Q.A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

Showing the 12 most recent of 69 on this concept.

1. 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

2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.There are three families $F_1, F_2, F_3$. $F_1$ has 2 boys and 1 girl; $F_2$ has 1 boy and 2 girls; $F_3$ has 1 boy and 1 girl. A family is randomly chosen and a child is chosen from that family randomly. If it is known that the child thus selected is a girl, then the probability that she is from $F_2$ is (A) $\dfrac{4}{9}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{3}{7}$ (D) $\dfrac{5}{7}$

   ›Reveal solutionSolution

   Applying Bayes' theorem across the three equally-likely families gives $P(F_2\mid\text{girl})=\dfrac49$.

   Concept and Intuition

   This is a textbook application of Bayes' theorem: we're given the reverse conditional probabilities (family $\to$ probability of picking a girl) and asked for the forward one (girl picked $\to$ probability she's from a specific family).

   Step-by-Step Solution

   1. Prior: $P(F_1)=P(F_2)=P(F_3)=\dfrac13$.
   2. $P(\text{girl}\mid F_1)=\dfrac13$ (1 girl out of 3 children), $P(\text{girl}\mid F_2)=\dfrac23$, $P(\text{girl}\mid F_3)=\dfrac12$.
   3. Total probability: $P(\text{girl})=\frac13\cdot\frac13+\frac13\cdot\frac23+\frac13\cdot\frac12=\frac13\left(\frac13+\frac23+\frac12\right)=\frac13\cdot\frac{2+4+3}{6}=\frac13\cdot\frac96=\frac12$
   4. $P(F_2\cap\text{girl})=\dfrac13\cdot\dfrac23=\dfrac29$.
   5. Bayes: $P(F_2\mid\text{girl})=\dfrac{P(F_2\cap\text{girl})}{P(\text{girl})}=\dfrac{2/9}{1/2}=\dfrac{4}{9}$.

   Common Mistakes

   • Treating the child-selection probability as uniform over all 9 children (2+1+1+2+1+1=9, wrong weighting) instead of first picking a family uniformly, then a child within it — the families have unequal sizes so this changes the answer.
   • Arithmetic slip adding $\frac13+\frac23+\frac12$ without a common denominator.

   ✓Final answer

   The correct option is (A) — $\dfrac49$.

   ANSWER: A

3. 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

4. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.An unbiased coin is tossed 3 times. If the third toss gets head, then the probability of getting at least one more head is (A) $3/4$ (B) $1/4$ (C) $1/2$ (D) $1/3$

   ›Reveal solutionSolution

   This tests recognizing conditional independence: given the third toss's outcome, the first two tosses remain independent unbiased flips, so their probability is unaffected by the condition. Answer: $3/4$.

   Concept and Intuition

   Since coin tosses are independent events, knowing the outcome of the third toss (head) gives us no information about the first two tosses. So the question really just asks: what's the probability of getting at least one head in two independent fair coin tosses? This is most easily computed via the complement (no heads at all in two tosses).

   Step-by-Step Solution

   1. The condition 'the third toss gets head' is independent of the outcomes of the first two tosses (each coin toss is independent of the others).
   2. So conditioning on the third toss being heads doesn't change the probabilities for the first two tosses — they remain two independent fair coin flips.
   3. We want $P(\text{at least one head among the first two tosses})$.
   4. Use the complement: $P(\text{no heads in first two tosses}) = P(\text{both tails}) = \dfrac12\times\dfrac12=\dfrac14$.
   5. So $P(\text{at least one head}) = 1-\dfrac14 = \dfrac34$.

   Common Mistakes

   • Overcomplicating this by trying to build a full conditional probability tree over all three tosses, when the key insight (independence removes any real conditioning effect) simplifies it immediately.
   • Misreading 'at least one MORE head' as requiring at least one head among ALL three tosses (which would be different, and trivially true anyway since the third is already head) — the question specifically asks about the first two, i.e., 'one more' beyond the given third-toss head.

   ✓Final answer

   The correct option is (A) — $3/4$.

   ANSWER: A

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 2022Set eng-2022-07-06-FN1 markMCQ
   Q.Two persons P and Q are considering to apply for a job. The probability that P applies for the job is 1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for the job is (A) $4/5$ (B) $5/6$ (C) $7/8$ (D) $11/12$

   ›Reveal solutionSolution

   Chain the given conditional probabilities to find $P(Q)$ and $P(P\cap Q)$, then use the complement rule — the answer is (A) $4/5$.

   Concept and Intuition

   Conditional probability definitions let us cross-multiply to recover the joint probability $P(P\cap Q)$ from either conditional. Once $P(P), P(Q), P(P\cap Q)$ are all known, De Morgan's law converts "neither event" into the complement of the union.

