Q.In a college, $30\%$ students fail in physics, $25\%$ fail in mathematics and $10\%$ fail in both. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is
(A) $\dfrac{1}{10}$
(B) $\dfrac{2}{5}$
(C) $\dfrac{9}{20}$
(D) $\dfrac{1}{3}$

Showing the 12 most recent of 45 on this concept.

1. 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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2. 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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3. 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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4. 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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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: $$$$

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6. COMEDK 2025Set 2025-E1 markMCQ
   Q.If A and B are two events such that $P(\bar{A})=0.3, P(B)=0.4, P(A \cap \bar{B})=0.5$, then find the value of $P(B / A \cup \bar{B})$ (A) 0.33 (B) 0.7 (C) 0.8 (D) 0.25

   ›Reveal solutionSolution

   The problem asks for a conditional probability $P(B \mid A \cup \bar{B})$. Using the definition of conditional probability and the given probabilities, we compute $P(B \cap (A \cup \bar{B}))$ and $P(A \cup \bar{B})$, then divide. The final value is $0.25$, so the correct option is (D).

   We are given:

   • $P(\bar{A}) = 0.3$, so $P(A) = 1 - 0.3 = 0.7$.
   • $P(B) = 0.4$.
   • $P(A \cap \bar{B}) = 0.5$.

   We need $P(B \mid A \cup \bar{B})$, which is the probability that $B$ occurs given that $A$ or $\bar{B}$ (or both) occurs.

   Concept and intuition:

   Conditional probability $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$. Here $X = B$ and $Y = A \cup \bar{B}$. So we need the probability that both $B$ and $(A \cup \bar{B})$ happen, divided by the probability that $A \cup \bar{B}$ happens. The key is to simplify the intersection $B \cap (A \cup \bar{B})$ using set algebra — it often collapses to something simpler.

   1. Simplify the numerator $P(B \cap (A \cup \bar{B}))$

      Using the distributive law:

      $B \cap (A \cup \bar{B}) = (B \cap A) \cup (B \cap \bar{B})$

      But $B \cap \bar{B} = \emptyset$, so this is just $B \cap A$.

      Hence $P(B \cap (A \cup \bar{B})) = P(A \cap B)$.

   2. Find $P(A \cap B)$

      We know $P(A) = 0.7$ and $P(A \cap \bar{B}) = 0.5$.

      Since $A = (A \cap B) \cup (A \cap \bar{B})$ and these are disjoint,

      $P(A) = P(A \cap B) + P(A \cap \bar{B})$

      $\Rightarrow 0.7 = P(A \cap B) + 0.5$

      $\Rightarrow P(A \cap B) = 0.2$.

   3. Find the denominator $P(A \cup \bar{B})$

      Use the inclusion-exclusion principle: …

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7. 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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8. 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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9. 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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10. 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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11. 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. …

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12. 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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