose which $k$ men are fixed.

      By inclusion–exclusion, the number of permutations with at least one fixed point is:

   $\sum_{k=1}^{5} (-1)^{k+1} \binom{5}{k} (5-k)!.$

   So the number with no fixed points (derangements) is:

   $!5 = 5! - \sum_{k=1}^{5} (-1)^{k+1} \binom{5}{k} (5-k)!.$

   Equivalently, the standard derangement formula:

   $!5 = 5! \sum_{k=0}^{5} \frac{(-1)^k}{k!}.$

   4. Compute the sum

   $\sum_{k=0}^{5} \frac{(-1)^k}{k!} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}.$

   Compute step by step: …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If 2 cards drawn at random from a well shuffled pack of 52 playing cards are from the same suit, then the probability of getting a face card and a card having a prime number is (A) $\dfrac{8}{13}$ (B) $\dfrac{2}{13}$ (C) $\dfrac{8}{221}$ (D) $\dfrac{32}{221}$

   ›Reveal solutionSolution

   Conditioned on both cards being from the same suit, within each $13$-card suit there are $3$ face cards and $4$ prime-number cards. The favourable-to-total ratio is $\dfrac{4\times 12}{4\times\binom{13}{2}}=\dfrac{2}{13}$, option (B).

   Set up the conditional experiment

   Two cards are drawn and we are told they are from the same suit. A single suit has $13$ cards, so the conditioned sample space is

   $4\ \text{suits}\times\binom{13}{2} = 4\times 78 = 312\ \text{equally-likely same-suit pairs.}$

   Within one suit:

   • Face cards: J, Q, K $\Rightarrow 3$ cards.
   • Prime-number cards: $2,3,5,7$ are prime $\Rightarrow 4$ cards (Ace $=1$ is not prime).

   Count the favourable pairs …

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9. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If two cards are drawn at random simultaneously from a well shuffled pack of 52 playing cards, then the probability of getting a card having a composite number and a card having a number which is a multiple of 3 is (A) $\dfrac{94}{663}$ (B) $\dfrac{62}{663}$ (C) $\dfrac{102}{663}$ (D) $\dfrac{64}{663}$

   ›Reveal solutionSolution

   Composite-number cards: $20$; multiple-of-$3$ cards: $12$; cards that are both: $8$. Counting unordered pairs with one card of each type gives $204$ favourable out of $\binom{52}{2}=1326$, i.e. $\dfrac{102}{663}$ — option (C).

   Classify the ranks (numbers $1$–$10$).

   • Composite: $4,6,8,9,10$ → $5$ ranks → $20$ cards.
   • Multiple of $3$: $3,6,9$ → $3$ ranks → $12$ cards.
   • Both ($6,9$): $2$ ranks → $8$ cards.
   • Composite-only ($4,8,10$): $12$ cards; multiple-of-3-only ($3$): $4$ cards.

   Favourable pairs — one composite card and one multiple-of-3 card (two distinct cards). …

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10. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.Bag ‘P’ contains 3 white, 2 red, 5 blue balls and bag ‘Q’ contains 2 white, 3 red, 5 blue balls. A ball is chosen at random from ‘P’ and is placed in ‘Q’. If a ball is chosen from bag ‘Q’ at random, then the probability that it is a red ball is (A) $\dfrac{9}{50}$ (B) $\dfrac{13}{45}$ (C) $\dfrac{16}{55}$ (D) $\dfrac{12}{35}$

    ›Reveal solutionSolution

    Transferring a ball from P into Q changes Q's composition, so apply the Law of Total Probability over the three possible transfer colours. The probability of then drawing red from Q is $\dfrac{16}{55}$, option (C).

    Setup. Bag P has 3 white, 2 red, 5 blue (10 balls). Bag Q has 2 white, 3 red, 5 blue (10 balls). One ball is moved from P to Q, so Q then holds $11$ balls, and its red count depends on the transferred colour.

    Transfer probabilities from P.

    $P(\text{white})=\tfrac{3}{10},\quad P(\text{red})=\tfrac{2}{10},\quad P(\text{blue})=\tfrac{5}{10}.$

    Probability of red from Q after each transfer (Q now has 11 balls):

    • White moved $\Rightarrow$ Q red $=3 \Rightarrow \tfrac{3}{11}$. …

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.A, B, C, D cut a pack of 52 well shuffled playing cards successively in the same order. If the person who cuts a spade first, wins the game and the game continues until this happens, then the probability that A wins the game is (A) $\dfrac{74}{175}$ (B) $\dfrac{44}{175}$ (C) $\dfrac{54}{175}$ (D) $\dfrac{64}{175}$

    ›Reveal solutionSolution

    The probability that A wins is the sum of an infinite geometric series where A wins on the 1st, 5th, 9th, … cut. The result is $\frac{64}{175}$, which corresponds to option (D).

    Concept and Intuition

    This is a classic “first success” problem with a twist: the players cut in a fixed order (A, then B, then C, then D, then back to A, and so on). The game stops as soon as someone cuts a spade. Since the pack is well-shuffled, each cut is independent with probability $\frac{13}{52} = \frac14$ of being a spade.

    We want the probability that A is the first to cut a spade. That means A must succeed on her turn, and all players before her in that round must have failed. Because the game can go through many full cycles, we sum over all possible rounds where A wins.

    Step-by-step reasoning

    1. Probability of a spade on any single cut There are 13 spades in a 52-card deck, so

    $p = \frac{13}{52} = \frac14, \quad q = 1-p = \frac34.$

    2. When does A win? A wins if she cuts a spade on her first turn (round 1), or if everyone fails in the first round, then everyone fails again in the second round, …, and then A succeeds on her turn in some later round.
       • Round 1: A cuts first. She wins immediately with probability $p = \frac14$.
       • Round 2: For A to win in round 2, all four players must fail in round 1 (probability $q^4$), then A must succeed on her turn in round 2 (probability $p$). So probability = $q^4 \cdot p$.
       • Round 3: All four fail in round 1, all four fail in round 2, then A succeeds in round 3: probability = $(q^4)^2 \cdot p$.
       • In general, A wins on round $k$ ($k = 1,2,3,\dots$) with probability

    $(q^4)^{k-1} \cdot p.$

    3. Sum over all rounds The total probability that A wins is the infinite geometric series: $P(A) = p + q^4 p + (q^4)^2 p + \cdots = p \sum_{k=0}^\infty (q^4)^k.$ …

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12. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.A bag contains four balls. Two balls are drawn randomly and found them to be white. The probability that all the balls in the bag are white is (A) $\dfrac{1}{2}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{2}{3}$

    ›Reveal solutionSolution

    Take the number of white balls $W\in\{0,1,2,3,4\}$ as equally likely a priori, update on the evidence "two drawn balls are white" via Bayes, and the posterior $P(W=4)=\tfrac{3}{5}$.

    Setup. A bag of $4$ balls; each is white or non-white. Let $W$ be the (unknown) count of white balls, with prior $P(W=w)=\tfrac{1}{5}$ for $w=0,1,2,3,4$. Two balls are drawn without replacement and both are white — call this event $E$.

    Step 1 — likelihoods. Total ways to choose $2$ of $4$ is $\binom{4}{2}=6$. Drawing two whites is impossible for $W<2$.

    $P(E\mid W=4)=\frac{\binom{4}{2}}{\binom{4}{2}}=1,\quad P(E\mid W=3)=\frac{\binom{3}{2}}{6}=\frac{1}{2},\quad P(E\mid W=2)=\frac{\binom{2}{2}}{6}=\frac{1}{6}.$

    Step 2 — total probability of the evidence. …

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