   Step-by-Step Solution

   1. Given: $P(P)=\tfrac14$, $P(P|Q)=\tfrac12$, $P(Q|P)=\tfrac13$.
   2. $P(P\cap Q) = P(Q|P)\cdot P(P) = \tfrac13\times\tfrac14 = \tfrac1{12}$.
   3. Also $P(P\cap Q) = P(P|Q)\cdot P(Q) \Rightarrow \tfrac1{12} = \tfrac12\cdot P(Q) \Rightarrow P(Q)=\tfrac16$.
   4. $P(P\cup Q) = P(P)+P(Q)-P(P\cap Q) = \tfrac14+\tfrac16-\tfrac1{12} = \tfrac{3}{12}+\tfrac{2}{12}-\tfrac{1}{12}=\tfrac{4}{12}=\tfrac13$.
   5. By De Morgan's law, "neither P nor Q applies" is the complement of $P\cup Q$: $P(\overline P\cap\overline Q) = 1-\tfrac13=\tfrac23$.
   6. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$.
   7. $P(\overline P|\overline Q) = \dfrac{P(\overline P\cap\overline Q)}{P(\overline Q)} = \dfrac{2/3}{5/6} = \dfrac23\times\dfrac65 = \dfrac{12}{15}=\dfrac45$.

   Common Mistakes

   • Mixing up which conditional probability to use for computing the joint probability first (must use $P(Q|P)P(P)$, not $P(P|Q)P(P)$, to directly match given data).
   • Forgetting to convert to complements at the final step (computing $P(P|Q)$-type instead of $P(\overline P|\overline Q)$).

   ✓Final answer

   The correct option is (A) — $4/5$.

   ANSWER: A

7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.For a biased die, the probabilities for different faces to turn up are given below

   Face	1	2	3	4	5	6
   Probability	0.1	0.32	0.21	0.15	0.05	0.17

   The die is tossed and you are told that either face 1 or 2 has turned up. Then the probability that it is face 1 is (A) $\dfrac{10}{33}$ (B) $\dfrac{5}{21}$ (C) $\dfrac{8}{21}$ (D) $\dfrac{1}{42}$

   ›Reveal solutionSolution

   Conditioning on "face 1 or face 2" just means renormalizing the two individual probabilities so they add to 1.

   Concept and Intuition

   $P(\text{face 1}\mid\text{face 1 or 2})=\dfrac{P(\text{face 1})}{P(\text{face 1})+P(\text{face 2})}$, since these two events are mutually exclusive and their union is the conditioning event.

   Step-by-Step Solution

   1. $P(1)=0.1$, $P(2)=0.32$.
   2. $P(1\text{ or }2)=0.1+0.32=0.42$.
   3. $P(1\mid1\text{ or }2)=\dfrac{0.1}{0.42}=\dfrac{10}{42}=\dfrac{5}{21}$.

   Common Mistakes

   • Dividing by the total probability $1$ instead of the restricted event's probability $0.42$.
   • Arithmetic slip simplifying $10/42$ (should reduce to $5/21$).

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{21}$.

   ANSWER: B

8. 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

9. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.A box contains $n$ coins, $m$ of which are fair and the rest are biased. When a biased coin is tossed, the probability of getting a head is twice as likely as tail. A coin is drawn from the box at random and is tossed twice. It is found that first time it shows head and the second time it shows tail. Then the probability that the coin drawn is fair is (A) $\frac{7m}{8n+m}$ (B) $\frac{9m}{8n+m}$ (C) $\frac{7m}{8m+n}$ (D) $\frac{9m}{8m+n}$

   ›Reveal solutionSolution

   This is a Bayes'-theorem problem; computing the likelihoods for fair vs. biased coins and combining with the prior $m/n$ gives $\frac{9m}{8n+m}$.

   Concept and Intuition

   Bayes' theorem updates our belief about which "type" of coin was drawn, given the observed outcome (head then tail), by weighing each type's prior probability by how likely it was to produce that exact outcome.

   Step-by-Step Solution

   1. Fair coin: $P(H)=P(T)=\frac12$, so $P(HT\mid\text{fair})=\frac12\cdot\frac12=\frac14$.
   2. Biased coin: head is twice as likely as tail, so $P(H)=\frac23,\,P(T)=\frac13$; $P(HT\mid\text{biased})=\frac23\cdot\frac13=\frac29$.
   3. Priors: $P(\text{fair})=\frac mn$, $P(\text{biased})=\frac{n-m}{n}$.
   4. Bayes: $P(\text{fair}\mid HT) = \dfrac{\frac mn\cdot\frac14}{\frac mn\cdot\frac14+\frac{n-m}{n}\cdot\frac29}$.
   5. Multiply numerator and denominator by $36n$: numerator $=9m$; denominator $=9m+8(n-m)=m+8n$.
   6. So $P(\text{fair}\mid HT)=\dfrac{9m}{8n+m}$.

   Common Mistakes

   • Using $P(H)=P(T)=\frac12$ for the biased coin too, or mixing up which outcome ($HT$ vs. $TH$) is asked.

