Q.An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$ $P(B \text{ fails alone}) = 0.15$ $P(A \text{ and B fail}) = 0.15$ Evaluate the following probabilities

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A, $B_1$, $B_2$, $B_3$ are the events in a random experiment. If $P(B_1) = 0.25$, $P(B_2) = 0.30$, $P(B_3) = 0.45$, $P\left(\dfrac{A}{B_1}\right) = 0.05$, $P\left(\dfrac{A}{B_2}\right) = 0.04$, $P\left(\dfrac{A}{B_3}\right) = 0.03$, then $P\left(\dfrac{B_2}{A}\right) =$ (A) $\dfrac{6}{19}$ (B) $\dfrac{8}{19}$ (C) $\dfrac{12}{19}$ (D) $\dfrac{5}{19}$

   ›Reveal solutionSolution

   This is a classic Bayes’ theorem problem: we are given prior probabilities for three mutually exclusive events and the conditional probabilities of A given each, and we need the posterior probability of $B_2$ given A. The answer is $\frac{6}{19}$, which corresponds to option (A).

   We start with the concept: Bayes’ theorem lets us “reverse” conditional probabilities. Here, we know $P(A \mid B_i)$ and want $P(B_2 \mid A)$. The key is that the $B_i$ form a partition of the sample space (they are the only possible “causes” of A), so we can compute $P(A)$ using the law of total probability, then apply Bayes’ formula.

   1. Identify the given data

      • $P(B_1) = 0.25$, $P(B_2) = 0.30$, $P(B_3) = 0.45$
      • $P(A \mid B_1) = 0.05$, $P(A \mid B_2) = 0.04$, $P(A \mid B_3) = 0.03$ The events $B_1, B_2, B_3$ are mutually exclusive and exhaustive (their probabilities sum to 1), so they form a partition.

   2. Compute the total probability of A

      By the law of total probability:

   $P(A) = P(B_1)P(A \mid B_1) + P(B_2)P(A \mid B_2) + P(B_3)P(A \mid B_3)$

   Substitute:

   $P(A) = (0.25)(0.05) + (0.30)(0.04) + (0.45)(0.03)$

   Calculate each term:

   • $0.25 \times 0.05 = 0.0125$
   • $0.30 \times 0.04 = 0.0120$
   • $0.45 \times 0.03 = 0.0135$ Sum:

   $P(A) = 0.0125 + 0.0120 + 0.0135 = 0.0380$

   3. Apply Bayes’ theorem for $P(B_2 \mid A)$ Bayes’ theorem states: $$$$

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2. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.If $A$ and $B$ are two events of a random experiment such that $P(\overline{A}) = \dfrac{2}{3}$, $P(B) = \dfrac{4}{15}$ and $P(A \cap \overline{B}) = \dfrac{1}{5}$, then $\sqrt{195 \left[ P\left(B \mid (A \cup \overline{B})\right) + P(A \cup B) \right]} =$ (A) 9 (B) 11 (C) 13 (D) 15

   ›Reveal solutionSolution

   This problem requires us to calculate probabilities of various event combinations (complement, intersection, union) and a conditional probability using fundamental set theory identities. We then substitute these values into the given expression to find the final numerical result. The final value is $\boxed{11}$.

   To solve this problem, we need to systematically break down the given expression and calculate each probability term using the fundamental rules of probability and set theory. The key is to correctly apply the formulas for complements, unions, intersections, and conditional probabilities, often using set identities to simplify complex event descriptions.

   Here's a step-by-step approach:

   1. Determine $P(A)$ from $P(\overline{A})$:

      The probability of an event $A$ and its complement $\overline{A}$ always sum to 1.

      $P(A) + P(\overline{A}) = 1$

      Given $P(\overline{A}) = \frac{2}{3}$, we can find $P(A)$:

      $P(A) = 1 - P(\overline{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.

   2. Determine $P(A \cap B)$ using $P(A \cap \overline{B})$:

      The event $A$ can be partitioned into two mutually exclusive events: $A \cap B$ (A and B both occur) and $A \cap \overline{B}$ (A occurs, but B does not).

