Q.Probability that A speaks truth is $\frac{4}{5}$. A coin is tossed. A reports that a head appears. The probability that actually there was head is (A) $\frac{4}{5}$ (B) $\frac{1}{2}$ (C) $\frac{1}{5}$ (D) $\frac{2}{5}$

Showing the 12 most recent of 69 on this concept.

1. 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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2. 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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3. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.Let $A$ and $B$ be two events with $P(A) = \dfrac{1}{7}$, $P(A/B) = \dfrac{2}{5}$ and $P(B) = \dfrac{2}{7}$. Then the value $P(B/A)$ is (A) $\dfrac{1}{5}$ (B) $\dfrac{5}{49}$ (C) $\dfrac{4}{5}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   Chain the conditional-probability definition twice: first get $P(A\cap B)$ from $P(A/B)$, then get $P(B/A)$ from that.

   Concept and Intuition

   Conditional probability is defined as $P(E/F)=\dfrac{P(E\cap F)}{P(F)}$. Given one conditional probability, we can back out the joint probability $P(A\cap B)$, and then use it (with the definition again, but conditioning the other way) to find the other conditional probability.

   Step-by-Step Solution

   1. $P(A/B) = \dfrac{P(A\cap B)}{P(B)} \Rightarrow P(A\cap B) = P(A/B)\cdot P(B) = \dfrac{2}{5}\times\dfrac{2}{7} = \dfrac{4}{35}$.
   2. Now use $P(B/A) = \dfrac{P(A\cap B)}{P(A)} = \dfrac{4/35}{1/7}$. …

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4. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.On every evening, a student either watches TV or reads a book. The probability of watching TV is $\dfrac{4}{5}$. If he watches TV, the probability that he will fall asleep is $\dfrac{3}{4}$ and it is $\dfrac{1}{4}$ when he reads a book. If the student is found to be asleep on an evening, the probability that he watched the TV is (A) $\dfrac{11}{13}$ (B) $\dfrac{12}{13}$ (C) $\dfrac{2}{13}$ (D) $\dfrac{4}{13}$

   ›Reveal solutionSolution

   This tests Bayes' theorem: finding the probability of a cause (watched TV) given an observed effect (fell asleep). The answer is $\dfrac{12}{13}$.

   Concept and Intuition

   When asked "given the observed outcome, what's the probability of a particular cause," that's a Bayes'-theorem setup: compute the joint probability of that cause with the outcome, and divide by the total probability of the outcome (summed over all causes).

   Step-by-Step Solution

   1. $P(\text{TV})=\tfrac45$, $P(\text{book})=\tfrac15$.
   2. $P(\text{asleep}\mid\text{TV})=\tfrac34$, $P(\text{asleep}\mid\text{book})=\tfrac14$.
   3. Total probability of falling asleep: $P(\text{asleep})=P(\text{TV})P(\text{asleep}\mid\text{TV})+P(\text{book})P(\text{asleep}\mid\text{book})=\tfrac45\cdot\tfrac34+\tfrac15\cdot\tfrac14=\tfrac35+\tfrac1{20}=\tfrac{12}{20}+\tfrac1{20}=\tfrac{13}{20}$. …

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5. 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 $=15+30+36+30+15=126$. …

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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. $P(\overline Q) = 1-P(Q) = 1-\tfrac16=\tfrac56$. …

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7. 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 …

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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). …

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9. 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, $P(A\cap B)=\dfrac3{36}$. …

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10. 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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11. 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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12. 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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