Q.Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is $2$-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?

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

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2. 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 …

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3. 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. $P(\text{white})=\frac12\cdot\frac23+\frac12\cdot\frac37=\frac13+\frac{3}{14}$. …

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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 …

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5. 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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6. 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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7. 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: …

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8. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Two urns identical in appearance contain respectively 3 green and 2 black balls and 2 green and 5 black balls. One urn is selected at random and a ball is drawn from it. The probability that it is black is ________ (A) $\dfrac{39}{70}$ (B) $\dfrac{37}{70}$ (C) $\dfrac{41}{70}$ (D) $\dfrac{33}{70}$

   ›Reveal solutionSolution

   Apply the law of total probability over the two equally-likely urns to get $P(\text{black}) = \dfrac{39}{70}$.

   Concept and Intuition

   When an experiment first randomly selects between two scenarios (here, two urns, each equally likely) and then performs a further random step (drawing a ball), the overall probability of an outcome is the weighted average of the outcome's probability under each scenario, weighted by the scenario's own probability.

   Step-by-Step Solution

   1. Urn 1 has $3$ green $+ 2$ black $=5$ balls, so $P(\text{black}\mid\text{Urn 1}) = \dfrac{2}{5}$.
   2. Urn 2 has $2$ green $+ 5$ black $=7$ balls, so $P(\text{black}\mid\text{Urn 2}) = \dfrac{5}{7}$.
   3. Each urn is chosen with probability $\dfrac12$. …

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9. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.Bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. If one ball is drawn at random from one of the bags and is found to be red, then the probability that it was drawn from the bag B is (A) $\dfrac{23}{54}$ (B) $\dfrac{25}{51}$ (C) $\dfrac{25}{52}$ (D) $\dfrac{27}{55}$

   ›Reveal solutionSolution

   A classic Bayes'-theorem problem: given the ball drawn is red, the probability it came from bag B is $\dfrac{25}{52}$.

   Concept and Intuition

   This is Bayes' theorem: we're given the outcome (red ball drawn) and want to find the probability of which "cause" (bag A or B) produced it, using the prior probability of choosing each bag (each $\frac12$) and each bag's own probability of yielding red.

   Step-by-Step Solution

   1. Bag A: 2 white, 3 red (5 total) $\Rightarrow P(\text{red}|A)=\dfrac35$.
   2. Bag B: 4 white, 5 red (9 total) $\Rightarrow P(\text{red}|B)=\dfrac59$.
   3. $P(A)=P(B)=\dfrac12$ (bag chosen at random).
   4. Total probability of red: $P(\text{red})=P(A)P(\text{red}|A)+P(B)P(\text{red}|B)=\dfrac12\cdot\dfrac35+\dfrac12\cdot\dfrac59=\dfrac3{10}+\dfrac5{18}$.
   5. Common denominator 90: $\dfrac3{10}=\dfrac{27}{90}$, $\dfrac5{18}=\dfrac{25}{90}$, sum $=\dfrac{52}{90}=\dfrac{26}{45}$.
   6. Bayes: $P(B|\text{red})=\dfrac{P(B)P(\text{red}|B)}{P(\text{red})}=\dfrac{(1/2)(5/9)}{26/45}=\dfrac{5/18}{26/45}$. …

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10. 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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11. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.A man is known to speak truth 7 out of 10 times. After throwing a die with 100 faces marked $1, 2, 3, \ldots, 100$ on its faces, the man reports that he got a prime number on the die. What is the probability that it is actually a prime? (A) $\frac{5}{16}$ (B) $\frac{7}{16}$ (C) $\frac{11}{16}$ (D) $\frac{10}{16}$

    ›Reveal solutionSolution

    A classic Bayes'-theorem "truth-telling" problem: combining the base rate of primes among 1–100 with the man's truthfulness gives $P(\text{actually prime}\mid\text{reports prime}) = \frac{7}{16}$.

    Concept and Intuition

    Even though the man is more likely to tell the truth than lie, the base rate of the event matters. Since primes are a minority (25 out of 100), a "prime" report is diluted by false reports from the much larger non-prime set.

    Step-by-Step Solution

    1. Number of primes from 1 to 100 is 25 (2, 3, 5, ..., 97), so $P(\text{prime})=\frac{25}{100}=\frac14$, and $P(\text{not prime})=\frac34$.
    2. He speaks truth with probability $\frac{7}{10}$: $P(\text{reports prime}\mid\text{prime})=\frac{7}{10}$, and (lying) $P(\text{reports prime}\mid\text{not prime})=\frac{3}{10}$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.$P(A|A\cap B) + P(B|A\cap B) =$ (A) $1$ (B) $P(A\cup B)$ (C) $P(A\cap B)$ (D) $2$

    ›Reveal solutionSolution

    Both conditional probabilities equal $1$ because $A\cap B$ is a subset of both $A$ and $B$, so their sum is $2$.

    Concept and Intuition

    Conditioning an event on its own subset (or superset relationship) often collapses to a probability of exactly $1$: if $E\subseteq F$, then $P(F|E)=1$, since knowing $E$ occurred guarantees $F$ occurred too.

    Step-by-Step Solution

    1. $P(A|A\cap B) = \dfrac{P\big(A\cap(A\cap B)\big)}{P(A\cap B)}$.
    2. Since $A\cap(A\cap B) = A\cap B$ (intersecting with $A$ again changes nothing, as $A\cap B\subseteq A$), this is $\dfrac{P(A\cap B)}{P(A\cap B)}=1$.
    3. Similarly, $P(B|A\cap B) = \dfrac{P\big(B\cap(A\cap B)\big)}{P(A\cap B)} = \dfrac{P(A\cap B)}{P(A\cap B)}=1$.
    4. Sum $=1+1=2$. …

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