Q.A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively $\frac{3}{10}$, $\frac{1}{5}$, $\frac{1}{10}$ and $\frac{2}{5}$. The probabilities that he will be late are $\frac{1}{4}$, $\frac{1}{3}$ and $\frac{1}{12}$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?

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

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3. 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 $P(A)$. …

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4. 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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5. 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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6. 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: $$$$

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7. COMEDK 2021Set 2021-B1 markMCQ
   Q.At a certain university 4% of male students are over 6 feet tall and 1% of female students are over 6 feet tall. The total student population is divided in the ratio 3 : 2, in favor of female students. If a student is selected at random from amongst all those over 6 feet tall, what is the probability that the student is a female? (A) 1/3 (B) 2/5 (C) 3/11 (D) 3/5

   ›Reveal solutionSolution

   Bayes' theorem gives $P(\text{female}\mid>6\text{ft})=\frac{0.006}{0.022}=\frac{3}{11}$.

   Population split $3:2$ in favour of females ⇒ $P(F)=\tfrac35=0.6,\ P(M)=\tfrac25=0.4$.

   Tall fractions: $P(T\mid F)=0.01,\ P(T\mid M)=0.04.$

   By Bayes' theorem: …

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8. COMEDK 2025Set 2025-A1 markMCQ
   Q.If for two events $A$ and $B, P(A-B)=\frac{1}{5}$ and $P(A)=\frac{3}{5}$ then $P(B / A)=$ (A) $\frac{2}{3}$ (B) $\frac{1}{2}$ (C) $\frac{3}{5}$ (D) $\frac{2}{5}$

   ›Reveal solutionSolution

   The key is to interpret $P(A-B)$ as $P(A \cap B^c)$ and use the definition of conditional probability. The result is $P(B/A) = \frac{2}{3}$, so option (A) is correct.

   We are asked for $P(B/A)$, the probability of $B$ given $A$. The definition is

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

   We know $P(A) = \frac{3}{5}$, so we need $P(A \cap B)$. The given $P(A-B) = \frac{1}{5}$ is the key: $A-B$ means “$A$ and not $B$,” i.e., $A \cap B^c$.

   1. Relate $P(A-B)$ to $P(A \cap B)$ Since $A$ is the union of the disjoint parts “$A$ and $B$” and “$A$ and not $B$,” we have

   $P(A) = P(A \cap B) + P(A \cap B^c).$

   Here $P(A \cap B^c) = P(A-B) = \frac{1}{5}$ and $P(A) = \frac{3}{5}$.

   2. Solve for $P(A \cap B)$

   $\frac{3}{5} = P(A \cap B) + \frac{1}{5} \quad\Rightarrow\quad P(A \cap B) = \frac{3}{5} - \frac{1}{5} = \frac{2}{5}.$

   3. Apply the conditional probability formula $P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{2/5}{3/5} = \frac{2}{3}.$ …

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9. 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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10. 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.

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

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

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

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11. 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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12. 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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