Q.A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Showing the 12 most recent of 69 on this concept.

1. 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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2. 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. $P(\text{red}) = \dfrac12\cdot\dfrac47+\dfrac12\cdot\dfrac25 = \dfrac{4}{14}+\dfrac{2}{10}=\dfrac{2}{7}+\dfrac15$. …

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3. 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): $P(B) = P(S_1)P(B\mid S_1)+P(S_2)P(B\mid S_2) = \frac13\cdot\frac58+\frac23\cdot\frac25 = \frac{5}{24}+\frac{4}{15}.$ …

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4. 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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5. 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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6. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. If one ball is drawn at random from one of the bags and is found to be red, then the probability that it was drawn from the bag B is (A) $\dfrac{23}{54}$ (B) $\dfrac{25}{51}$ (C) $\dfrac{25}{52}$ (D) $\dfrac{27}{55}$

   ›Reveal solutionSolution

   A classic Bayes'-theorem problem: given the ball drawn is red, the probability it came from bag B is $\dfrac{25}{52}$.

   Concept and Intuition

   This is Bayes' theorem: we're given the outcome (red ball drawn) and want to find the probability of which "cause" (bag A or B) produced it, using the prior probability of choosing each bag (each $\frac12$) and each bag's own probability of yielding red.

   Step-by-Step Solution

   1. Bag A: 2 white, 3 red (5 total) $\Rightarrow P(\text{red}|A)=\dfrac35$.
   2. Bag B: 4 white, 5 red (9 total) $\Rightarrow P(\text{red}|B)=\dfrac59$.
   3. $P(A)=P(B)=\dfrac12$ (bag chosen at random).
   4. Total probability of red: $P(\text{red})=P(A)P(\text{red}|A)+P(B)P(\text{red}|B)=\dfrac12\cdot\dfrac35+\dfrac12\cdot\dfrac59=\dfrac3{10}+\dfrac5{18}$.
   5. Common denominator 90: $\dfrac3{10}=\dfrac{27}{90}$, $\dfrac5{18}=\dfrac{25}{90}$, sum $=\dfrac{52}{90}=\dfrac{26}{45}$.
   6. Bayes: $P(B|\text{red})=\dfrac{P(B)P(\text{red}|B)}{P(\text{red})}=\dfrac{(1/2)(5/9)}{26/45}=\dfrac{5/18}{26/45}$. …

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7. 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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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: …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.An urn A contains 4 white and 1 black ball; urn B contains 3 white and 2 black balls and urn C contains 2 white and 3 black balls. One ball is transferred randomly from A to B; later one ball is transferred randomly from B to C. Finally, if a ball is drawn randomly from C, then the probability that it is a black ball is (A) $\dfrac{7}{12}$ (B) $\dfrac{89}{180}$ (C) $\dfrac{101}{180}$ (D) $\dfrac{17}{36}$

   ›Reveal solutionSolution

   A two-stage transfer-then-draw problem, solved by branching over all four possible transfer outcomes: probability of black $=\dfrac{101}{180}$.

   Concept and Intuition

   This is a sequential conditional-probability problem: each transfer changes the composition of the receiving urn, so we must branch over every possible outcome of each transfer and weight the final draw accordingly (total probability theorem, applied twice).

   Step-by-Step Solution

   1. Urn A: 4W,1B. Transfer to B: $P(\text{W})=\frac45$, $P(\text{B})=\frac15$.
   2. If white moved to B: B becomes 4W,2B (6 balls). If black moved to B: B becomes 3W,3B (6 balls).
   3. From B (4W,2B): transfer white to C with $P=\frac{4}{6}=\frac23$ (C becomes 3W,3B), or black with $P=\frac26=\frac13$ (C becomes 2W,4B).
   4. From B (3W,3B): transfer white to C with $P=\frac12$ (C becomes 3W,3B), or black with $P=\frac12$ (C becomes 2W,4B).
   5. Final draw from C: $P(\text{black}\mid 3W3B)=\frac12$; $P(\text{black}\mid 2W4B)=\frac{4}{6}=\frac23$.
   6. Combine all four branches:
      • A-white, B-white: $\frac45\cdot\frac23\cdot\frac12=\frac{8}{30}=\frac{4}{15}$
      • A-white, B-black: $\frac45\cdot\frac13\cdot\frac23=\frac{8}{45}$ …

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10. 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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11. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.Bag A contains 3 white and 4 black balls. Bag B contains 4 white and 3 black balls. Bag C contains 2 white and 5 black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag C is (A) $\dfrac{1}{6}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{2}{13}$

    ›Reveal solutionSolution

    This is a direct Bayes'-theorem (inverse probability) question. Answer: $\dfrac{2}{9}$.

    Concept and Intuition

    We're given the outcome (a white ball was drawn) and asked for the probability of a particular cause (it came from bag C). This is exactly Bayes' theorem: $P(C\mid W) = \dfrac{P(C)P(W\mid C)}{\sum_i P(\text{bag}_i)P(W\mid \text{bag}_i)}$.

    Step-by-Step Solution

    1. Each bag is chosen with probability $\frac13$.
    2. $P(W\mid A) = \dfrac{3}{7}$ (3 white out of 7 total in bag A).
    3. $P(W\mid B) = \dfrac{4}{7}$ (4 white out of 7 in bag B).
    4. $P(W\mid C) = \dfrac{2}{7}$ (2 white out of 7 in bag C).
    5. Total probability of white: $P(W) = \frac13\left(\dfrac37+\dfrac47+\dfrac27\right) = \frac13\cdot\dfrac{9}{7} = \dfrac{9}{21} = \dfrac{3}{7}$. …

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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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