Q.There are two bags, one of which contains $3$ black and $4$ white balls while the other contains $4$ black and $3$ white balls. A die is thrown. If it shows up $1$ or $3$, a ball is taken from the Ist bag; but if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2024Set 2024-M1 markMCQ
   Q.An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is (A) $\frac{17}{30}$ (B) $\frac{37}{60}$ (C) $\frac{31}{60}$ (D) $\frac{23}{30}$

   ›Reveal solutionSolution

   Conditioning on the first two draws (white -> not replaced; black -> replaced plus one extra black) and using total probability gives $P(\text{3rd black}) = \tfrac{23}{30}$ — option (D).

   Rules. Start with 2 white, 2 black (total 4). Drawing white removes it (whites $-1$, total $-1$). Drawing black puts it back and adds one more black (blacks $+1$, total $+1$).

   First two draws and the composition just before the 3rd draw

   • WW: $\tfrac24\cdot\tfrac13 = \tfrac16$; urn becomes W0, B2 (total 2) $\Rightarrow P(\text{black}) = 1$.
   • WB: $\tfrac24\cdot\tfrac23 = \tfrac13$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BW: $\tfrac24\cdot\tfrac25 = \tfrac15$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BB: $\tfrac24\cdot\tfrac35 = \tfrac{3}{10}$; urn becomes W2, B4 (total 6) $\Rightarrow P(\text{black}) = \tfrac23$. …

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2. COMEDK 2025Set 2025-A1 markMCQ
   Q.Three bags contain a number of red and white balls are as follows. Bag I: 3 red balls Bag II: 2 red balls and 1 white ball Bag III: 3 White balls The probability that bag $i$ will be chosen and a ball is selected from it is $\frac{i}{6}, i=1,2,3$. If a white ball is selected, what is the probablity that it came from Bag III (A) $\frac{9}{11}$ (B) $\frac{2}{11}$ (C) $0$ (D) $\frac{1}{11}$

   ›Reveal solutionSolution

   This is a classic Bayes’ theorem problem: we are given prior probabilities for choosing each bag and the conditional probabilities of drawing a white ball from each bag; after observing a white ball, we want the posterior probability that it came from Bag III. The answer is $\frac{9}{11}$.

   We start by noting that the problem gives us three bags with different compositions, and a rule for choosing a bag: the probability of picking bag $i$ is $\frac{i}{6}$. Then, from that bag, we pick a ball uniformly at random. We are told that a white ball was selected, and we need the probability that it came from Bag III. This is a textbook case for Bayes’ theorem, which lets us “reverse” conditional probabilities: we know $P(\text{white} \mid \text{bag } i)$, and we want $P(\text{bag } i \mid \text{white})$.

   Let’s work through it step by step.

   1. Define events and write down given probabilities. Let $B_i$ be the event that bag $i$ is chosen ($i=1,2,3$). We are told:

   $P(B_1) = \frac{1}{6},\quad P(B_2) = \frac{2}{6},\quad P(B_3) = \frac{3}{6}.$

   Let $W$ be the event that a white ball is drawn.

   From the bag compositions:

   • Bag I: 3 red, 0 white → $P(W \mid B_1) = 0$.
   • Bag II: 2 red, 1 white → $P(W \mid B_2) = \frac{1}{3}$.
   • Bag III: 0 red, 3 white → $P(W \mid B_3) = 1$.

   2. Compute the total probability of drawing a white ball. By the law of total probability:

   $P(W) = P(B_1)P(W\mid B_1) + P(B_2)P(W\mid B_2) + P(B_3)P(W\mid B_3).$

   Substitute:

   $P(W) = \frac{1}{6}\cdot 0 + \frac{2}{6}\cdot\frac{1}{3} + \frac{3}{6}\cdot 1 = 0 + \frac{2}{18} + \frac{3}{6} = \frac{1}{9} + \frac{1}{2}.$

   Get a common denominator (18):