      $P(A) = P(A \cap B) + P(A \cap \overline{B})$

      We are given $P(A \cap \overline{B}) = \frac{1}{5}$ and we found $P(A) = \frac{1}{3}$.

      So, $P(A \cap B) = P(A) - P(A \cap \overline{B}) = \frac{1}{3} - \frac{1}{5}$.

      To subtract these fractions, we find a common denominator, which is 15:

      $P(A \cap B) = \frac{5}{15} - \frac{3}{15} = \frac{2}{15}$.

   3. Calculate $P(A \cup B)$:

      The probability of the union of two events $A$ and $B$ is given by the addition rule.

      $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

      We have $P(A) = \frac{1}{3}$, $P(B) = \frac{4}{15}$ (given), and $P(A \cap B) = \frac{2}{15}$ (from Step 2).

      $P(A \cup B) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15}$.

      Using a common denominator of 15:

      $P(A \cup B) = \frac{5}{15} + \frac{4}{15} - \frac{2}{15} = \frac{5+4-2}{15} = \frac{7}{15}$.

      This is the first part of the sum inside the square root.

   4. Prepare for $P(B \mid (A \cup \overline{B}))$: Identify the intersection term:

      The conditional probability $P(X \mid Y)$ is defined as $\frac{P(X \cap Y)}{P(Y)}$. Here, $X=B$ and $Y=(A \cup \overline{B})$.

      So, we need to find $P(B \cap (A \cup \overline{B}))$.

      Using the distributive property of set intersection over union:

      $B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B})$.

      The event $B \cap \overline{B}$ means that event $B$ occurs AND event $B$ does NOT occur, which is impossible. Thus, $B \cap \overline{B} = \emptyset$.

      So, $B \cap (A \cup \overline{B}) = (B \cap A) \cup \emptyset = B \cap A$.

      Therefore, $P(B \cap (A \cup \overline{B})) = P(A \cap B)$.

      From Step 2, we know $P(A \cap B) = \frac{2}{15}$. This is the numerator for our conditional probability.

   5. Prepare for $P(B \mid (A \cup \overline{B}))$: Calculate $P(\overline{B})$:

      Similar to Step 1, we use the complement rule for event $B$.

      $P(\overline{B}) = 1 - P(B)$.

      Given $P(B) = \frac{4}{15}$:

      $P(\overline{B}) = 1 - \frac{4}{15} = \frac{15-4}{15} = \frac{11}{15}$.

   6. Prepare for $P(B \mid (A \cup \overline{B}))$: Calculate $P(A \cup \overline{B})$:

      This is the denominator for our conditional probability. We use the addition rule for $A$ and $\overline{B}$.

      $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})$

      We have $P(A) = \frac{1}{3}$ (from Step 1), $P(\overline{B}) = \frac{11}{15}$ (from Step 5), and $P(A \cap \overline{B}) = \frac{1}{5}$ (given).

      $P(A \cup \overline{B}) = \frac{1}{3} + \frac{11}{15} - \frac{1}{5}$.

      Using a common denominator of 15: …

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3. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $P(A) = \dfrac{3}{8}$, $P(\overline{A}|B) = P(\overline{B}|A) = \dfrac{3}{5}$, then $P(A \cap B) + P(B) =$ (A) $\dfrac{21}{40}$ (B) $\dfrac{2}{13}$ (C) $\dfrac{3}{14}$ (D) $\dfrac{5}{12}$

   ›Reveal solutionSolution

   We use the given conditional probabilities to set up equations for $P(A \cap B)$ and $P(B)$, then solve and sum them. The result is $\frac{21}{40}$, which corresponds to option (A).

   We are told:

   • $P(A) = \frac{3}{8}$
   • $P(\overline{A} \mid B) = \frac{3}{5}$
   • $P(\overline{B} \mid A) = \frac{3}{5}$

   We need $P(A \cap B) + P(B)$.

