Q.Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Showing the 12 most recent of 69 on this concept.

1. 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

2. 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

3. 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

4. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.Three boxes $B_1$, $B_2$ and $B_3$ contain balls with different colors as follows:

   	White	Black	Red
   $B_1$	2	1	2
   $B_2$	3	2	4
   $B_3$	4	3	2

   A die is thrown. Box $B_1$ is chosen if either 1 or 2 turns up. Box $B_2$ is chosen if 3 or 4 turns up and box $B_3$ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is drawn at random from that box. If the ball drawn is found to be Red, then the probability that it is drawn from box $B_2$ is (A) $\dfrac{7}{12}$ (B) $\dfrac{5}{12}$ (C) $\dfrac{1}{12}$ (D) $\dfrac{3}{26}$

   ›Reveal solutionSolution

   A Bayes'-theorem problem: reverse the conditional probability "ball is Red, which box?" using the total-probability formula.

   Concept and Intuition

   Bayes' theorem lets us "invert" a conditional probability. We know $P(\text{Red}\mid B_i)$ for each box and the prior $P(B_i)$ (each $1/3$ since the die is fair and each box corresponds to two faces); we want the posterior $P(B_2 \mid \text{Red})$.

   Step-by-Step Solution

   1. Totals in each box: $B_1$: $2+1+2=5$; $B_2$: $3+2+4=9$; $B_3$: $4+3+2=9$.
   2. $P(\text{Red}\mid B_1)=\dfrac{2}{5}$, $P(\text{Red}\mid B_2)=\dfrac{4}{9}$, $P(\text{Red}\mid B_3)=\dfrac{2}{9}$.
   3. Each box has prior probability $\dfrac13$ (each box corresponds to two die faces out of six).
   4. Total probability: $P(\text{Red}) = \dfrac13\left(\dfrac25+\dfrac49+\dfrac29\right) = \dfrac13\left(\dfrac25+\dfrac23\right) = \dfrac13\cdot\dfrac{16}{15}=\dfrac{16}{45}$.
   5. Bayes: $P(B_2\mid \text{Red}) = \dfrac{P(B_2)P(\text{Red}\mid B_2)}{P(\text{Red})} = \dfrac{\frac13\cdot\frac49}{\frac{16}{45}} = \dfrac{\frac{4}{27}}{\frac{16}{45}} = \dfrac{4}{27}\times\dfrac{45}{16}=\dfrac{5}{12}$.

   Common Mistakes

   • Forgetting to weight each box's conditional probability by its prior $1/3$.
   • Arithmetic slip when adding the three fractions with different denominators (5 and 9).

   ✓Final answer

   The correct option is (B) — $\dfrac{5}{12}$.

   ANSWER: B

5. 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

6. 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

7. 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

8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Bag $B_1$ contains 4 white and 2 black balls. Bag $B_2$ contains 3 white and 4 black balls. A bag is chosen at random and a ball is drawn from it at random, then the probability that the ball drawn is white, is (A) $\dfrac{1}{42}$ (B) $\dfrac{42}{32}$ (C) $\dfrac{33}{42}$ (D) $\dfrac{23}{42}$

   ›Reveal solutionSolution

   Total-probability rule over the two equally likely bags gives $P(\text{white})=\frac12\cdot\frac46+\frac12\cdot\frac37=\frac{23}{42}$.

   Concept and Intuition

   The ball drawn depends on which bag was chosen first. Since the bag choice is random with $P(B_1)=P(B_2)=\frac12$, and the draw is conditionally independent given the bag, the Law of Total Probability adds the two conditional probabilities weighted by how likely each bag is.

   Step-by-Step Solution

   1. $B_1$: 4 white, 2 black out of 6 $\Rightarrow P(\text{white}\mid B_1)=\frac{4}{6}=\frac23$.
   2. $B_2$: 3 white, 4 black out of 7 $\Rightarrow P(\text{white}\mid B_2)=\frac37$.
   3. $P(\text{white})=\frac12\cdot\frac23+\frac12\cdot\frac37=\frac13+\frac{3}{14}$.
   4. Common denominator 42: $\frac13=\frac{14}{42}$, $\frac{3}{14}=\frac{9}{42}$, sum $=\frac{23}{42}$.

