Q.Consider the experiment of throwing a die, if a multiple of $3$ comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a $3$'.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2024Set 2024-E1 markMCQ
   Q.A coin is tossed until a head appears or until the coin has been tossed three times. Given that 'head' does not appear on the first toss, what is the probability that the coin is tossed thrice? (A) $\frac{1}{2}$ (B) $\frac{3}{8}$ (C) $\frac{1}{8}$ (D) $\frac{1}{4}$

   ›Reveal solutionSolution

   Given the first toss is a tail, the coin is tossed a third time only if the second toss is also a tail — probability $\tfrac12$.

   The coin stops the moment a head appears, or after three tosses. We are told the first toss is a tail (no head on toss 1).

   Now the second toss decides:

   • Head on toss 2 $\Rightarrow$ stop after 2 tosses.
   • Tail on toss 2 $\Rightarrow$ a third toss is made. …

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2. 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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3. 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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4. COMEDK 2026Set 2026-M1 markMCQ
   Q.Samhita faces a three-headed dragon. She wins a "Tactical medal" if she manages to defeat exactly one of the three heads. The battle proceeds head-by-head under the following conditions: The probability of defeating the first head is $\frac{\mathbf{1}}{\mathbf{3}}$. After a win: if she defeats a head, the probability of defeating the next head is $\frac{2}{3}$. After a loss: if she fails to defeat a head, the probability of defeating the next head is $\frac{\mathbf{1}}{\mathbf{4}}$. What is the probability that Samhita earns the "Tactical medal"? (A) $\frac{23}{72}$ (B) $\frac{5}{36}$ (C) $\frac{17}{72}$ (D) $\frac{19}{72}$

   ›Reveal solutionSolution

   Summing the three disjoint "exactly one win" paths gives $\dfrac{1}{12}+\dfrac{1}{18}+\dfrac{1}{8}=\dfrac{19}{72}$ — option (D).

   Set up the conditional probabilities. Let $W_i$ mean "defeats head $i$" and $L_i$ mean "fails":

   • $P(W_1) = \dfrac{1}{3}$, so $P(L_1) = \dfrac{2}{3}$.
   • After a win: next-head win probability $=\dfrac{2}{3}$, so next-head loss probability $=\dfrac{1}{3}$.
   • After a loss: next-head win probability $=\dfrac{1}{4}$, so next-head loss probability $=\dfrac{3}{4}$.

   Earning the medal means exactly one of the three heads is defeated. The three disjoint sequences are $W_1L_2L_3$, $L_1W_2L_3$, and $L_1L_2W_3$.

   Path 1 — $W_1 L_2 L_3$ (win, then loss after a win, then loss after a loss):

   $P = \frac{1}{3}\cdot\frac{1}{3}\cdot\frac{3}{4} = \frac{3}{36} = \frac{1}{12}$

   Path 2 — $L_1 W_2 L_3$ (loss, then win after a loss, then loss after a win): …

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5. 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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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.Three fair dice are thrown. What is the probability of getting a total of 15 given that they exhibit three different numbers that are in arithmetic progression? (A) $\frac{1}{8}$ (B) $\frac{1}{6}$ (C) $\frac{1}{4}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The problem asks for the conditional probability that three dice sum to 15, given that the three numbers shown are distinct and form an arithmetic progression. By listing all such triples and counting those summing to 15, the probability is found to be 1/2.

   Concept and intuition

   We need $P(\text{sum}=15 \mid \text{distinct AP})$. The condition restricts the outcomes to only those triples $(a,b,c)$ with $a<b<c$ (since dice are fair, order matters but we can treat ordered rolls; however the condition “three different numbers” and “arithmetic progression” is symmetric). The key is to enumerate all possible ordered triples of distinct numbers from $\{1,2,3,4,5,6\}$ that are in arithmetic progression, then count how many of those sum to 15. Because the dice are fair, each ordered triple is equally likely, so the conditional probability is just the ratio of favorable ordered triples to total ordered triples satisfying the condition.

