Q.Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Two digits are selected at random from the digits 1 through 9. If their sum is even, then the probability that both are odd is (A) $\dfrac{3}{8}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{5}{8}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   Conditioning on 'sum even' restricts to same-parity pairs; among those, the odd-odd pairs make up $5/8$.

   Concept and Intuition

   Two digits sum to an even number exactly when they have the same parity (both odd or both even) — an odd plus an even is always odd. This turns the conditional-probability question into a simple counting problem over same-parity pairs only.

   Step-by-Step Solution

   1. Digits $1$ through $9$: odd $=\{1,3,5,7,9\}$ (5 digits), even $=\{2,4,6,8\}$ (4 digits).
   2. Sum is even $\iff$ both digits are odd, or both are even.
   3. Number of ways to choose 2 odd digits: $\binom52=10$.
   4. Number of ways to choose 2 even digits: $\binom42=6$.
   5. Total ways with even sum: $10+6=16$.
   6. $P(\text{both odd}\mid\text{sum even}) = \dfrac{\text{both odd}}{\text{sum even}} = \dfrac{10}{16}=\dfrac58$.

   Common Mistakes

   • Forgetting that "sum even" excludes the mixed odd-even case entirely, so the sample space for the conditional probability is only the 16 same-parity pairs, not all $\binom92=36$ pairs.
   • Miscounting the number of odd digits in $1$-$9$ (it's 5, not 4).

   ✓Final answer

   The correct option is (C) — $\dfrac58$.

   ANSWER: C

2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Box I contains 30 cards numbered 1 to 30 and Box II contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it randomly. If the number on the card is found to be a non-prime number, the probability that the card was drawn from Box I is (A) $\dfrac{4}{17}$ (B) $\dfrac{8}{17}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Bayes' theorem with the counts of non-prime numbers in each box gives $P(\text{Box I}\mid\text{non-prime})=8/17$.

   Concept and Intuition

   This is a classic "which urn/box did it come from" Bayes' problem: we're given the outcome (a non-prime card) and asked to find the probability of the cause (which box). We need $P(\text{non-prime}\mid\text{box})$ for each box, weighted by the prior $P(\text{box})=1/2$.

   Step-by-Step Solution

   1. Primes from 1 to 30: $2,3,5,7,11,13,17,19,23,29$ — that's $10$ primes, so $30-10=20$ non-primes in Box I.
   2. Primes from 31 to 50: $31,37,41,43,47$ — that's $5$ primes, so $20-5=15$ non-primes in Box II.
   3. $P(\text{non-prime}\mid\text{Box I})=\dfrac{20}{30}=\dfrac23$, $P(\text{non-prime}\mid\text{Box II})=\dfrac{15}{20}=\dfrac34$.
   4. $P(\text{Box I})=P(\text{Box II})=\dfrac12$.
   5. $P(\text{Box I and non-prime})=\dfrac12\cdot\dfrac23=\dfrac13$; $P(\text{Box II and non-prime})=\dfrac12\cdot\dfrac34=\dfrac38$.
   6. $P(\text{non-prime})=\dfrac13+\dfrac38=\dfrac{8}{24}+\dfrac{9}{24}=\dfrac{17}{24}$.
   7. $P(\text{Box I}\mid\text{non-prime})=\dfrac{1/3}{17/24}=\dfrac13\cdot\dfrac{24}{17}=\dfrac{8}{17}$.

   Common Mistakes

   • Forgetting that $1$ is not prime (it's non-prime), which would miscount Box I's non-primes.
   • Skipping the proper Bayes' normalization and just comparing raw counts across boxes of different sizes.

   ✓Final answer

   The correct option is (B) — $\dfrac{8}{17}$.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If two cards are drawn at a time at random from a well shuffled pack of 52 playing cards and they are found to be a king card and a card with prime number, then the probability that they are a black king card and a card with an odd prime number is (A) $\dfrac{32}{663}$ (B) $\dfrac{12}{663}$ (C) $\dfrac{3}{8}$ (D) $\dfrac{5}{8}$

   ›Reveal solutionSolution

   A conditional-probability counting problem: restrict the sample space to (King, prime-numbered-card) pairs, then count how many of those pairs are (black King, odd-prime card).