   ✓Final answer

   The correct option is (B) — $\frac{9m}{8n+m}$.

   ANSWER: B

10. 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

11. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.In a school there are 3 sections A, B and C. Section A contains 20 girls and 30 boys, section B contains 40 girls and 20 boys and section C contains 10 girls and 30 boys. The probabilities of selecting the section A, B and C are 0.2, 0.3 and 0.5 respectively. If a student selected at random from the school is a girl, then the probability that she belongs to section A is (A) $\dfrac{121}{200}$ (B) $\dfrac{16}{121}$ (C) $\dfrac{14}{81}$ (D) $\dfrac{16}{81}$

    ›Reveal solutionSolution

    A three-section Bayes'-theorem problem; the answer is $\dfrac{16}{81}$, option (D).

    Concept and Intuition

    Given a randomly selected student is a girl, we want to reverse-condition on which section she came from. This requires the total probability of "selecting a girl" (summed over all three sections weighted by section-selection probability), then applying Bayes' theorem to isolate section A's contribution.

    Step-by-Step Solution

    1. Conditional probabilities of picking a girl within each section: $P(\text{girl}\mid A)=\dfrac{20}{50}=\dfrac{2}{5}$; $P(\text{girl}\mid B)=\dfrac{40}{60}=\dfrac{2}{3}$; $P(\text{girl}\mid C)=\dfrac{10}{40}=\dfrac14$.
    2. Section-selection probabilities: $P(A)=\dfrac15,\ P(B)=\dfrac{3}{10},\ P(C)=\dfrac12$.
    3. Joint terms: $P(A)P(\text{girl}\mid A)=\dfrac15\cdot\dfrac25=\dfrac{2}{25}$; $P(B)P(\text{girl}\mid B)=\dfrac{3}{10}\cdot\dfrac23=\dfrac15$; $P(C)P(\text{girl}\mid C)=\dfrac12\cdot\dfrac14=\dfrac18$.
    4. Common denominator 200: $\dfrac{2}{25}=\dfrac{16}{200}$, $\dfrac15=\dfrac{40}{200}$, $\dfrac18=\dfrac{25}{200}$. Sum $=\dfrac{81}{200}=P(\text{girl})$.
    5. Bayes: $P(A\mid\text{girl})=\dfrac{16/200}{81/200}=\dfrac{16}{81}$.

    Common Mistakes

    • Using the section sizes directly as weights instead of the given selection probabilities $0.2,0.3,0.5$.
    • Arithmetic slips when combining fractions with different denominators — using a common denominator (200) avoids this.

    ✓Final answer

    The correct option is (D) — $\dfrac{16}{81}$.

    ANSWER: D

12. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.A bag contains 6 balls. If three balls are drawn at a time and all of them are found to be green, then the probability that exactly 5 of the balls in the bag are green is: (A) $\dfrac{4}{35}$ (B) $\dfrac{5}{35}$ (C) $\dfrac{2}{7}$ (D) $\dfrac{1}{7}$

    ›Reveal solutionSolution

    This is a Bayes'-theorem problem: given a uniform prior on the unknown number of green balls in the bag, update the belief after observing 3 green balls drawn. Answer: $2/7$.

    Concept and Intuition

    Before drawing, any number of green balls from $0$ to $6$ is equally likely (7 equally likely hypotheses). Observing "all 3 drawn balls are green" is much more likely under hypotheses with more green balls, so Bayes' theorem reweights the prior toward higher $k$. We want the posterior probability that $k=5$.

    Step-by-Step Solution

    1. Prior: $P(k \text{ green balls})=\dfrac17$ for $k=0,1,\dots,6$.
    2. Likelihood of drawing 3 balls, all green, given $k$ green balls out of 6: $P(\text{all green}\mid k)=\dfrac{\binom{k}{3}}{\binom{6}{3}}=\dfrac{\binom{k}{3}}{20}$ (zero for $k<3$).
    3. Compute: $k=3:\binom33=1\Rightarrow1/20$; $k=4:\binom43=4\Rightarrow4/20$; $k=5:\binom53=10\Rightarrow10/20$; $k=6:\binom63=20\Rightarrow20/20=1$.
    4. Total probability of observing "all green" (law of total probability, prior cancels as common factor $1/7$): proportional to $1/20+4/20+10/20+20/20=35/20$.
    5. Posterior $P(k=5\mid\text{all green})=\dfrac{10/20}{35/20}=\dfrac{10}{35}=\dfrac27$.

    Common Mistakes

    • Forgetting that $k=0,1,2$ contribute zero likelihood (can't draw 3 green balls if fewer than 3 exist).
    • Confusing this posterior-probability question with a plain hypergeometric-probability question (i.e., computing $P(\text{all green}\mid k=5)$ instead of $P(k=5\mid\text{all green})$).

    ✓Final answer

    The correct option is (C) — $\dfrac{2}{7}$.

    ANSWER: C