Q.A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from TATA NAGAR.

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.70% of the total employees of a factory are men. Among the employees of that factory, 30% of men and 15% of women are technical assistants. If an employee chosen at random is found to be a technical assistant, then the probability that this employee is a man is (A) $\dfrac{9}{23}$ (B) $\dfrac{3}{17}$ (C) $\dfrac{14}{17}$ (D) $\dfrac{14}{23}$

   ›Reveal solutionSolution

   A direct Bayes'/total-probability computation using a convenient total of 100 employees. The answer is $\dfrac{14}{17}$.

   Concept and Intuition

   We want $P(\text{man}\mid\text{technical assistant})$. Since we're given the proportion of men and women, and what fraction of each group are technical assistants, the cleanest approach is to work with actual counts (out of a convenient total like 100) rather than abstract probabilities — it avoids fraction juggling.

   Step-by-Step Solution

   1. Assume 100 employees: Men $=70$, Women $=30$.
   2. Technical assistants among men: $30\%$ of $70=21$.
   3. Technical assistants among women: $15\%$ of $30=4.5$.
   4. Total technical assistants $=21+4.5=25.5$.
   5. $P(\text{man}\mid\text{TA})=\dfrac{21}{25.5}=\dfrac{210}{255}=\dfrac{42}{51}=\dfrac{14}{17}$. …

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2. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.One ticket is selected at random from 50 tickets numbered $00, 01, 02, \ldots 49$. The probability that sum of the digits is 10, given that product of the digits is 9 is (A) $\dfrac{9}{10}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{2}{25}$

   ›Reveal solutionSolution

   Only two tickets (19 and 33) have digit-product 9, and only one of them (19) also has digit-sum 10, giving conditional probability $1/2$.

   Concept and Intuition

   Conditional probability $P(\text{sum}=10\mid\text{product}=9)$ restricts attention entirely to the tickets satisfying the "given" condition (product $=9$), then asks what fraction of those also satisfy the target condition (sum $=10$).

   Step-by-Step Solution

   1. List all two-digit combinations $(d_1,d_2)$ with $d_1\in\{0,1,2,3,4\}$ (tens digit, since tickets run 00–49) and $d_2\in\{0,\dots,9\}$ (units digit) whose product is 9:
      • $d_1=1,d_2=9$: ticket $19$.
      • $d_1=3,d_2=3$: ticket $33$.
      • No other integer pairs with $d_1\le4$ give product 9.
   2. So the "product = 9" event has exactly 2 tickets: $\{19, 33\}$.
   3. Check digit sums: $19\to1+9=10$ ✓; $33\to3+3=6$ ✗.
   4. Only 1 out of these 2 tickets also has digit-sum 10. …

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3. 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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4. 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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5. 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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6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.The probability that a person goes to college by car is $\dfrac{1}{5}$; by bus $\dfrac{2}{5}$ and by train is $\dfrac{3}{5}$ respectively. The probabilities that he reaches the college late if he takes car, bus, train are $\dfrac{2}{7}, \dfrac{4}{7}$ and $\dfrac{1}{7}$ respectively. If he reaches the college in time, the probability that he travelled by car is (A) $\dfrac{6}{29}$ (B) $\dfrac{24}{29}$ (C) $\dfrac{5}{29}$ (D) $\dfrac{23}{29}$

   ›Reveal solutionSolution

   Bayes' theorem: weight each mode's "on-time" probability by its usage probability, then take the car-share of the total.

   Concept and Intuition

   This is a direct application of Bayes' theorem / total probability: to find $P(\text{cause}\mid\text{effect})$, build the denominator as the weighted sum over every possible cause of reaching the observed effect (here, arriving on time), then take the numerator's share of it.

   Step-by-Step Solution

   1. On-time probabilities per mode: car $1-\tfrac27=\tfrac57$; bus $1-\tfrac47=\tfrac37$; train $1-\tfrac17=\tfrac67$.
   2. Total on-time probability: $P(\text{ontime})=\tfrac15\cdot\tfrac57+\tfrac25\cdot\tfrac37+\tfrac35\cdot\tfrac67=\tfrac{5}{35}+\tfrac{6}{35}+\tfrac{18}{35}=\tfrac{29}{35}$.
   3. Joint probability of taking the car AND being on time: $\tfrac15\cdot\tfrac57=\tfrac{5}{35}$. …

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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 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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9. 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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10. 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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11. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.2 aero planes I and II bond a target in succession. The probabilities of I and II scoring a hit correctly is 0.3 and 0.2 respectively. The second plane will bomb only if first misses the target. The probability that the target is hit by the $2^{nd}$ plane is (A) 0.06 (B) 0.14 (C) 0.32 (D) 0.7

    ›Reveal solutionSolution

    The 2nd plane gets a chance only after the 1st fails, so $P = 0.7\times 0.2 = 0.14$. Answer: (B).

    Concept and Intuition

    This is a sequential (conditional) experiment: the second trial happens only when the first fails. The event 'the target is hit by the 2nd plane' is therefore a compound event —

    $\{\text{I misses}\}\cap\{\text{II hits}\}$

    and because the two planes' performances are independent, the probability of the intersection is the product of the probabilities.

    A useful picture is a probability tree:

             ┌── I hits (0.3)  ────────────────► target hit by plane I  (0.3)
    Start ───┤
             └── I misses (0.7) ─┬── II hits (0.2) ──► hit by plane II  (0.7 × 0.2 = 0.14)
                                 └── II misses (0.8) ► target not hit   (0.7 × 0.8 = 0.56)
    

    The three leaves sum to $0.3+0.14+0.56=1$ ✓ — a good check that the model is complete.

    Step-by-Step Solution

    1. Let $H_1$ = plane I hits, with $P(H_1)=0.3$, so $P(\overline{H_1})=1-0.3=0.7$.
    2. Let $H_2$ = plane II hits (given it bombs), with $P(H_2)=0.2$.
    3. Plane II bombs only if plane I missed. Hence

    $P(\text{target hit by 2nd plane}) = P(\overline{H_1}\cap H_2) = P(\overline{H_1})\,P(H_2)$

    4. Substitute: $= 0.7\times 0.2 = 0.14$ …

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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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