   Step-by-step reasoning

   1. Characterize arithmetic progressions of three distinct numbers from 1 to 6. Three numbers $x, y, z$ (with $x<y<z$) are in AP iff $2y = x+z$. Since they are distinct and from 1 to 6, the possible triples (unordered) are:

   $(1,2,3),\ (1,3,5),\ (2,3,4),\ (2,4,6),\ (3,4,5),\ (4,5,6).$

   Check: For (1,2,3): $2\cdot2=1+3$; (1,3,5): $2\cdot3=1+5$; etc. That’s all.

   2. Count ordered triples satisfying the condition.

      For each unordered triple, the three numbers are distinct, so they can appear in any order on the three dice. That gives $3! = 6$ permutations per triple.

      Total ordered triples satisfying the condition = $6 \times 6 = 36$.

   3. Identify which of these triples sum to 15.

      Compute sums of the unordered triples:

      $\begin{aligned} (1,2,3) &\to 6,\\ (1,3,5) &\to 9,\\ (2,3,4) &\to 9,\\ (2,4,6) &\to 12,\\ (3,4,5) &\to 12,\\ (4,5,6) &\to 15. \end{aligned}$ …

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7. 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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8. 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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9. 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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10. KCET 2025Set A-11 markMCQ
    Q.Meera visits only one of the two temples A and B in her locality. Probability that she visits temple A is $\dfrac{2}{5}$. If she visits temple A, $\dfrac{1}{3}$ is the probability that she meets her friend, whereas it is $\dfrac{2}{7}$ if she visits temple B. Meera met her friend at one of the two temples. The probability that she met her at temple B is (A) $\dfrac{7}{16}$ (B) $\dfrac{5}{16}$ (C) $\dfrac{3}{16}$ (D) $\dfrac{9}{16}$

    ›Reveal solutionSolution

    The friend has already been met (the effect); we want the probability of the cause (temple B) — that reversal of conditioning is exactly Bayes' theorem.

    Step 1 — Name the events.

    Let $A$ = "Meera visits temple A", $B$ = "Meera visits temple B", $F$ = "she meets her friend".

    She visits only one of the two temples, so $A$ and $B$ are mutually exclusive and exhaustive:

    $P(A)=\frac25\quad\Longrightarrow\quad P(B)=1-\frac25=\frac35 .$

    Given: $P(F\mid A)=\dfrac13$, $P(F\mid B)=\dfrac27$.

    Step 2 — Why Bayes.

    We are told the outcome ($F$ happened) and asked for the probability of a cause ($B$). Bayes' theorem inverts the conditioning:

    $P(B\mid F)=\frac{P(B)\,P(F\mid B)}{P(A)\,P(F\mid A)+P(B)\,P(F\mid B)} .$

    The denominator is $P(F)$ by the law of total probability — the friend can be met on either branch.

    Step 3 — Compute the two branch probabilities.

    $P(A\cap F)=\frac25\times\frac13=\frac{2}{15},\qquad P(B\cap F)=\frac35\times\frac27=\frac{6}{35} .$

    Step 4 — Total probability of meeting the friend. …

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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 2018Set A-11 markMCQ
    Q.In a simultaneous throw of a pair of dice, the probability of getting a total more than $7$ is (A) $\dfrac{7}{12}$ (B) $\dfrac{5}{36}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{7}{36}$

    ›Reveal solutionSolution

    Count the ordered outcomes whose sum exceeds 7 out of the 36 equally likely outcomes of a pair of dice.

    Step 1 — The sample space.

    Each die shows one of $6$ faces, and the two dice are independent, so

    $n(S) = 6 \times 6 = 36$

    All $36$ ordered pairs $(a,b)$ are equally likely, so classical probability applies:

    $P(E) = \frac{n(E)}{n(S)}$

    Step 2 — Count the favourable outcomes (sum $> 7$, i.e. sum $= 8,9,10,11,12$).

    Sum	Ordered pairs	Count
    $8$	$(2,6),(3,5),(4,4),(5,3),(6,2)$	$5$
    $9$	$(3,6),(4,5),(5,4),(6,3)$	$4$
    $10$	$(4,6),(5,5),(6,4)$	$3$
    $11$	$(5,6),(6,5)$	$2$
    $12$	$(6,6)$	$1$

    $n(E) = 5+4+3+2+1 = 15$

    Step 3 — Probability.

    $P(\text{sum} > 7) = \frac{15}{36} = \frac{5}{12}$ …

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