   $P(W) = \frac{2}{18} + \frac{9}{18} = \frac{11}{18}.$

   3. Apply Bayes’ theorem to find $P(B_3 \mid W)$. Bayes’ theorem says:

   $P(B_3 \mid W) = \frac{P(B_3)\,P(W\mid B_3)}{P(W)}.$

   Plug in the numbers:

   $$$$

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3. COMEDK 2023Set 2023-E1 markMCQ
   Q.Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from bag A and put into bag B. However if tail appears then 2 balls are drawn at random from bag A and put into bag B. Now one ball is drawn at random from bag B. Given that the drawn ball from B is white, the probability that head appeared on the coin is (A) $\frac{23}{30}$ (B) $\frac{12}{23}$ (C) $\frac{11}{23}$ (D) $\frac{19}{30}$

   ›Reveal solutionSolution

   Computing $P(\text{white drawn}\mid H)=\frac45$ and $P(\text{white}\mid T)=\frac{11}{15}$, Bayes' theorem gives $P(H\mid\text{white})=\frac{12}{23}$.

   Head (transfer 1 ball from A$(3W,2R)$ to B$(1W)$, then draw from B's 2 balls):

   $P(\text{white})=\tfrac35(1)+\tfrac25\left(\tfrac12\right)=\tfrac35+\tfrac15=\tfrac45.$

   Tail (transfer 2 balls from A to B$(1W)$, then draw from B's 3 balls):

   $P(WW)=\tfrac{3}{10}\Rightarrow P(\text{white})=1,\quad P(WR)=\tfrac{6}{10}\Rightarrow \tfrac23,\quad P(RR)=\tfrac{1}{10}\Rightarrow \tfrac13.$ …

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4. COMEDK 2024Set 2024-E1 markMCQ
   Q.Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black. (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 0 (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

   We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

   ---

   Concept and Intuition

   The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

   ---

   Step-by-step reasoning

   1. Label the sides.

      Let the cards be:

      • Card A: both sides red (sides $R_1, R_2$)
      • Card B: both sides black (sides $B_1, B_2$)
      • Card C: one red, one black (sides $R_3, B_3$)

   2. Count equally likely outcomes.

      When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

   3. Condition on seeing a red side.

      The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities. …

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5. COMEDK 2021Set 2021-B1 markMCQ
   Q.An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3

   ›Reveal solutionSolution

   By symmetry the second-draw red probability averages to $\tfrac12$.

   Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

   • First ball red ($P=\tfrac{5}{10}=\tfrac12$): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$. …

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6. KCET 2021Set A-11 markMCQ
   Q.Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is (A) $\dfrac{1}{18}$ (B) $\dfrac{5}{18}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   We use conditional probability: reduce the sample space to only those outcomes where the sum is less than 6, then find the fraction of those that give a sum of exactly 3. The answer is $\frac{1}{5}$.

   The key idea here is conditional probability — the probability of one event given that another event has already occurred. When we say "if it is known that the sum was less than 6", we are no longer considering all 36 possible outcomes of throwing two dice. Instead, we restrict ourselves to only those outcomes that satisfy the condition, and then ask: within this smaller set, what fraction gives a sum of 3?

   Let’s work it out step by step.

   1. List all possible outcomes when two dice are thrown.

      Each die shows a number from 1 to 6. The total number of ordered pairs $(a,b)$ is $6 \times 6 = 36$.

   2. Identify the condition: sum less than 6.

      The possible sums are 2, 3, 4, and 5 (since sum = 1 is impossible with two dice). Let’s list all ordered pairs that give each sum:

      • Sum = 2: $(1,1)$ → 1 outcome
      • Sum = 3: $(1,2), (2,1)$ → 2 outcomes
      • Sum = 4: $(1,3), (2,2), (3,1)$ → 3 outcomes
      • Sum = 5: $(1,4), (2,3), (3,2), (4,1)$ → 4 outcomes

      Total outcomes with sum < 6: $1 + 2 + 3 + 4 = 10$.

      Watch out

      A common mistake is to forget that $(1,2)$ and $(2,1)$ are different outcomes. Dice are distinct, so order matters. Always count ordered pairs unless the problem explicitly says "identical dice" (which is rare in such problems).