   Concept and Intuition

   A standard deck has 4 suits (2 black: spades, clubs; 2 red: hearts, diamonds), each with 13 ranks: A, 2–10, J, Q, K. "Cards with a prime number" means cards whose rank value is a prime, i.e. rank 2, 3, 5, or 7 — four ranks × 4 suits = 16 cards. Kings are a separate rank (not a "number" card), 4 total, 2 of them black (spade, club). Since we are told the two drawn cards are exactly one King and one prime-numbered card, we treat every (King, prime-card) pairing as equally likely and count favourable outcomes among them.

   Step-by-Step Solution

   1. Total King cards = 4; total prime-numbered cards (ranks 2,3,5,7) = $4\times4=16$.
   2. Sample space size (ways to have one King and one prime card) = $4\times16=64$.
   3. "Odd prime number" cards are ranks 3, 5, 7 (2 is the only even prime, excluded): $3\times4=12$ cards.
   4. Black Kings = 2 (spade King, club King).
   5. Favourable outcomes = (black King) × (odd-prime card) = $2\times12=24$.
   6. Required probability $=\dfrac{24}{64}=\dfrac{3}{8}$.

   Common Mistakes

   • Including the Ace as a "number 1" and mistakenly treating 1 as prime — 1 is not prime and Ace is not a numbered card in this context.
   • Including 2 among the "odd primes" — 2 is prime but even, so it is excluded from "odd prime."
   • Forgetting only 2 of the 4 Kings are black.

   ✓Final answer

   The correct option is (C) — $\dfrac{3}{8}$.

   ANSWER: C

4. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.One ticket is selected at random from 50 tickets numbered $00, 01, 02, \ldots 49$. The probability that sum of the digits is 10, given that product of the digits is 9 is (A) $\dfrac{9}{10}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{2}{25}$

   ›Reveal solutionSolution

   Only two tickets (19 and 33) have digit-product 9, and only one of them (19) also has digit-sum 10, giving conditional probability $1/2$.

   Concept and Intuition

   Conditional probability $P(\text{sum}=10\mid\text{product}=9)$ restricts attention entirely to the tickets satisfying the "given" condition (product $=9$), then asks what fraction of those also satisfy the target condition (sum $=10$).

   Step-by-Step Solution

   1. List all two-digit combinations $(d_1,d_2)$ with $d_1\in\{0,1,2,3,4\}$ (tens digit, since tickets run 00–49) and $d_2\in\{0,\dots,9\}$ (units digit) whose product is 9:
      • $d_1=1,d_2=9$: ticket $19$.
      • $d_1=3,d_2=3$: ticket $33$.
      • No other integer pairs with $d_1\le4$ give product 9.
   2. So the "product = 9" event has exactly 2 tickets: $\{19, 33\}$.
   3. Check digit sums: $19\to1+9=10$ ✓; $33\to3+3=6$ ✗.
   4. Only 1 out of these 2 tickets also has digit-sum 10.
   5. $P(\text{sum}=10\mid\text{product}=9)=1/2$.

   Common Mistakes

   • Missing that the tens digit is capped at 4 (tickets only go up to 49), which rules out some product-9 pairs that would exist for a full two-digit range.
   • Computing an unconditional probability (over all 50 tickets) instead of restricting to the given-condition subset.

   ✓Final answer

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

   ANSWER: C

5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Two balls are drawn at random from a box containing 4 white, 6 black balls one after the other without replacement. If it is known that second ball drawn is black, then the probability that the first ball drawn is also black is (A) $\dfrac{5}{11}$ (B) $\dfrac{5}{9}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{5}{13}$

   ›Reveal solutionSolution

   Conditional probability with sampling without replacement, solved with Bayes' theorem; answer is $\frac{5}{9}$.

