Q.A and B are two events such that $P(A) \neq 0$. Find $P(B|A)$, if

Showing the 12 most recent of 45 on this concept.

1. 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

2. 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

3. 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

4. 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).

5. 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

6. 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

7. 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

8. KCET 2020Set A-11 markMCQ
   Q.Events $E_1$ and $E_2$ form a partition of the sample space S. A is any event such that $P(E_1) = P(E_2) = \frac{1}{2}$, $P(E_2/A) = \frac{1}{2}$ and $P(A/E_2) = \frac{2}{3}$, then $P(E_1/A)$ is (A) $\frac{1}{2}$ (B) $\frac{2}{3}$ (C) $1$ (D) $\frac{1}{4}$

   ›Reveal solutionSolution

   A partition's posterior probabilities must add to 1, so $P(E_1|A) = 1 - P(E_2|A) = 1 - \tfrac12 = \tfrac12$.

   Step 1 — What "partition of the sample space" means.

   $E_1$ and $E_2$ form a partition of $S$ if they are mutually exclusive ($E_1 \cap E_2 = \varnothing$) and exhaustive ($E_1 \cup E_2 = S$), with non-zero probabilities. So exactly one of them must occur.

   Step 2 — Conditioning preserves the partition.

   Conditioning on an event $A$ (with $P(A) > 0$) just renormalises probabilities inside $A$; it does not destroy the partition. Formally,

   $A = (A \cap E_1) \cup (A \cap E_2), \qquad (A\cap E_1)\cap(A\cap E_2) = \varnothing$

   Dividing by $P(A)$:

   $\frac{P(A\cap E_1)}{P(A)} + \frac{P(A\cap E_2)}{P(A)} = \frac{P(A)}{P(A)} = 1$

   which is precisely

   $P(E_1|A) + P(E_2|A) = 1$

   The posterior probabilities of a partition must still sum to 1 — a fact worth remembering, since it turns many Bayes questions into a one-line subtraction.

   Step 3 — Substitute the given value.

   $P(E_1|A) = 1 - P(E_2|A) = 1 - \frac{1}{2} = \frac{1}{2}$

   Step 4 — Note the redundant data (a deliberate distraction).

   $P(E_1) = P(E_2) = \tfrac12$ and $P(A|E_2) = \tfrac23$ are not needed. They are consistent, though — Bayes' theorem gives

   $P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} = \frac12$

   which forces $P(A|E_1) = P(A|E_2) = \tfrac23$. Intuitively: if $A$ is equally likely under either hypothesis, observing $A$ tells you nothing new, so the posteriors stay at the priors, $\tfrac12$ each. ✓

   ✓Final answer

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

   ANSWER: A

9. 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

10. 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

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 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:

    $P(B_3 \mid W) = \frac{\frac{3}{6} \cdot 1}{\frac{11}{18}} = \frac{\frac{1}{2}}{\frac{11}{18}} = \frac{1}{2} \cdot \frac{18}{11} = \frac{9}{11}.$

    Tip

    Notice that Bag I contributes zero probability of a white ball, so it can be ignored entirely in the numerator — it only affects the denominator by adding zero. This often simplifies Bayes’ calculations: only bags that can produce the observed outcome matter.

    Watch out

    A common mistake is to forget that the prior probabilities $\frac{i}{6}$ are not equal — they favor higher-numbered bags. If you mistakenly treated all bags as equally likely, you would get $\frac{1}{2}$ instead of $\frac{9}{11}$.

    Thus, given that a white ball was drawn, the probability it came from Bag III is $\frac{9}{11}$.

    ✓Final answer

    The correct option is (A).

    ANSWER: A