Q.A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from TATA NAGAR.

Showing the 12 most recent of 45 on this concept.

1. 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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2. 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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3. 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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4. KCET 2021Set A-11 markMCQ
   Q.A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of cars and plant Y manufactures 30% of cars. 80% of cars at plant X and 90% of cars at plant Y are rated as standard quality. A car is chosen at random and is found to be of standard quality. The probability that it has come from plant X is (A) $\dfrac{56}{73}$ (B) $\dfrac{56}{84}$ (C) $\dfrac{56}{83}$ (D) $\dfrac{56}{79}$

   ›Reveal solutionSolution

   This is a reverse-probability question (effect → cause), so use Bayes' theorem with the total probability of a standard-quality car in the denominator.

   Step 1 — Define the events.

   • $X$: the car came from plant X — $P(X) = 0.70$
   • $Y$: the car came from plant Y — $P(Y) = 0.30$
   • $S$: the car is of standard quality

   The conditional (likelihood) data given:

   $P(S\mid X) = 0.80, \qquad P(S\mid Y) = 0.90$

   Note $X$ and $Y$ are mutually exclusive and exhaustive ($0.7 + 0.3 = 1$), which is exactly what Bayes' theorem needs.

   Step 2 — Why Bayes and not simple conditioning.

   We are told the effect (the chosen car is standard) and asked for the probability of the cause (it came from X). That inversion — $P(X\mid S)$ from $P(S\mid X)$ — is precisely Bayes' theorem:

   $P(X \mid S) = \frac{P(X)\,P(S\mid X)}{P(X)\,P(S\mid X) + P(Y)\,P(S\mid Y)}$

   Step 3 — Compute the numerator.

   $P(X)P(S\mid X) = 0.70 \times 0.80 = 0.56$

   Step 4 — Compute the denominator (total probability of a standard car).

   $P(S) = 0.70\times0.80 + 0.30\times0.90 = 0.56 + 0.27 = 0.83$

   Step 5 — Divide. …

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.In an entrance test, there are multiple choice questions. There are four possible answers to each question of which only one is correct. The probability that a student knows the answer to a question is $90 \%$. If he gets the correct answer to a question, then the probability that he was guessing is (A) $\frac{36}{37}$ (B) $\frac{1}{9}$ (C) $\frac{1}{37}$ (D) $\frac{37}{40}$

   ›Reveal solutionSolution

   We use Bayes’ theorem to reverse the conditional probability: given that the student answered correctly, we want the probability that he was guessing. The answer is $\frac{1}{37}$, so the correct option is (C).

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   Concept and intuition

   This is a classic Bayes’ theorem problem. We have two ways a student can get a correct answer: either he knows the answer (and thus answers correctly for sure), or he guesses (and has a 1 in 4 chance of being correct). We are told the prior probability that he knows the answer is 90%. The question asks: given that he got it right, what’s the chance he was actually guessing? Bayes’ theorem lets us “flip” the conditional probability, using the known likelihoods.

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   Step-by-step reasoning

   1. Define events clearly Let $K$ = “student knows the answer”, and $G$ = “student guesses” (so $G$ is the complement of $K$). Let $C$ = “student gives the correct answer”. We are given:

   $P(K) = 0.9, \quad P(G) = 0.1.$

   2. Conditional probabilities for answering correctly

      • If the student knows the answer, he is certain to be correct: $P(C \mid K) = 1$.
      • If he guesses, he picks randomly among 4 options, so $P(C \mid G) = \frac{1}{4}$.

   3. Find the total probability of a correct answer

      By the law of total probability:

   $P(C) = P(C \mid K)P(K) + P(C \mid G)P(G) = (1)(0.9) + \left(\frac{1}{4}\right)(0.1) = 0.9 + 0.025 = 0.925.$

   4. Apply Bayes’ theorem We want $P(G \mid C)$, the probability that he was guessing given that he answered correctly: $$$$

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6. 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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7. COMEDK 2024Set 2024-M1 markMCQ
   Q.A and B each have a calculator which can generate a single digit random number from the set $\{1,2,3,4,5,6,7,8\}$. They can generate a random number on their calculator. Given that the sum of the two numbers is 12 , then the probability that the two numbers are equal is (A) $\frac{5}{64}$ (B) $\frac{1}{5}$ (C) $\frac{1}{16}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   We are asked for the conditional probability that two numbers are equal given their sum is 12.

   The only equal pair summing to 12 is (6,6), and there are 5 total pairs summing to 12.

   So the probability is $\frac{1}{5}$, which corresponds to option (B).

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   Concept and intuition

   This is a classic conditional probability problem: we are not interested in all possible outcomes, only those where the sum is exactly 12. The phrase “given that the sum is 12” means we restrict our universe to those pairs. Then we count how many of those restricted outcomes have the two numbers equal. The trap is to forget to restrict the denominator — many students mistakenly use the total number of all possible pairs (64) instead of only the favorable-sum pairs.

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   Step-by-step solution

   1. Identify the sample space

      Each of A and B picks a digit from $\{1,2,\dots,8\}$.

      Total possible ordered pairs $(a,b)$: $8 \times 8 = 64$.

   2. List all pairs with sum 12

      We need $a + b = 12$, with $1 \le a,b \le 8$.

      Possible values for $a$:

      • If $a = 4$, then $b = 8$
      • If $a = 5$, then $b = 7$
      • If $a = 6$, then $b = 6$
      • If $a = 7$, then $b = 5$
      • If $a = 8$, then $b = 4$

      So the pairs are:

   $(4,8),\ (5,7),\ (6,6),\ (7,5),\ (8,4)$

   That’s 5 ordered pairs.

   3. Count the favorable outcomes “The two numbers are equal” means $a = b$. …

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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 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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10. COMEDK 2021Set 20211 markMCQ
    Q.A student answers a multiple choice question with 5 alternatives, of which exactly one is correct. The probability that he knows the correct answer is $p, 0 < p < 1$. If he knows the answer, he answers it correctly; if he does not know the answer, he guesses it, and the probability that he guesses correctly is $\frac{1}{5}$. Given that he has answered the question correctly, the probability that he knew the correct answer is (A) $\frac{3p}{4p+3}$ (B) $\frac{5p}{3p+2}$ (C) $\frac{5p}{4p+1}$ (D) $\frac{4p}{3p+1}$

    ›Reveal solutionSolution

    By Bayes: P(K | correct) = P(K) P(correct | K) / P(correct) = p / [ (4p + 1)/5 ] = 5p/(4p + 1).

    Concept: Bayes' theorem. (The stem is truncated, but this is the standard question: given that he answered correctly, find the probability that he actually knew the answer.)

    Let K = he knows the answer, P(K) = p, so P(not K) = 1 - p.

    If he knows it, he is certainly correct: P(correct | K) = 1.

    If he does not know it, he guesses among 5 alternatives: P(correct | not K) = 1/5. …

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11. 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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12. 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.

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    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.

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    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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