   ---

   Concept and intuition

   Conditional probabilities like $P(\overline{A} \mid B)$ relate the probability of the complement of $A$ given $B$ to the joint probability $P(\overline{A} \cap B)$. Since $P(\overline{A} \mid B) = \frac{P(\overline{A} \cap B)}{P(B)}$, we can write an equation linking $P(B)$ and $P(A \cap B)$. Similarly, $P(\overline{B} \mid A)$ gives a relation between $P(A)$ and $P(A \cap B)$. This lets us solve for the unknowns.

   ---

   Step-by-step solution

   1. Use $P(\overline{B} \mid A)$ to find $P(A \cap B)$ By definition:

   $P(\overline{B} \mid A) = \frac{P(\overline{B} \cap A)}{P(A)} = \frac{3}{5}$

   Since $P(\overline{B} \cap A) = P(A) - P(A \cap B)$, we have:

   $\frac{P(A) - P(A \cap B)}{P(A)} = \frac{3}{5}$

   Substitute $P(A) = \frac{3}{8}$:

   $\frac{\frac{3}{8} - P(A \cap B)}{\frac{3}{8}} = \frac{3}{5}$

   Multiply both sides by $\frac{3}{8}$:

   $\frac{3}{8} - P(A \cap B) = \frac{3}{5} \cdot \frac{3}{8} = \frac{9}{40}$

   So:

   $P(A \cap B) = \frac{3}{8} - \frac{9}{40} = \frac{15}{40} - \frac{9}{40} = \frac{6}{40} = \frac{3}{20}$

   2. Use $P(\overline{A} \mid B)$ to find $P(B)$ By definition:

   $P(\overline{A} \mid B) = \frac{P(\overline{A} \cap B)}{P(B)} = \frac{3}{5}$

   Now $P(\overline{A} \cap B) = P(B) - P(A \cap B)$. Substitute $P(A \cap B) = \frac{3}{20}$:

   $\frac{P(B) - \frac{3}{20}}{P(B)} = \frac{3}{5}$

   Multiply both sides by $P(B)$: …

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4. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If $P\left(\dfrac{A}{B}\right) = \dfrac{3}{10}$, $P\left(\dfrac{B}{A}\right) = \dfrac{4}{5}$ and $P(A \cup B) = K P(B)$, then $\dfrac{1}{K} =$ (A) $\dfrac{40}{49}$ (B) $\dfrac{40}{43}$ (C) $\dfrac{100}{101}$ (D) $1$

   ›Reveal solutionSolution

   The key idea is to use the definitions of conditional probability to relate $P(A \cap B)$ to $P(A)$ and $P(B)$, then express $P(A \cup B)$ in terms of $P(B)$ alone. The result is $\frac{1}{K} = \frac{40}{43}$.

   We are given two conditional probabilities and a relation involving the union. The goal is to find $\frac{1}{K}$, where $P(A \cup B) = K P(B)$. This is a problem about linking conditional probabilities to the basic probability of events, so we start by writing down what each conditional means.

   Recall: $P(A/B) = \frac{P(A \cap B)}{P(B)}$ and $P(B/A) = \frac{P(A \cap B)}{P(A)}$. These are not symmetric — each gives a different ratio. Our job is to use them to find $P(A)$ and $P(A \cap B)$ in terms of $P(B)$, then compute $P(A \cup B)$.

   1. From $P(A/B)$ we get $P(A \cap B)$ in terms of $P(B)$.

   $P(A/B) = \frac{3}{10} \quad\Rightarrow\quad \frac{P(A \cap B)}{P(B)} = \frac{3}{10}$

   So

   $P(A \cap B) = \frac{3}{10} P(B).$

   2. From $P(B/A)$ we get $P(A)$ in terms of $P(A \cap B)$.

   $P(B/A) = \frac{4}{5} \quad\Rightarrow\quad \frac{P(A \cap B)}{P(A)} = \frac{4}{5}$

   Hence

   $P(A) = \frac{5}{4} P(A \cap B).$

   3. Substitute the expression for $P(A \cap B)$ from step 1 into step 2.

   $P(A) = \frac{5}{4} \cdot \frac{3}{10} P(B) = \frac{15}{40} P(B) = \frac{3}{8} P(B).$

   So $P(A)$ is $\frac{3}{8}$ of $P(B)$.