   Common Mistakes

   • Simplifying $\frac46$ incorrectly or forgetting to weight by $\frac12$ for each bag.
   • Adding numerators/denominators directly instead of finding a common denominator (a classic fraction-addition slip).

   ✓Final answer

   The correct option is (D) — $\dfrac{23}{42}$.

   ANSWER: D

9. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.It is given that in a random experiment events A and B are such that $P(A) = \dfrac{1}{4}$, $P(A|B) = \dfrac{1}{2}$ and $P(B|A) = \dfrac{2}{3}$ then $P(B) =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   Using $P(B|A)$ to find $P(A\cap B)$, then dividing by $P(A|B)$, gives $P(B)=\dfrac13$.

   Concept and Intuition

   The conditional probability definitions $P(A|B)=\dfrac{P(A\cap B)}{P(B)}$ and $P(B|A)=\dfrac{P(A\cap B)}{P(A)}$ share the common quantity $P(A\cap B)$ — computing it from one equation lets us solve the other for the unknown probability.

   Step-by-Step Solution

   1. From $P(B|A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac23$, and $P(A)=\dfrac14$: $P(A\cap B)=\dfrac23\times\dfrac14=\dfrac16$.
   2. From $P(A|B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac12$: $P(B)=\dfrac{P(A\cap B)}{P(A|B)}=\dfrac{1/6}{1/2}=\dfrac13$.

   Common Mistakes

   • Confusing which conditional probability to use to find $P(A\cap B)$ first (must use the one whose conditioning event's total probability, $P(A)$, is already known).
   • Dividing instead of multiplying (or vice versa) when rearranging the conditional probability formula.

   ✓Final answer

   The correct option is (A) — $\dfrac13$.

   ANSWER: A

10. 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

11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.In a test a student either guesses or copies or knows the answer to answer a multiple choice question with four choices. The probability that he makes a guess is $1/3$ and the probability that he copies the answer is $1/6$. The probability that his answer is correct, given that he copied it is $1/8$. The probability that he knew the answer to the question, given that he answered it correctly is (A) $\dfrac{29}{24}$ (B) $\dfrac{22}{29}$ (C) $\dfrac{24}{29}$ (D) $\dfrac{23}{29}$

    ›Reveal solutionSolution

    A classic Bayes'-theorem problem with three mutually exclusive causes (guess/copy/know) for a correct answer. Answer: $24/29$.

    Concept and Intuition

    This is a textbook application of Bayes' theorem: we're given the prior probabilities of three exhaustive, mutually exclusive ways of answering (guessing, copying, knowing) along with the conditional probability of being correct under each, and asked to reverse the conditioning — find the probability the student knew the answer, GIVEN that they got it right.

    Step-by-Step Solution

    1. The three ways of answering are exhaustive: $P(\text{know})=1-P(\text{guess})-P(\text{copy})=1-\tfrac13-\tfrac16=1-\tfrac12=\tfrac12$.
    2. Conditional correctness probabilities: $P(C\mid\text{guess})=\tfrac14$ (one of four choices), $P(C\mid\text{copy})=\tfrac18$ (given), $P(C\mid\text{know})=1$.
    3. Total probability of being correct: $P(C)=\tfrac13\cdot\tfrac14+\tfrac16\cdot\tfrac18+\tfrac12\cdot1=\tfrac1{12}+\tfrac1{48}+\tfrac12$.
    4. Common denominator 48: $\tfrac{4}{48}+\tfrac1{48}+\tfrac{24}{48}=\tfrac{29}{48}$.
    5. Bayes' theorem: $P(\text{know}\mid C)=\dfrac{P(\text{know})P(C\mid\text{know})}{P(C)}=\dfrac{\tfrac12\cdot1}{29/48}=\dfrac{24/48}{29/48}=\dfrac{24}{29}$.

    Common Mistakes

    • Forgetting to compute $P(\text{know})$ as the complement of guess+copy.
    • Arithmetic slip converting to a common denominator when summing the total probability.

    ✓Final answer

    The correct option is (C) — $\dfrac{24}{29}$.

    ANSWER: C

12. 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