   Concept and Intuition

   When balls are drawn one after another without replacement, the marginal probability that any particular draw (say the 2nd) is black equals the overall proportion of black balls, $\frac{6}{10}$ — position doesn't matter for the marginal event by symmetry. To find the conditional probability of the first draw given information about the second, we use Bayes' theorem: we need the joint probability of both events and divide by the marginal probability of the conditioning event.

   Step-by-Step Solution

   1. Total balls: 4 white (W) + 6 black (B) = 10.
   2. $P(\text{2nd is black}) = P(\text{1st B, 2nd B}) + P(\text{1st W, 2nd B})$ $= \frac{6}{10}\cdot\frac{5}{9} + \frac{4}{10}\cdot\frac{6}{9} = \frac{30}{90}+\frac{24}{90} = \frac{54}{90} = \frac{3}{5}$.
   3. $P(\text{1st black and 2nd black}) = \frac{6}{10}\cdot\frac{5}{9} = \frac{30}{90} = \frac{1}{3}$.
   4. By Bayes' theorem: $P(\text{1st black}\mid \text{2nd black}) = \dfrac{P(\text{1st B, 2nd B})}{P(\text{2nd B})} = \dfrac{1/3}{3/5} = \dfrac{5}{9}$.

   Common Mistakes

   • Assuming $P(\text{2nd black}) = \frac{5}{9}$ (the conditional probability given 1st was black) instead of computing the correct marginal $\frac{3}{5}$.
   • Forgetting to include both cases (1st W/2nd B and 1st B/2nd B) when finding the joint denominator event.

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{9}$.

   ANSWER: B

6. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Two dice are rolled. If A denote the event that the same number shows on each die and B denote the event that the sum of the numbers on both dice is greater than 7, then $P(A|B)$ and $P(B|A)$ respectively are (A) $\dfrac{2}{5}, \dfrac{1}{4}$ (B) $\dfrac{1}{5}, \dfrac{1}{2}$ (C) $\dfrac{1}{5}, \dfrac{1}{4}$ (D) $\dfrac{1}{2}, \dfrac{3}{5}$

   ›Reveal solutionSolution

   Counting the 36 equally likely dice outcomes directly gives $P(A|B)=\frac15$ and $P(B|A)=\frac12$.

   Concept and Intuition

   With two fair dice there are $36$ equally likely outcomes, so every probability here reduces to simple counting: enumerate the outcomes in $A$, in $B$, and in $A\cap B$, then apply the conditional probability formula $P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$ directly as a ratio of counts.

   Step-by-Step Solution

   1. $A$ = "same number on each die": outcomes $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$ — $6$ outcomes, so $P(A)=\dfrac6{36}$.
   2. $B$ = "sum $>7$", i.e. sum $\in\{8,9,10,11,12\}$. Counting pairs per sum: sum 8 has 5 pairs, 9 has 4, 10 has 3, 11 has 2, 12 has 1 — total $5+4+3+2+1=15$ outcomes, so $P(B)=\dfrac{15}{36}$.
   3. $A\cap B$: doubles with sum $>7$ — check each double: $(4,4)$ sum 8 ✓, $(5,5)$ sum 10 ✓, $(6,6)$ sum 12 ✓; $(1,1),(2,2),(3,3)$ have sums $2,4,6$, all $\le7$ — excluded. So $A\cap B$ has $3$ outcomes, $P(A\cap B)=\dfrac3{36}$.
   4. $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{3/36}{15/36}=\dfrac3{15}=\dfrac15$.
   5. $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{3/36}{6/36}=\dfrac36=\dfrac12$.

   Common Mistakes

   • Miscounting the number of outcomes with sum $>7$ (forgetting sum $=8$ is included since $>7$ means $\ge8$).
   • Swapping $P(A|B)$ and $P(B|A)$ in the final answer.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{5}, \dfrac{1}{2}$.

