Q.There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?

Showing the 12 most recent of 69 on this concept.

1. 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 $P(\text{at least one head}) = 1-\dfrac14 = \dfrac34$.

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2. 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: $P(\text{fair})=\frac mn$, $P(\text{biased})=\frac{n-m}{n}$. …

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3. 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 $P(E)=\dfrac34$. …

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4. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.In a toy factory, the machines A, B and C are used to manufacture 30%, 40% and 30% of the output, respectively. The probabilities of toys made by machines A, B, and C to be defective are respectively 2%, 3% and 1%. A toy is taken from the factory and is found to be defective. The probability that it was manufactured by the machine B is (A) $4/5$ (B) $2/9$ (C) $3/4$ (D) $4/7$

   ›Reveal solutionSolution

   This tests Bayes' theorem for finding the probability of a specific cause given an observed

   effect (a defective toy); the answer is $4/7$.

   Concept and Intuition

   When an outcome (a defective toy) could have come from several sources, each with its own prior

   probability and its own conditional probability of producing that outcome, Bayes' theorem lets us

   "invert" the conditioning: given that the outcome occurred, what's the probability it came from a

   particular source? The key is to first find the total probability of the outcome by summing over

   all sources (the law of total probability), then take the one source's contribution as a fraction of

   that total.

   Step-by-Step Solution

   1. Prior probabilities: $P(A)=0.3,\ P(B)=0.4,\ P(C)=0.3$.
   2. Conditional defect rates: $P(D\mid A)=0.02,\ P(D\mid B)=0.03,\ P(D\mid C)=0.01$.
   3. Total probability of a defective toy (law of total probability):

   $P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C) = 0.3(0.02)+0.4(0.03)+0.3(0.01)$

   $= 0.006+0.012+0.003 = 0.021.$

   4. By Bayes' theorem: …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Bag A contains 3 white and 4 black balls. Bag B contains 4 white and 3 black balls. Bag C contains 2 white and 5 black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag C is (A) $\dfrac{1}{6}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{2}{13}$

   ›Reveal solutionSolution

   This is a direct Bayes'-theorem (inverse probability) question. Answer: $\dfrac{2}{9}$.

   Concept and Intuition

   We're given the outcome (a white ball was drawn) and asked for the probability of a particular cause (it came from bag C). This is exactly Bayes' theorem: $P(C\mid W) = \dfrac{P(C)P(W\mid C)}{\sum_i P(\text{bag}_i)P(W\mid \text{bag}_i)}$.

   Step-by-Step Solution

   1. Each bag is chosen with probability $\frac13$.
   2. $P(W\mid A) = \dfrac{3}{7}$ (3 white out of 7 total in bag A).
   3. $P(W\mid B) = \dfrac{4}{7}$ (4 white out of 7 in bag B).
   4. $P(W\mid C) = \dfrac{2}{7}$ (2 white out of 7 in bag C).
   5. Total probability of white: $P(W) = \frac13\left(\dfrac37+\dfrac47+\dfrac27\right) = \frac13\cdot\dfrac{9}{7} = \dfrac{9}{21} = \dfrac{3}{7}$. …

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6. 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: $P(C)=\tfrac13\cdot\tfrac14+\tfrac16\cdot\tfrac18+\tfrac12\cdot1=\tfrac1{12}+\tfrac1{48}+\tfrac12$. …

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7. 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. $P(F_2\cap\text{girl})=\dfrac13\cdot\dfrac23=\dfrac29$. …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.A person is known to speak the truth in 3 out of 4 occasions. If he throws a die and reports that it is six, then the probability that it is actually six is (A) $\dfrac{3}{8}$ (B) $\dfrac{2}{7}$ (C) $\dfrac{1}{9}$ (D) $\dfrac{4}{5}$

   ›Reveal solutionSolution

   A classic Bayes'-theorem problem: combine the prior $P(\text{six})=1/6$ with the witness's reliability $3/4$ to get the posterior probability it's actually six, given he reports six. The answer is $3/8$.

   Concept and Intuition

   Before he speaks, the die has $P(\text{six})=1/6$. His report is evidence, but he isn't perfectly reliable (truthful $3/4$ of the time). Bayes' theorem updates the prior probability using how likely the report "six" is under each of the two possibilities (six actually occurred vs. it didn't).

   Step-by-Step Solution

   1. Let $E_1$ = the die actually shows six, $E_2$ = it doesn't. $P(E_1)=\dfrac16$, $P(E_2)=\dfrac56$.
   2. Let $A$ = he reports "six". If $E_1$ occurred, he reports six truthfully with probability $\dfrac34$: $P(A\mid E_1)=\dfrac34$.
   3. If $E_2$ occurred, he reports "six" only if he lies, with probability $\dfrac14$: $P(A\mid E_2)=\dfrac14$.
   4. By Bayes' theorem:

   $P(E_1\mid A)=\frac{P(A\mid E_1)P(E_1)}{P(A\mid E_1)P(E_1)+P(A\mid E_2)P(E_2)}.$

   5. Numerator: $\dfrac34\cdot\dfrac16=\dfrac{3}{24}=\dfrac18$.
   6. Denominator's second term: $\dfrac14\cdot\dfrac56=\dfrac{5}{24}$.
   7. Sum $=\dfrac{3}{24}+\dfrac{5}{24}=\dfrac{8}{24}=\dfrac13$. …

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9. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.There are 2 bags each containing 3 white and 5 black balls and 4 bags each containing 6 white and 4 black balls. If a ball drawn randomly from a bag is found to be black, then the probability that this ball is from the first set of bags is (A) $\dfrac{25}{57}$ (B) $\dfrac{25}{41}$ (C) $\dfrac{2}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   A classic Bayes'-theorem (inverse-probability) question: given the ball drawn is black, find the probability it came from the first group of bags. Answer: $\dfrac{25}{57}$.

   Concept and Intuition

   When an experiment happens in two stages — first a bag is picked (from one of two groups of bags), then a ball is drawn from it — and we're told the result of the second stage (a black ball came out), Bayes' theorem lets us reverse the direction of reasoning and find the probability about the first stage (which group the bag was from). The prior probability of picking from a group is proportional to how many bags are in that group (each individual bag is equally likely to be picked), and then we weight by how likely that group is to produce the observed outcome.

   Step-by-Step Solution

   1. Groups and priors. Group 1 (call it $S_1$) has 2 bags (each 3 white, 5 black — 8 balls). Group 2 ($S_2$) has 4 bags (each 6 white, 4 black — 10 balls). Total bags $=6$, each equally likely to be chosen, so

   $P(S_1)=\frac{2}{6}=\frac13,\qquad P(S_2)=\frac{4}{6}=\frac23.$

   2. Likelihoods of drawing black. From a group-1 bag: $P(B\mid S_1)=\dfrac{5}{8}$. From a group-2 bag: $P(B\mid S_2)=\dfrac{4}{10}=\dfrac25$.
   3. Total probability of drawing a black ball (law of total probability): $P(B) = P(S_1)P(B\mid S_1)+P(S_2)P(B\mid S_2) = \frac13\cdot\frac58+\frac23\cdot\frac25 = \frac{5}{24}+\frac{4}{15}.$ …

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

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11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.A person is known to speak false once out of 4 times. If that person picks a card at random from a pack of 52 cards and reports that it is a king, then the probability that it is actually a king is (A) $\dfrac{1}{37}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{12}{37}$ (D) $\dfrac{25}{37}$

    ›Reveal solutionSolution

    Classic Bayes'-theorem problem: weigh the prior probability of drawing a king against the person's truth-telling reliability. Answer: $1/5$.

    Concept and Intuition

    The report 'it is a king' can arise two ways: the card really is a king and the person tells the truth, or the card is NOT a king and the person lies (falsely claims king). Bayes' theorem combines these into the posterior probability that it's actually a king.

    Step-by-Step Solution

    1. Prior: $P(K)=4/52=1/13$ (king drawn), $P(\bar K)=12/13$ (not a king).
    2. Truth-telling: $P(\text{truth})=3/4$, $P(\text{lie})=1/4$.
    3. $P(\text{reports king}\mid K) = P(\text{truth}) = 3/4$ (truthfully reports the actual king).
    4. $P(\text{reports king}\mid \bar K) = P(\text{lie}) = 1/4$ (lies about a non-king, falsely calling it a king). …

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12. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.In a bolt factory, machines A, B, C manufacture 25%, 35%, 40% of the total output respectively. There is a chance of having 5%, 4%, 2% defective bolts manufactured by A, B, C respectively. If a bolt is drawn at random from the output, then the probability that it is defective is (A) $\dfrac{69}{2000}$ (B) $\dfrac{59}{2000}$ (C) $\dfrac{79}{2000}$ (D) $\dfrac{89}{2000}$

    ›Reveal solutionSolution

    Classic total-probability ("law of total probability") setup: weight each machine's defect rate by its share of output and sum. The answer is (A).

    Concept and Intuition

    When an item can come from several mutually exclusive, exhaustive sources (here, machines A, B, C), and each source has its own conditional probability of producing a defect, the overall probability of a defect is the weighted average:

    $P(D) = \sum_i P(\text{source}_i)\,P(D\mid \text{source}_i).$

    This is the law of total probability — it's the natural way to combine "how much each machine contributes" with "how likely each machine's own output is defective."

    Step-by-Step Solution

    1. Let $A, B, C$ denote drawing a bolt from each machine: $P(A)=0.25$, $P(B)=0.35$, $P(C)=0.40$ (these sum to 1, as they must).
    2. Conditional defect probabilities: $P(D\mid A)=0.05$, $P(D\mid B)=0.04$, $P(D\mid C)=0.02$.
    3. By the law of total probability:

    $P(D) = P(A)P(D\mid A) + P(B)P(D\mid B) + P(C)P(D\mid C).$

    4. Compute each term: $0.25\times0.05=0.0125$; $0.35\times0.04=0.014$; $0.40\times0.02=0.008$. …

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