Q.Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Showing the 12 most recent of 45 on this concept.

1. COMEDK 2024Set 2024-E1 markMCQ
   Q.Suppose we have three cards identical in form except that both sides of the first card are coloured red, both sides of the second are coloured black, and one side of the third card is coloured red and the other side is coloured black. The three cards are mixed and a card is picked randomly. If the upper side of the chosen card is coloured red, what is the probability that the other side is coloured black. (A) $\frac{1}{6}$ (B) $\frac{1}{2}$ (C) 0 (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   This is a classic conditional probability problem (often called the "three cards" or "Bertrand's box" variant). The key is that seeing a red side updates the probability space to only the red sides, and among those, only one belongs to the mixed card. The answer is 1/3.

   We are asked: given that the visible side is red, what is the probability that the other side is black? This is not simply "one of the two remaining cards has a black other side" because the cards are not equally likely once we condition on the observation.

   ---

   Concept and Intuition

   The pitfall is to think: "We see red, so the card is either the all-red or the mixed card. That's two possibilities, so the chance is 1/2." But this ignores that the all-red card has two red sides, while the mixed card has only one red side. When we pick a card at random and then look at a random side, each of the six sides is equally likely to be the one we see. Seeing red eliminates the three black sides, leaving only the three red sides. Among those three red sides, two belong to the all-red card and only one belongs to the mixed card. So the probability that the other side is black is 1/3.

   ---

   Step-by-step reasoning

   1. Label the sides.

      Let the cards be:

      • Card A: both sides red (sides $R_1, R_2$)
      • Card B: both sides black (sides $B_1, B_2$)
      • Card C: one red, one black (sides $R_3, B_3$)

   2. Count equally likely outcomes.

      When we pick a card uniformly at random and then look at a random side, there are $3 \times 2 = 6$ equally likely side-views. Each of the six sides has probability $1/6$ of being the one we see.

   3. Condition on seeing a red side.

      The red sides are: $R_1, R_2$ (from card A) and $R_3$ (from card C). That's 3 red sides. So the conditional space has 3 equally likely possibilities.

   4. Identify which of these have the other side black.

      Only the red side $R_3$ (from card C) has a black other side. The other two red sides ($R_1, R_2$) have red on the other side.

   5. Compute the conditional probability.

   $P(\text{other side black} \mid \text{top side red}) = \frac{\text{number of favorable red sides}}{\text{total number of red sides}} = \frac{1}{3}.$

   Watch out

   The common mistake is to forget that the all-red card contributes two red sides to the sample space, making it twice as likely as the mixed card once we condition on seeing red.

   Tip

   A neat way to think: "Probability = (number of red sides with black opposite) / (total number of red sides)." This avoids the card-counting trap entirely.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. COMEDK 2024Set 2024-M1 markMCQ
   Q.An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is replaced along with another ball of the same colour. The process is repeated. The probability that the third ball drawn is black is (A) $\frac{17}{30}$ (B) $\frac{37}{60}$ (C) $\frac{31}{60}$ (D) $\frac{23}{30}$

   ›Reveal solutionSolution

   Conditioning on the first two draws (white -> not replaced; black -> replaced plus one extra black) and using total probability gives $P(\text{3rd black}) = \tfrac{23}{30}$ — option (D).

   Rules. Start with 2 white, 2 black (total 4). Drawing white removes it (whites $-1$, total $-1$). Drawing black puts it back and adds one more black (blacks $+1$, total $+1$).

   First two draws and the composition just before the 3rd draw

   • WW: $\tfrac24\cdot\tfrac13 = \tfrac16$; urn becomes W0, B2 (total 2) $\Rightarrow P(\text{black}) = 1$.
   • WB: $\tfrac24\cdot\tfrac23 = \tfrac13$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BW: $\tfrac24\cdot\tfrac25 = \tfrac15$; urn becomes W1, B3 (total 4) $\Rightarrow P(\text{black}) = \tfrac34$.
   • BB: $\tfrac24\cdot\tfrac35 = \tfrac{3}{10}$; urn becomes W2, B4 (total 6) $\Rightarrow P(\text{black}) = \tfrac23$.

   (The four path probabilities sum to $\tfrac16+\tfrac13+\tfrac15+\tfrac{3}{10} = 1$.)

   Total probability

   $P(\text{3rd black}) = \tfrac16(1) + \tfrac13\cdot\tfrac34 + \tfrac15\cdot\tfrac34 + \tfrac{3}{10}\cdot\tfrac23$

   $= \frac{10}{60} + \frac{15}{60} + \frac{9}{60} + \frac{12}{60} = \frac{46}{60} = \frac{23}{30}$

   ✓Final answer

   $P(\text{third ball is black}) = \dfrac{23}{30}$ — option (D).

   ANSWER: D

3. COMEDK 2021Set 2021-B1 markMCQ
   Q.An urn contains 5 red and 5 black coloured balls. A ball is picked at random, its colour noted and then is put back into the urn. Now, additional two balls of the same colour are put into the urn and then a ball is drawn at random. What is the probability that the ball now chosen is red. (A) 7/24 (B) 1/2 (C) 5/8 (D) 2/3

   ›Reveal solutionSolution

   By symmetry the second-draw red probability averages to $\tfrac12$.