   ANSWER: B

7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.A family consists of 8 persons. If 4 persons are chosen at random and they are found to be 2 men and 2 women, then the probability that there are equal number of men and women in that family is (A) $\dfrac{1}{5}$ (B) $\dfrac{3}{7}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{2}{7}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem: given the observed sample (2 men, 2 women out of 4 drawn), find the posterior probability that the family itself is evenly split (4 men, 4 women), by weighting each possible family composition's likelihood of producing that sample.

   Concept and Intuition

   Before drawing, we don't know how many of the 8 family members are men — call it $k$ ($k$ can range from 0 to 8, but only $k=2,3,4,5,6$ can possibly yield a sample of 2 men and 2 women out of 4 drawn, since we need at least 2 men and at least 2 women in the family). Treating each feasible value of $k$ as equally likely a priori, Bayes' theorem says: $P(k=4\mid\text{observed 2M,2W}) = \dfrac{P(\text{observed}\mid k=4)}{\sum_{k} P(\text{observed}\mid k)}$, since the priors cancel when they're equal.

   Step-by-Step Solution

   1. For a family with $k$ men and $8-k$ women, the probability of drawing exactly 2 men and 2 women in a sample of 4 (hypergeometric) is proportional to $\binom{k}{2}\binom{8-k}{2}$ (the $\binom{8}{4}$ denominator is common to all $k$ and cancels in the ratio).
   2. Only $k=2,3,4,5,6$ give a nonzero value (need $k\ge2$ and $8-k\ge2$).
   3. Compute $L(k)=\binom{k}{2}\binom{8-k}{2}$ for each:
      • $k=2$: $\binom22\binom62 = 1\times15=15$
      • $k=3$: $\binom32\binom52 = 3\times10=30$
      • $k=4$: $\binom42\binom42 = 6\times6=36$
      • $k=5$: $\binom52\binom32 = 10\times3=30$
      • $k=6$: $\binom62\binom22 = 15\times1=15$
   4. Sum of likelihoods $=15+30+36+30+15=126$.
   5. With equal priors on each feasible $k$, the posterior probability of $k=4$ (equal numbers of men and women in the family) is $\dfrac{L(4)}{\sum L(k)} = \dfrac{36}{126}=\dfrac{2}{7}$.

   Common Mistakes

   • Confusing this with a simple hypergeometric probability calculation for a KNOWN family composition, rather than recognising it needs Bayesian updating over the UNKNOWN composition.
   • Weighting the prior by $\binom{8}{k}$ (as if each person's gender were an independent fair coin flip) instead of treating each feasible composition as equally likely — the latter is what reproduces one of the given options exactly.

   ✓Final answer

   The correct option is (D) — $\dfrac27$.

   ANSWER: D

8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.A die is thrown twice. Let A be the event of getting a prime number when the die is thrown first time and B be the event of getting an even number when the die is thrown second time. Then $P(A/\bar B)=$ (A) $\dfrac{1}{2}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Since $A$ and $B$ (hence $\bar B$) come from two independent throws of the die, conditioning on $\bar B$ does not change $P(A)$; the answer is $1/2$.

   Concept and Intuition

   When two events are determined by physically independent trials (first throw vs second throw of a die), any conditional probability between them collapses to the unconditional probability — conditioning on an independent event changes nothing.

   Step-by-Step Solution

   1. $A$: prime number on the first throw, i.e. outcome in $\{2,3,5\}$, so $P(A)=\dfrac{3}{6}=\dfrac{1}{2}$.
   2. $B$: even number on the second throw, i.e. outcome in $\{2,4,6\}$, so $P(B)=\dfrac{1}{2}$, and $P(\bar B)=\dfrac{1}{2}$ (odd on the second throw).
   3. Because the first and second throws are independent, $A$ (a first-throw event) and $\bar B$ (a second-throw event) are independent of each other.
   4. Hence $P(A\mid \bar B) = P(A) = \dfrac{1}{2}$.

