Q.$A$ and $B$ are two students. Their chances of solving a problem correctly are $\dfrac{1}{3}$ and $\dfrac{1}{4}$, respectively. If the probability of their making a common error is $\dfrac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is
(A) $\dfrac{1}{12}$
(B) $\dfrac{1}{40}$
(C) $\dfrac{13}{120}$
(D) $\dfrac{10}{13}$

Showing the 12 most recent of 45 on this concept.

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

   🔒 Start your 14-day free trial to see the full solution →
2. KCET 2022Set C-41 markMCQ
   Q.If A and B are two events such that $P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$ and $P(A|B) = \dfrac{1}{4}$, then $P(A' \cap B')$ is (A) $3/16$ (B) $1/12$ (C) $3/4$ (D) $1/4$

   ›Reveal solutionSolution

   Get $P(A\cap B)$ from the conditional probability, use the addition rule for $P(A\cup B)$, then apply De Morgan: $P(A'\cap B')=1-P(A\cup B)$.

   Step 1 — Recover the joint probability from the conditional.

   Conditional probability is defined by $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ — it re-scales the probability of $A$ to the reduced sample space $B$. Rearranging gives the multiplication rule:

   $P(A\cap B)=P(A|B)\,P(B)=\frac14\times\frac13=\frac1{12}.$

   Step 2 — Addition rule for the union.

   $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac12+\frac13-\frac1{12}.$

   Taking LCM $12$: …

   🔒 Start your 14-day free trial to see the full solution →
3. COMEDK 2024Set 2024-A1 markMCQ
   Q.P and Q are considering to apply for a job. The probability that P applies for the job is $\frac{1}{4}$. The probability that $\mathrm{P}$ applies for the job given that $\mathrm{Q}$ applies for the job is $\frac{1}{2}$, and the probability that Q applies for the job given that P applies for the job is $\frac{1}{3}$. Then the probability that $\mathrm{P}$ does not apply for the job given that $\mathrm{Q}$ does not apply for the job is (A) $\frac{4}{5}$ (B) $\frac{7}{8}$ (C) $\frac{5}{6}$ (D) $\frac{11}{12}$

   ›Reveal solutionSolution

   We are given conditional probabilities and need to find $P(\overline{P} \mid \overline{Q})$. Using the definitions of conditional probability and the law of total probability, we compute $P(Q)$ and $P(\overline{Q})$, then apply Bayes' theorem to get $\frac{4}{5}$, which corresponds to option (A).

   Concept and intuition:

   This is a classic problem of working backwards from conditional probabilities to find a joint probability table. We know $P(P)$, $P(P \mid Q)$, and $P(Q \mid P)$. From these, we can find $P(P \cap Q)$ in two ways, which lets us solve for $P(Q)$. Then we can compute the desired conditional probability $P(\overline{P} \mid \overline{Q})$ using the complement rule and the definition of conditional probability.

   Step-by-step solution:

   1. Write down what is given. Let $P$ = event that P applies, $Q$ = event that Q applies. We have:

   $P(P) = \frac{1}{4}, \quad P(P \mid Q) = \frac{1}{2}, \quad P(Q \mid P) = \frac{1}{3}.$

   2. Use the definition of conditional probability to express $P(P \cap Q)$ in two ways. From $P(P \mid Q) = \frac{P(P \cap Q)}{P(Q)}$, we get

   $P(P \cap Q) = P(P \mid Q) \cdot P(Q) = \frac{1}{2} P(Q).$

   From $P(Q \mid P) = \frac{P(P \cap Q)}{P(P)}$, we get

   $P(P \cap Q) = P(Q \mid P) \cdot P(P) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}.$

   3. Equate the two expressions for $P(P \cap Q)$ to find $P(Q)$.

   $\frac{1}{2} P(Q) = \frac{1}{12} \quad \Rightarrow \quad P(Q) = \frac{1}{6}.$

   4. Find $P(\overline{Q})$ and $P(P \cap \overline{Q})$.

   $P(\overline{Q}) = 1 - P(Q) = 1 - \frac{1}{6} = \frac{5}{6}.$

   Also, $P(P \cap \overline{Q}) = P(P) - P(P \cap Q) = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}.$

   5. Find $P(\overline{P} \cap \overline{Q})$. Since $\overline{P} \cap \overline{Q}$ is the complement of $P \cup Q$, we can use:

   $P(\overline{P} \cap \overline{Q}) = 1 - P(P \cup Q).$

   First, $P(P \cup Q) = P(P) + P(Q) - P(P \cap Q) = \frac{1}{4} + \frac{1}{6} - \frac{1}{12}.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. KCET 2020Set A-11 markMCQ
   Q.If A and B are two events such that $P(A) = \frac{1}{3}$, $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{6}$, then $P(A'/B)$ is (A) $\frac{2}{3}$ (B) $\frac{1}{3}$ (C) $\frac{1}{2}$ (D) $\frac{1}{12}$

   ›Reveal solutionSolution

   Use $P(A'|B) = 1 - P(A|B)$ (equivalently $\frac{P(B)-P(A\cap B)}{P(B)}$), which gives $\frac{2}{3}$.

   Step 1 — The definition of conditional probability.

   For $P(B) > 0$,

   $P(A'|B) = \frac{P(A' \cap B)}{P(B)}$

   Conditioning on $B$ means we restrict the sample space to $B$ and ask what fraction of $B$ lies outside $A$.

   Step 2 — Split $B$ into the part inside $A$ and the part outside $A$.

   The events $A \cap B$ and $A' \cap B$ are disjoint and their union is exactly $B$:

   $P(B) = P(A \cap B) + P(A' \cap B)$

   $\Rightarrow P(A' \cap B) = P(B) - P(A \cap B) = \frac{1}{2} - \frac{1}{6} = \frac{3-1}{6} = \frac{2}{6} = \frac{1}{3}$

   Step 3 — Divide by $P(B)$.

