Q.A person has undertaken a construction job. The probabilities are 0.65 that there will be a strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.

Showing the 12 most recent of 45 on this concept.

1. 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

2. 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).

3. 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.

   $P(X\mid S) = \frac{0.56}{0.83} = \frac{56}{83}$

   Step 6 — Sanity check. The posterior $56/83 \approx 0.675$ is a little below the prior $0.70$ — which makes sense, because plant Y has the higher standard-quality rate ($90\% > 80\%$), so learning the car is standard shifts a little belief towards Y. ✓

   (The distractors 73, 84, 79 are what you get by mis-adding $0.56 + 0.27$; the denominator must be the total probability $0.83$.)

   ✓Final answer

   The correct option is (C) $\frac{56}{83}$.

   ANSWER: C

4. 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

5. 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

6. 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):

   $P = \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{1}{3} = \frac{2}{36} = \frac{1}{18}$

   Path 3 — $L_1 L_2 W_3$ (loss, then loss after a loss, then win after a loss):

   $P = \frac{2}{3}\cdot\frac{3}{4}\cdot\frac{1}{4} = \frac{6}{48} = \frac{1}{8}$

   Total (common denominator $72$):

   $\frac{1}{12}+\frac{1}{18}+\frac{1}{8} = \frac{6}{72}+\frac{4}{72}+\frac{9}{72} = \frac{19}{72}$

   ✓Final answer

   $P(\text{Tactical medal}) = \dfrac{19}{72}$ — option (D).

7. 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

8. 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

9. COMEDK 2026Set 2026-M1 markMCQ
   Q.A teacher has two jars of candy on her desk: Jar 1: Contains 3 Strawberry candies and 2 Orange candies. Jar 2: Contains 1 Strawberry candy and 4 Orange candies. The teacher randomly picks two candies from Jar 1 and drops them into Jar 2. Then, a student reaches into Jar 2 and picks two candies. What is the probability that the student picks two Strawberry candies? (A) $\frac{6}{35}$ (B) $\frac{4}{21}$ (C) $\frac{3}{70}$ (D) $\frac{1}{14}$

   ›Reveal solutionSolution

   Condition on how many strawberries move from Jar 1 to Jar 2, then compute the chance of drawing two strawberries from the now 7-candy Jar 2. Total $=\frac{1}{14}$.

   Setup. Jar 1 has $3$ Strawberry (S) and $2$ Orange (O). Two candies are moved into Jar 2, which started with $1$ S and $4$ O. After the transfer Jar 2 holds $7$ candies. Let $k$ = number of strawberries transferred.

   Transfer probabilities (choosing $2$ of $5$ from Jar 1, $\binom{5}{2}=10$):

   $P(k{=}2)=\frac{\binom{3}{2}}{10}=\frac{3}{10},\quad P(k{=}1)=\frac{\binom{3}{1}\binom{2}{1}}{10}=\frac{6}{10},\quad P(k{=}0)=\frac{\binom{2}{2}}{10}=\frac{1}{10}.$

   Draw two S from Jar 2 (which now has $1+k$ strawberries out of $7$, $\binom{7}{2}=21$):

   • $k=2$: Jar 2 has $3$ S $\Rightarrow \dfrac{\binom{3}{2}}{21}=\dfrac{3}{21}=\dfrac{1}{7}$.
   • $k=1$: Jar 2 has $2$ S $\Rightarrow \dfrac{\binom{2}{2}}{21}=\dfrac{1}{21}$.
   • $k=0$: Jar 2 has $1$ S $\Rightarrow \dfrac{\binom{1}{2}}{21}=0$.

   Total probability.

   $P=\frac{3}{10}\cdot\frac{1}{7}+\frac{6}{10}\cdot\frac{1}{21}+\frac{1}{10}\cdot 0=\frac{3}{70}+\frac{1}{35}=\frac{3}{70}+\frac{2}{70}=\frac{5}{70}=\frac{1}{14}.$

   ✓Final answer

   The probability is $\dfrac{1}{14}$ — option (D).

10. COMEDK 2025Set 2025-E1 markMCQ
    Q.A pot contains 5 red and 2 green balls. A ball is drawn at random from this pot. If a drawn ball is green, then a red ball is added to the pot. If a drawn ball is red, then a green ball is added to the pot, while the original ball drawn is not replaced in the pot. Now a second ball is drawn at random from the pot, what is the probability that the second ball drawn is a red ball? (A) $\frac{12}{49}$ (B) $\frac{32}{49}$ (C) $\frac{3}{7}$ (D) $\frac{27}{49}$

    ›Reveal solutionSolution

    Condition on the first draw: $P(\text{2nd red}) = \tfrac{2}{7}\cdot\tfrac{6}{7} + \tfrac{5}{7}\cdot\tfrac{4}{7} = \tfrac{32}{49}$.

    Start: $5$ red, $2$ green, total $7$. The first ball is not replaced, and a ball of the other colour is added.

    Case 1 — first ball is green. $P = \dfrac{2}{7}$.

    Remove that green ($5R,1G$), then add a red $\Rightarrow 6R,1G$ (total $7$).

    $P(\text{2nd red}\mid \text{green first}) = \frac{6}{7}$

    Case 2 — first ball is red. $P = \dfrac{5}{7}$.

    Remove that red ($4R,2G$), then add a green $\Rightarrow 4R,3G$ (total $7$).

    $P(\text{2nd red}\mid \text{red first}) = \frac{4}{7}$

    Total probability:

    $P(\text{2nd red}) = \frac{2}{7}\cdot\frac{6}{7} + \frac{5}{7}\cdot\frac{4}{7} = \frac{12}{49} + \frac{20}{49} = \frac{32}{49}$

    ✓Final answer

    $P(\text{second ball red}) = \dfrac{32}{49}$ — option (B).

11. 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.

    $P(E\mid R)=\frac{P(R\mid E)P(E)}{P(R\mid E)P(E)+P(R\mid E^c)P(E^c)} =\frac{\frac{2}{3}\cdot\frac{3}{7}}{\frac{2}{3}\cdot\frac{3}{7}+\frac{1}{3}\cdot\frac{4}{7}} =\frac{\frac{6}{21}}{\frac{6}{21}+\frac{4}{21}}=\frac{6}{10}=\frac{3}{5}.$

    ✓Final answer

    The probability that the number is actually even is $\dfrac{3}{5}$ — option (D).

12. 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).

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    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.

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    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 $a = b$.

       Among the sum-12 pairs, the only equal pair is $(6,6)$.

       So there is exactly 1 favorable outcome.

    4. Apply conditional probability

    $P(\text{equal} \mid \text{sum}=12) = \frac{\text{number of equal pairs with sum 12}}{\text{total pairs with sum 12}} = \frac{1}{5}.$

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    Watch out

    A common mistake is to compute $\frac{1}{64}$ (since only one equal pair overall sums to 12 out of all 64 pairs). But the condition “given sum = 12” changes the denominator to 5, not 64.

    Tip

    Always re-read: “given that the sum is 12” means we only care about the 5 outcomes listed. Conditional probability shrinks the universe.

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    ✓Final answer

    The correct option is (B).

    ANSWER: B