   Common Mistakes

   • Assuming some dependence between the two throws and trying to compute a joint sample space unnecessarily.
   • Confusing $P(A/\bar B)$ with $P(\bar B/A)$ or with $P(A\cap \bar B)$.

   ✓Final answer

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

   ANSWER: A

9. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.In a class consisting of 40 boys and 30 girls, 30% of the boys and 40% of the girls are good at Mathematics. If a student selected at random from that class is found to be a girl, then the probability that she is not good at Mathematics is (A) $\dfrac{3}{5}$ (B) $\dfrac{2}{5}$ (C) $\dfrac{3}{10}$ (D) $\dfrac{7}{10}$

   ›Reveal solutionSolution

   Conditioning on "the student is a girl" restricts the sample space to the 30 girls only; the answer is $\frac35$.

   Concept and Intuition

   This is conditional probability with a twist: the condition ("selected student is a girl") is given as a fact, not something to be computed via Bayes' theorem. Once we know the student is a girl, the boys' statistics become irrelevant — we simply work within the group of girls.

   Step-by-Step Solution

   1. Total girls $=30$.
   2. Girls good at Mathematics $= 40\%$ of $30 = 12$.
   3. Girls not good at Mathematics $= 30-12=18$.
   4. Since we are told the selected student is a girl, the relevant sample space is just these 30 girls.
   5. $P(\text{not good at maths} \mid \text{girl}) = \frac{18}{30} = \frac{3}{5}.$

   Common Mistakes

   • Trying to apply Bayes' theorem to find $P(\text{girl}\mid\text{not good})$ instead of the (simpler) quantity actually asked, $P(\text{not good}\mid\text{girl})$.
   • Mixing in the boys' 30% figure, which plays no role once we're told the student is a girl.

   ✓Final answer

   The correct option is (A) — $\dfrac{3}{5}$.

   ANSWER: A

10. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.If it is known that a woman has two children and she has at least one girl child, the probability that the woman has both girl children is (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    This is a classic conditional-probability trap: conditioning on "at least one girl" (not "the elder/first child is a girl") leaves 3 equally likely outcomes, of which 1 is both-girls, giving $\dfrac13$.

    Concept and Intuition

    With two children, birth order matters for counting equally likely outcomes: $BB, BG, GB, GG$, each with probability $1/4$. The event "at least one girl" is a set of outcomes, not a statement about a specific (e.g. first) child, so it correctly removes only $BB$ and keeps three outcomes, not two. This distinguishes it from the simpler (and different) question "given the elder child is a girl," which would leave only $\{GB,GG\}$ and give probability $1/2$.

    Step-by-Step Solution

    1. Sample space (ordered by birth, say elder-younger): $\{BB,BG,GB,GG\}$, each with probability $\dfrac14$.
    2. Event $E$ = "at least one girl" = $\{BG,GB,GG\}$, so $P(E)=\dfrac34$.
    3. Event $F$ = "both girls" = $\{GG\}$, and $F\subset E$, so $P(F\cap E)=P(F)=\dfrac14$.
    4. $P(F\mid E)=\dfrac{P(F\cap E)}{P(E)}=\dfrac{1/4}{3/4}=\dfrac13$.

    Common Mistakes

    • Confusing "at least one girl" with "a specific (e.g. the first) child is a girl," which would wrongly give $\dfrac12$.
    • Forgetting that $BG$ and $GB$ are distinct outcomes (birth order matters), undercounting the conditioning event.

    ✓Final answer

    The correct option is (B) $\dfrac13$.

    ANSWER: B

11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.A pair of dice is thrown. Then the probability that either of the dice shows 2 when their sum is 6 is (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{3}{5}$

    ›Reveal solutionSolution

    This is a conditional probability: restrict the sample space to pairs summing to 6, then count how many of those have a 2 on either die.

    Concept and Intuition

    "Given that the sum is 6" restricts attention to only those (die1, die2) pairs whose total is 6. Within that restricted, equally-likely set, we simply count the favourable outcomes.