   Start: 5 red, 5 black (10 total). First ball is drawn, replaced, then 2 more of the same colour are added, so the urn always has 12 balls before the second draw.

   • First ball red ($P=\tfrac{5}{10}=\tfrac12$): urn now 7R, 5B ⇒ $P(\text{red})=\tfrac{7}{12}$.
   • First ball black ($P=\tfrac12$): urn now 5R, 7B ⇒ $P(\text{red})=\tfrac{5}{12}$.

   Total probability:

   $\frac12\cdot\frac{7}{12}+\frac12\cdot\frac{5}{12}=\frac{7+5}{24}=\frac{12}{24}=\frac12.$

   ✓Final answer

   The correct option is (B) — 1/2

4. 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).

5. COMEDK 2025Set 2025-E1 markMCQ
   Q.Two numbers are selected at random from integers 1 to 9 . If their sum is even, what is the probability that both the numbers are odd? (A) $\frac{4}{9}$ (B) $\frac{5}{8}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   This is a conditional probability problem: given that the sum of two numbers from 1–9 is even, we want the probability both are odd. The answer is 5/8, option (B).

   We are selecting two numbers from 1 to 9 without replacement (since "selected at random" from distinct integers usually implies no repetition). The sum is even only if both numbers are odd or both are even. So the condition restricts us to those pairs. The question asks: among those pairs with an even sum, what fraction consists of two odd numbers?

   ---

   1. Count total possible pairs (without replacement)

   From 1 to 9, there are 9 numbers. The number of ways to choose any two distinct numbers is

   $\binom{9}{2} = 36.$

   2. Count pairs with an even sum

   A sum is even when both numbers have the same parity.

   • Odd numbers from 1 to 9: 1, 3, 5, 7, 9 → 5 odds.

     Number of odd–odd pairs: $\binom{5}{2} = 10$.

   • Even numbers from 1 to 9: 2, 4, 6, 8 → 4 evens.

     Number of even–even pairs: $\binom{4}{2} = 6$.

   So total pairs with an even sum:

   $10 + 6 = 16.$

   3. Apply conditional probability

   We want

   $P(\text{both odd} \mid \text{sum even}) = \frac{\text{Number of odd–odd pairs}}{\text{Number of even-sum pairs}} = \frac{10}{16} = \frac{5}{8}.$

   Tip

   A common mistake is to treat this as an unconditional probability and compute $\frac{10}{36}$, which gives $\frac{5}{18}$ — not even among the options. The condition "given sum is even" changes the denominator from 36 to 16.

   Watch out

   Another pitfall: forgetting that selection is without replacement. If replacement were allowed, the counts would differ, but here the problem implies distinct integers.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

6. 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).

   ---

   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.

   ---

   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

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

   ---

   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.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. COMEDK 2026Set 2026-A1 markMCQ
   Q.Vishnu has two jars of marbles, Jar A and Jar B. Jar A contains 3 yellow marbles and 2 green marbles. Jar B contains 4 yellow marbles and 3 green marbles. Vishnu flips a fair coin. If it lands heads, he picks two marbles at random without replacement from Jar A. If it lands tails, he picks two marbles at random with replacement from Jar B. Given that Vishnu picked one yellow and one green marble, what is the probability that they came from Jar B? (A) $\frac{21}{41}$ (B) $\frac{49}{89}$ (C) $\frac{40}{89}$ (D) $\frac{20}{41}$

   ›Reveal solutionSolution

   $P(E\mid A)=\tfrac35$ (without replacement), $P(E\mid B)=\tfrac{24}{49}$ (with replacement); Bayes gives $P(B\mid E)=\dfrac{40}{89}$ — option (C).

   Likelihoods of drawing one yellow and one green ($E$).

   Jar A (3 yellow, 2 green; two draws without replacement):

   $P(E\mid A)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{6}{10}=\frac{3}{5}.$

   Jar B (4 yellow, 3 green; two draws with replacement):

   $P(E\mid B)=2\cdot\frac{4}{7}\cdot\frac{3}{7}=\frac{24}{49}.$

   Bayes' theorem with $P(A)=P(B)=\tfrac12$ (the $\tfrac12$ cancels):

   $P(B\mid E)=\frac{P(E\mid B)}{P(E\mid A)+P(E\mid B)}=\frac{\frac{24}{49}}{\frac{3}{5}+\frac{24}{49}}.$

   Combine the denominator over $245$:

   $\frac{3}{5}+\frac{24}{49}=\frac{147+120}{245}=\frac{267}{245}.$

   $P(B\mid E)=\frac{24}{49}\cdot\frac{245}{267}=\frac{24\cdot 5}{267}=\frac{120}{267}=\frac{40}{89}.$

   ✓Final answer

   $P(B\mid E)=\dfrac{40}{89}$, which is option (C).

8. 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

9. 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

   $(2,6),(3,5),(4,4),(5,3),(6,2)\quad(5\text{ outcomes}).$

   Those in which $5$ appears at least once: $(3,5)$ and $(5,3)$ — $2$ outcomes.

   $P(\text{5 appears}\mid \text{sum}=8)=\frac{2}{5}.$

   ✓Final answer

   The correct option is (B) — $\frac{2}{5}$

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.

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

    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

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. 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