   $P(A'|B) = \frac{P(A' \cap B)}{P(B)} = \frac{\tfrac{1}{3}}{\tfrac{1}{2}} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}$

   Step 4 — Cross-check with the complement rule. …

   🔒 Start your 14-day free trial to see the full solution →
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).

   ---

   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.

   ---

   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: $$$$

   🔒 Start your 14-day free trial to see the full solution →
6. COMEDK 2026Set 2026-M1 markMCQ
   Q.Advika chooses one of three scarves every morning: Red, Blue, or Green. The probability she chooses Red is $20 \%$. The probability she chooses Blue is twice the probability of choosing Red. On the remaining days she wears a Green scarf. Once a scarf is chosen, she decides whether to wear a Hat (H) and Sunglasses (S). These choices are independent of each other but depend on the scarf colour: $\begin{array}{|l|l|l|} \hline \text { Scarf colour } & \mathbf{P ( H )} & \mathbf{P ( S )} \\ \hline \text { Red } & 0.5 & 0.8 \\ \hline \text { Blue } & 0.4 & 0.5 \\ \hline \text { Green } & 0.1 & 0.5 \\ \hline \end{array}$ Advika is spotted outdoors wearing both a Hat and Sunglasses. What is the probability that she is wearing the Red scarf? (A) $\frac{13}{313}$ (B) $\frac{8}{21}$ (C) $\frac{4}{9}$ (D) $\frac{8}{13}$

   ›Reveal solutionSolution

   Bayes' theorem on scarf colour given that both a hat and sunglasses are worn. Priors $P(R)=0.2,\ P(B)=0.4,\ P(G)=0.4$; likelihoods $P(H\cap S\mid\text{colour})=P(H)\,P(S)$. The posterior $P(R\mid H\cap S)=\frac{4}{9}$ — option (C).

   Concept. Hat and sunglasses are independent given the scarf, so $P(H\cap S\mid\text{colour})=P(H\mid\text{colour})\cdot P(S\mid\text{colour})$. Bayes' theorem then reverses the conditioning to give the probability of the scarf colour from the observed accessories.

   Step 1 — Priors.

   $P(R)=20\%=0.2,\qquad P(B)=2P(R)=0.4,\qquad P(G)=1-0.2-0.4=0.4.$

   Step 2 — Likelihood of wearing both accessories for each colour.

   $P(H\cap S\mid R)=0.5\times0.8=0.40,$

   $P(H\cap S\mid B)=0.4\times0.5=0.20,$

   $P(H\cap S\mid G)=0.1\times0.5=0.05.$ …

   🔒 Start your 14-day free trial to see the full solution →
7. 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. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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): …

   🔒 Start your 14-day free trial to see the full solution →
9. 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).

   ---

   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.

   ---

   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 $a = b$. …

   🔒 Start your 14-day free trial to see the full solution →
10. COMEDK 2024Set 2024-E1 markMCQ
    Q.Let $\mathrm{A}$ and ${B}$ be two events such that $P(A / B)=\frac{1}{2}$ and $P(B / A)=\frac{1}{3}$ and $P(A \cap B)=\frac{1}{6}$ then, which one of the following is not true? (A) $\mathrm{A} \text { and } \mathrm{B} \text { are not independent }$ (B) $P(A \cup B)=\frac{2}{3}$ (C) $P\left(A^{\prime} \cap B\right)=\frac{1}{6}$ (D) $A \text { and } B \text { are independent }$

    ›Reveal solutionSolution

    $P(A)=\tfrac12,\ P(B)=\tfrac13$, and $P(A)P(B)=\tfrac16=P(A\cap B)$, so $A,B$ ARE independent; the false statement is that they are not independent.

    From $P(A\cap B)=P(A\mid B)\,P(B)$:

    $\frac{1}{6}=\frac{1}{2}\,P(B)\implies P(B)=\frac{1}{3}$

    From $P(A\cap B)=P(B\mid A)\,P(A)$:

    $\frac{1}{6}=\frac{1}{3}\,P(A)\implies P(A)=\frac{1}{2}$

    Independence test:

    $P(A)\,P(B)=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}=P(A\cap B)$

    So $A$ and $B$ are independent. Checking the other options:

    $P(A\cup B)=\frac12+\frac13-\frac16=\frac{2}{3}\quad(\text{B true})$ …

    🔒 Start your 14-day free trial to see the full solution →
11. KCET 2021Set A-11 markMCQ
    Q.Given that A and B are two events such that $P(B) = \dfrac{3}{5}$, $P(A/B) = \dfrac{1}{2}$ and $P(A \cup B) = \dfrac{4}{5}$, then $P(A) =$ (A) $\dfrac{3}{10}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

    ›Reveal solutionSolution

    Turn the conditional probability into $P(A\cap B)$, then substitute into the addition theorem and solve for $P(A)$.

    Step 1 — Extract $P(A\cap B)$ from the conditional probability.

    By definition,

    $P(A/B)=\frac{P(A\cap B)}{P(B)}\qquad (P(B)\neq0).$

    Rearranging (the multiplication theorem):

    $P(A\cap B)=P(A/B)\cdot P(B)=\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}.$

    This is the key move — the conditional probability is not directly usable in the union formula, but $P(A\cap B)$ is.

    Step 2 — Apply the addition theorem.

    $P(A\cup B)=P(A)+P(B)-P(A\cap B).$

    The intersection is subtracted because the elements common to $A$ and $B$ would otherwise be counted twice.

    Step 3 — Substitute the known values.

    $\frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}.$

    Step 4 — Solve for $P(A)$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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): …

    🔒 Start your 14-day free trial to see the full solution →