    Step-by-Step Solution

    1. Pairs $(d_1,d_2)$ with $d_1+d_2=6$: $(1,5),(2,4),(3,3),(4,2),(5,1)$ — 5 outcomes, each equally likely given the sum is 6.
    2. Favourable outcomes (either die shows 2): $(2,4)$ and $(4,2)$ — 2 outcomes.
    3. Required probability $=\dfrac{2}{5}$.

    Common Mistakes

    • Forgetting to restrict to the sum-6 sample space and instead computing an unconditional probability over all 36 outcomes.
    • Missing that $(2,4)$ and $(4,2)$ are distinct ordered outcomes (two dice are distinguishable).

    ✓Final answer

    The correct option is (C) — $\dfrac25$.

    ANSWER: C

12. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.A box P contains 3 white and 7 red balls. A bag Q contains 4 green and 5 blue balls. Two balls are randomly drawn from box P. If both are of same color, one ball is drawn from bag Q and if the two balls are of different color, 2 balls are drawn from the bag Q. If it is known that there is exactly one green ball among the ball or balls drawn from bag Q, then the probability that the two balls drawn from box P are of different colors is (A) $\dfrac{35}{67}$ (B) $\dfrac{21}{62}$ (C) $\dfrac{20}{43}$ (D) $\dfrac{32}{67}$

    ›Reveal solutionSolution

    A two-stage Bayes' theorem problem; conditioning on "exactly one green ball from Q" gives $P(\text{different colors from P}) = \frac{35}{67}$.

    Concept and Intuition

    This is a compound experiment: the outcome in box P (same-color vs different-color) determines how many balls are drawn from bag Q, and hence changes the probability model for "exactly one green." We must compute, for each P-outcome, the probability of observing exactly one green from Q, then combine via Bayes' theorem using the P-outcome's prior probability.

    Step-by-Step Solution

    1. Box P has 3 white + 7 red = 10 balls. $P(\text{same color}) = \dfrac{{}^3C_2+{}^7C_2}{{}^{10}C_2} = \dfrac{3+21}{45}=\dfrac{24}{45}=\dfrac{8}{15}$. $P(\text{different colors}) = 1-\dfrac{8}{15}=\dfrac{7}{15}$ (check: $\dfrac{{}^3C_1{}^7C_1}{{}^{10}C_2}=\dfrac{21}{45}=\dfrac{7}{15}$ ✓).
    2. If same color (S): draw 1 ball from Q (4 green, 5 blue, 9 total). "Exactly one green" among 1 ball drawn just means that ball is green: $P(\text{1 green}\mid S)=\dfrac{4}{9}$.
    3. If different colors (D): draw 2 balls from Q. $P(\text{exactly one green}\mid D) = \dfrac{{}^4C_1\cdot{}^5C_1}{{}^9C_2} = \dfrac{20}{36}=\dfrac{5}{9}$.
    4. By Bayes' theorem:

    $P(D\mid \text{one green}) = \dfrac{P(D)P(\text{one green}\mid D)}{P(S)P(\text{one green}\mid S)+P(D)P(\text{one green}\mid D)} = \dfrac{\frac{7}{15}\cdot\frac{5}{9}}{\frac{8}{15}\cdot\frac{4}{9}+\frac{7}{15}\cdot\frac{5}{9}}$

    5. Numerator $=\dfrac{35}{135}$; denominator $=\dfrac{32}{135}+\dfrac{35}{135}=\dfrac{67}{135}$. So $P(D\mid\text{one green}) = \dfrac{35}{67}$.

    Common Mistakes

    • Using the same probability model for "one green" regardless of whether 1 or 2 balls were drawn from Q — the two cases have genuinely different sample spaces.
    • Forgetting to weight each conditional probability by the correct prior $P(S)$ or $P(D)$ before combining.

    ✓Final answer

    The correct option is (A) — $\dfrac{35}{67}$.

    ANSWER: A