Q.If $P(A) = 0.8$, $P(B) = 0.5$ and $P(B|A) = 0.4$, find

Showing the 12 most recent of 45 on this concept.

1. 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:

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

      We have $P(\bar{B}) = 1 - P(B) = 1 - 0.4 = 0.6$.

      So $P(A \cup \bar{B}) = 0.7 + 0.6 - 0.5 = 0.8$.

   4. Compute the conditional probability

      $P(B \mid A \cup \bar{B}) = \frac{P(A \cap B)}{P(A \cup \bar{B})} = \frac{0.2}{0.8} = 0.25$.

   Tip

   A common mistake is to forget that $B \cap \bar{B} = \emptyset$ and try to compute $P(B \cap (A \cup \bar{B}))$ directly — but the distributive trick saves time and avoids errors.

   Watch out

   Do not confuse $P(A \cup \bar{B})$ with $P(A) + P(\bar{B})$ — they are not disjoint because $A$ and $\bar{B}$ can overlap (and indeed do, since $P(A \cap \bar{B}) = 0.5$).

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. 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 $P(A)$.

   $P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}=\frac{1}{5}+\frac{3}{10}=\frac{2}{10}+\frac{3}{10}=\frac{5}{10}=\frac{1}{2}.$

   Consistency check. With $P(A)=\tfrac12$, $P(B)=\tfrac35$, $P(A\cap B)=\tfrac{3}{10}$: note $P(A)P(B)=\tfrac12\cdot\tfrac35=\tfrac{3}{10}=P(A\cap B)$, so $A$ and $B$ are independent — entirely consistent with $P(A/B)=\tfrac12=P(A)$. And $P(A\cup B)=\tfrac12+\tfrac35-\tfrac3{10}=\tfrac{5+6-3}{10}=\tfrac{8}{10}=\tfrac45$. ✓

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{2}$.

   ANSWER: B

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

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

   Watch out

   A common mistake is to confuse $P(A-B)$ with $P(B^c)$ or to think $P(A-B) = P(A) - P(B)$. Remember: $A-B$ is only the part of $A$ that excludes $B$, not the whole complement of $B$.

   Tip

   Visualize a Venn diagram: $A$ is a circle split into the $B$ overlap and the rest. $P(A-B)$ is the “crescent” of $A$ outside $B$. Subtracting that from $P(A)$ gives the overlap directly.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. COMEDK 2023Set 2023-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } P(B)=\frac{3}{5} \quad P(A / B)=\frac{1}{2} \text { and } P(A \cup B)=\frac{4}{5} \text { then } P(A \cup B)^{\prime}+P\left(A^{\prime} \cup B\right)=$

   (A) $\frac{4}{5}$ (B) $\frac{1}{2}$ (C) 1 (D) $\frac{1}{5}$

   ›Reveal solutionSolution

   Step 5: sum = 1/5 + 4/5 = 1.

   Concept: conditional probability, addition theorem, complements, De Morgan.

   Given P(B) = 3/5, P(A|B) = 1/2, P(A U B) = 4/5.

   Step 1: P(A ^ B) = P(A|B) P(B) = (1/2)(3/5) = 3/10.

   Step 2: from P(A U B) = P(A) + P(B) - P(A ^ B),

   P(A) = 4/5 - 3/5 + 3/10 = 1/5 + 3/10 = 1/2.

   Step 3: P((A U B)') = 1 - P(A U B) = 1 - 4/5 = 1/5.

   Step 4: P(A' U B) = 1 - P((A' U B)') = 1 - P(A ^ B') [De Morgan]

   P(A ^ B') = P(A) - P(A ^ B) = 1/2 - 3/10 = 1/5.

   So P(A' U B) = 1 - 1/5 = 4/5.

   Step 5: sum = 1/5 + 4/5 = 1.

   ✓Final answer

   The correct option is (C) — 1

   ANSWER: C

5. 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 $12$:

   $P(A\cup B)=\frac{6}{12}+\frac{4}{12}-\frac{1}{12}=\frac{9}{12}=\frac34.$

   Step 3 — De Morgan's law.

   The event "neither $A$ nor $B$ occurs" is exactly the complement of "$A$ or $B$ occurs":

   $A'\cap B'=(A\cup B)'.$

   Hence

   $P(A'\cap B')=1-P(A\cup B)=1-\frac34=\frac14.$

   ✓Final answer

   The correct option is (D) — $1/4$.

   ANSWER: D

6. KCET 2025Set A-11 markMCQ
   Q.If A and B are two non-mutually exclusive events such that $P(A|B) = P(B|A)$, then (A) $A \subset B$ but $A \neq B$ (B) $A = B$ (C) $A \cap B = \phi$ (D) $P(A) = P(B)$

   ›Reveal solutionSolution

   Write both conditional probabilities with the same numerator $P(A \cap B)$, cancel it (legal because the events are not mutually exclusive), and the denominators must be equal.

   Step 1 — Definition of conditional probability.

   $P(A\mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B \mid A) = \frac{P(B \cap A)}{P(A)}.$

   Note $A \cap B = B \cap A$, so the two fractions share the same numerator.

   Step 2 — Impose the given condition.

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

   Step 3 — Cancel — and see why we are allowed to.

   The events are non-mutually-exclusive, i.e. $A \cap B \ne \phi$ and $P(A \cap B) \ne 0$. (This hypothesis is exactly what makes the cancellation valid — if $P(A\cap B)$ were 0, both sides would be 0 for any $P(A), P(B)$ and nothing would follow.) Dividing both sides by $P(A\cap B)$:

   $\frac{1}{P(B)} = \frac{1}{P(A)} \;\Longrightarrow\; P(A) = P(B).$

   Step 4 — Why the other options are not forced.