   3. Identify the event of interest: sum = 3. …

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7. COMEDK 2023Set 2023-E1 markMCQ
   Q.A die is thrown twice and the sum of numbers appearing is observed to be 8 . What is the conditional probability that the number 5 has appeared atleast once? (A) $\frac{5}{36}$ (B) $\frac{2}{5}$ (C) $\frac{1}{18}$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   Given the sum is $8$, the sample space is the $5$ ordered pairs summing to $8$; two of them include a $5$, giving probability $\frac{2}{5}$.

   The ordered outcomes with sum $8$ are

   $(2,6),(3,5),(4,4),(5,3),(6,2)\quad(5\text{ outcomes}).$ …

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8. COMEDK 2026Set 2026-A1 markMCQ
   Q.Vishnu has two jars of marbles, Jar A and Jar B. Jar A contains 3 yellow marbles and 2 green marbles. Jar B contains 4 yellow marbles and 3 green marbles. Vishnu flips a fair coin. If it lands heads, he picks two marbles at random without replacement from Jar A. If it lands tails, he picks two marbles at random with replacement from Jar B. Given that Vishnu picked one yellow and one green marble, what is the probability that they came from Jar B? (A) $\frac{21}{41}$ (B) $\frac{49}{89}$ (C) $\frac{40}{89}$ (D) $\frac{20}{41}$

   ›Reveal solutionSolution

   $P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

   Likelihoods of drawing one yellow and one green ($E$).

   Jar A (3 yellow, 2 green; two draws without replacement):

   $P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$

   Jar B (4 yellow, 3 green; two draws with replacement):

   $P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$

   Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels): …

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9. COMEDK 2025Set 2025-A1 markMCQ
   Q.A bag contains $(n+1)$ coins. It is known that one of these coins has a head on both sides, whereas the other coins are fair. One of these coins is selected at random and tossed. If the probability that the toss results in heads is $\frac{7}{12}$, then the value of $n$ is : (A) 5 (B) 3 (C) 2 (D) 4

   ›Reveal solutionSolution

   The key idea is to treat the coin selection as a two‑case partition (two‑headed coin vs. fair coins) and apply the law of total probability. Solving the resulting equation gives $n = 5$, so the correct option is (A).

   ---

   Concept and intuition

   We have a mixed bag: one trick coin that always lands heads, and $n$ fair coins that land heads with probability $\frac12$. When we pick a coin at random and toss it, the overall chance of heads is a weighted average of the two cases. The weight for the trick coin is $\frac{1}{n+1}$ (since there are $n+1$ coins total), and for any fair coin it’s $\frac{n}{n+1}$. The problem gives that overall probability as $\frac{7}{12}$, so we set up an equation and solve for $n$.

   ---

   Step‑by‑step solution

   1. Define the events Let $T$ be the event that the two‑headed coin is selected, and $F$ the event that a fair coin is selected. Since selection is random,

   $P(T) = \frac{1}{n+1}, \qquad P(F) = \frac{n}{n+1}.$

   2. Conditional probabilities for heads If the trick coin is chosen, heads is certain:

   $P(\text{heads} \mid T) = 1.$

   If a fair coin is chosen, the chance of heads is $\frac12$:

   $P(\text{heads} \mid F) = \frac12.$

   3. Apply the law of total probability The overall probability of heads is

   $P(\text{heads}) = P(T) \cdot P(\text{heads}\mid T) + P(F) \cdot P(\text{heads}\mid F).$

   Substituting the values:

   $P(\text{heads}) = \frac{1}{n+1} \cdot 1 + \frac{n}{n+1} \cdot \frac12.$

   4. Simplify the expression

   $P(\text{heads}) = \frac{1}{n+1} + \frac{n}{2(n+1)} = \frac{2 + n}{2(n+1)}.$

   5. Set equal to the given probability The problem states this equals $\frac{7}{12}$: $$$$

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10. COMEDK 2025Set 2025-E1 markMCQ
    Q.A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball? (A) $\frac{12}{49}$ (B) $\frac{32}{49}$ (C) $\frac{3}{7}$ (D) $\frac{27}{49}$

    ›Reveal solutionSolution

    Condition on the first draw: $P(\text{2nd red}) = \tfrac{2}{7}\cdot\tfrac{6}{7} + \tfrac{5}{7}\cdot\tfrac{4}{7} = \tfrac{32}{49}$.

    Start: $5$ red, $2$ green, total $7$. The first ball is not replaced, and a ball of the other colour is added.

    Case 1 — first ball is green. $P = \dfrac{2}{7}$.

    Remove that green ($5R,1G$), then add a red $\Rightarrow 6R,1G$ (total $7$).

    $P(\text{2nd red}\mid \text{green first}) = \frac{6}{7}$

    Case 2 — first ball is red. $P = \dfrac{5}{7}$.

    Remove that red ($4R,2G$), then add a green $\Rightarrow 4R,3G$ (total $7$). …

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11. KCET 2020Set A-11 markMCQ
    Q.The probability of solving a problem by three persons A, B and C independently is $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is (A) $\frac{1}{12}$ (B) $\frac{1}{4}$ (C) $\frac{1}{24}$ (D) $\frac{1}{8}$

    ›Reveal solutionSolution

    The problem asks for the probability that exactly two of the three persons solve it. We compute the sum of probabilities for each pair solving while the third fails, giving $\frac{1}{8}$.

    The key idea: "solved by any two of them" means exactly two solve it, not at least two. The third person must fail. Since A, B, and C work independently, we multiply their individual probabilities for each specific outcome and then add the three possible cases.

    A common mistake is to include the case where all three solve it. That would be "at least two", not "any two". The phrase "any two" in probability problems almost always means exactly two.

    Let’s denote:

    • $P(A) = \frac{1}{2}$, so $P(\text{A fails}) = 1 - \frac{1}{2} = \frac{1}{2}$
    • $P(B) = \frac{1}{4}$, so $P(\text{B fails}) = 1 - \frac{1}{4} = \frac{3}{4}$
    • $P(C) = \frac{1}{3}$, so $P(\text{C fails}) = 1 - \frac{1}{3} = \frac{2}{3}$

    We want exactly two successes. There are three mutually exclusive ways this happens:

    1. A and B solve, C fails

       Probability = $P(A) \times P(B) \times P(\text{C fails})$

       $= \frac{1}{2} \times \frac{1}{4} \times \frac{2}{3} = \frac{2}{24} = \frac{1}{12}$

    2. A and C solve, B fails

       Probability = $P(A) \times P(C) \times P(\text{B fails})$

       $= \frac{1}{2} \times \frac{1}{3} \times \frac{3}{4} = \frac{3}{24} = \frac{1}{8}$

    3. B and C solve, A fails

       Probability = $P(B) \times P(C) \times P(\text{A fails})$ …

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12. KCET 2019Set A-11 markMCQ
    Q.A man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set $S = \{1, 2, 3, 4, 5, 6, 7\}$ and reports that it is even. The probability that it is actually even is (A) $\frac{2}{5}$ (B) $\frac{1}{5}$ (C) $\frac{1}{10}$ (D) $\frac{3}{5}$

    ›Reveal solutionSolution

    By Bayes' theorem the probability is $\frac{3}{5}$ — option (D).

    The set $S=\{1,2,3,4,5,6,7\}$ has $3$ even numbers $(2,4,6)$ and $4$ odd numbers, and one number is picked at random.

    Let $E$ = "the number is even" and $R$ = "the man reports it as even."

    Priors.

    $P(E)=\frac{3}{7},\qquad P(E^c)=\frac{4}{7}.$

    Likelihoods. He tells the truth with probability $\frac{2}{3}$ and lies with probability $\frac{1}{3}$.

    • If the number really is even, a truthful report says "even": $P(R\mid E)=\frac{2}{3}$.
    • If the number is odd, only a lie reports "even": $P(R\mid E^c)=\frac{1}{3}$.

    Bayes' theorem. …

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