   Equality of probabilities does not force equality of sets. Example: toss a fair coin twice; let $A$ = "first toss is head", $B$ = "second toss is head". Then $P(A) = P(B) = 1/2$, $P(A\cap B) = 1/4 \neq 0$, and indeed $P(A|B) = P(B|A) = 1/2$ — yet $A \ne B$ and neither is a subset of the other. So (A) and (B) are not implied. (C) contradicts the given non-mutual-exclusivity.

   ✓Final answer

   The correct option is (D) — $P(A) = P(B)$.

   ANSWER: D

7. 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.

   First find $P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{1/6}{1/2} = \dfrac{1}{3}$. Since $A$ and $A'$ partition the space, $P(A|B) + P(A'|B) = 1$, so

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

   Both routes agree. (Note $P(A) = \tfrac13$ was not even needed — a useful reminder that a conditional probability given $B$ depends only on how $B$ is split.)

   ✓Final answer

   The correct option is (A) — $\dfrac{2}{3}$.

   ANSWER: A

8. KCET 2025Set A-11 markMCQ
   Q.If A and B are two events such that $A \subset B$ and $P(B) \neq 0$, then which of the following is correct? (A) $P(A|B) = \dfrac{P(B)}{P(A)}$ (B) $P(A|B) < P(A)$ (C) $P(A|B) \geq P(A)$ (D) $P(A) = P(B)$

   ›Reveal solutionSolution

   Use $A \subset B \Rightarrow A \cap B = A$, then note that dividing $P(A)$ by $P(B) \le 1$ cannot make it smaller.

   Step 1 — Simplify the intersection.

   If every element of $A$ lies in $B$, then $A \cap B = A$. Hence

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

   Step 2 — Compare with $P(A)$.

   Every probability satisfies $0 < P(B) \le 1$ (we are told $P(B) \ne 0$). Therefore $\dfrac{1}{P(B)} \ge 1$, and multiplying the non-negative number $P(A)$ by a factor $\ge 1$ gives

   $P(A\mid B) = P(A)\cdot\frac{1}{P(B)} \;\ge\; P(A).$

   Equality holds exactly when $P(B) = 1$ (or when $P(A) = 0$).

   Intuition: conditioning on $B$ throws away all outcomes outside $B$ — but none of $A$ lies outside $B$. So $A$'s share of the shrunken sample space can only grow.

   Step 3 — Eliminate.

   • (A) $P(A|B) = P(B)/P(A)$ — wrong; the ratio is upside down (and can exceed 1).
   • (B) $P(A|B) < P(A)$ — the inequality points the wrong way.
   • (D) $P(A) = P(B)$ — not implied; $A$ can be a strictly smaller subset, e.g. rolling a die with $A = \{2\}$, $B = \{2,4,6\}$: $P(A) = 1/6$, $P(B) = 1/2$, and $P(A|B) = 1/3 \ge 1/6$. ✓ consistent only with (C).

   ✓Final answer

   The correct option is (C) — $P(A \mid B) \ge P(A)$.

   ANSWER: C

9. 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:

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

   $P(A'\cap B)=P(B)-P(A\cap B)=\frac13-\frac16=\frac{1}{6}\quad(\text{C true})$

   Option (D) "$A$ and $B$ are independent" is true. Therefore the statement that is not true is (A) "$A$ and $B$ are not independent".

   ✓Final answer

   The statement that is not true is "$A$ and $B$ are not independent" — option (A).

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

       $= \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{24}$

    Since these three cases cannot happen together, we add them:

    $\frac{1}{12} + \frac{1}{8} + \frac{1}{24}$

    Convert to denominator 24:

    $\frac{2}{24} + \frac{3}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$

    Watch out

    If you mistakenly included the case where all three solve it, you would add $\frac{1}{2} \times \frac{1}{4} \times \frac{1}{3} = \frac{1}{24}$, giving $\frac{7}{24}$ — which is not even among the options. The phrase "any two" excludes the all-three case.

    ✓Final answer

    The probability is $\boxed{\frac{1}{4}}$, which corresponds to option (B).

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

    Common denominator 12: $\frac{3}{12} + \frac{2}{12} - \frac{1}{12} = \frac{4}{12} = \frac{1}{3}.$

    So $P(\overline{P} \cap \overline{Q}) = 1 - \frac{1}{3} = \frac{2}{3}.$

    6. Compute the desired conditional probability.

    $P(\overline{P} \mid \overline{Q}) = \frac{P(\overline{P} \cap \overline{Q})}{P(\overline{Q})} = \frac{\frac{2}{3}}{\frac{5}{6}} = \frac{2}{3} \cdot \frac{6}{5} = \frac{12}{15} = \frac{4}{5}.$

    Watch out

    A common mistake is to assume $P(P \mid Q)$ and $P(Q \mid P)$ are reciprocals or that $P(P \cap Q)$ can be found by multiplying $P(P)$ and $P(Q)$ directly — that only works for independent events, which is not the case here.

    Tip

    You can also solve this by constructing a 2×2 probability table. From $P(P)=\frac14$ and $P(P \cap Q)=\frac1{12}$, fill in the rest systematically. The final answer is the same.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. 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.

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

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

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

    Step 3 — Total probability of both accessories (denominator).

    $P(H\cap S)=(0.2)(0.40)+(0.4)(0.20)+(0.4)(0.05)=0.08+0.08+0.02=0.18.$

    Step 4 — Posterior for Red.

    $P(R\mid H\cap S)=\frac{P(R)\,P(H\cap S\mid R)}{P(H\cap S)}=\frac{0.08}{0.18}=\frac{8}{18}=\frac{4}{9}.$

    ✓Final answer

    $P(\text{Red}\mid H\cap S)=\dfrac{4}{9}$ — option (C).

    ANSWER: C