   4. Now write $P(A \cup B)$ using the inclusion-exclusion formula.

   $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

   Substitute the expressions we have:

   $P(A \cup B) = \frac{3}{8} P(B) + P(B) - \frac{3}{10} P(B).$

   5. Combine the terms over a common denominator. …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Let $P = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ be a matrix. Three elements of this matrix P are selected at random. A is the event of having the three elements whose sum is odd. B is the event of selecting the three elements which are in a row or column. Then $P(A) + P\left(\dfrac{A}{B}\right) =$ (A) $\dfrac{221}{420}$ (B) $\dfrac{17}{21}$ (C) $\dfrac{21}{20}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   We compute the probability that three randomly chosen entries sum to an odd number, then the conditional probability of that event given they lie in a single row or column, and add them. The result simplifies to $\frac{17}{21}$, which is option (B).

   ---

   Concept & Intuition

   The matrix has 9 entries. We pick 3 of them uniformly at random.

   • For event A (odd sum), we need to count how many triples have an odd total. Since odd/even depends only on parity, we first classify the 9 numbers by parity: 1,3,5,7,9 are odd (5 odds); 2,4,6,8 are even (4 evens). The sum of three numbers is odd iff we have an odd number of odd entries among them — i.e., 1 or 3 odds.
   • For event B (all in one row or one column), we count triples that lie entirely in a single row (3 rows, each row has 3 entries → 1 triple per row) or entirely in a single column (3 columns, each column has 3 entries → 1 triple per column). That gives $3+3=6$ triples.
   • Then $P(A/B)$ is the fraction of those 6 triples that also have an odd sum.

   We compute both probabilities and add them.

   ---

   Step-by-step

   1. Total number of ways to choose 3 entries from 9

   $\binom{9}{3} = 84.$

   2. Count triples with odd sum (event A)
      • Case 1: exactly 1 odd, 2 evens Choose 1 odd from 5 odds: $\binom{5}{1}=5$ Choose 2 evens from 4 evens: $\binom{4}{2}=6$ Total: $5 \times 6 = 30$ triples.
      • Case 2: exactly 3 odds, 0 evens Choose 3 odds from 5 odds: $\binom{5}{3}=10$ Total: $10$ triples.
      • So $|A| = 30 + 10 = 40$. Hence

   $P(A) = \frac{40}{84} = \frac{10}{21}.$

   3. Count triples in a row or column (event B)
      • Rows: 3 rows, each row has exactly 1 triple (all three entries). So 3 triples.
      • Columns: 3 columns, each column has exactly 1 triple. So 3 triples.
      • Total: $|B| = 6$. Hence

   $P(B) = \frac{6}{84} = \frac{1}{14}.$

   4. Count triples that are both in a row/column AND have odd sum (event $A \cap B$) Check each row and column for odd sum: …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.A card is drawn randomly from a well shuffled pack of 52 cards. If A is the event of getting a diamond card and B is the event of getting an ace card, then the probability that exactly one of the events among A and B to occur is (A) $\dfrac{15}{52}$ (B) $\dfrac{4}{13}$ (C) $\dfrac{17}{52}$ (D) $\dfrac{5}{13}$

   ›Reveal solutionSolution

   The probability that exactly one of the events A (diamond) or B (ace) occurs is the sum of their individual probabilities minus twice the probability of both occurring. The result is $\frac{15}{52}$, which corresponds to option (A).

   We want the probability that exactly one of the two events happens — that is, either we draw a diamond that is not an ace, or we draw an ace that is not a diamond. This is a classic "exclusive or" (XOR) situation.

   Why this approach works:

   If we simply add $P(A) + P(B)$, we count the case where both occur (the ace of diamonds) twice. To get exactly one, we subtract that double-counted overlap once more than usual — hence $P(A) + P(B) - 2P(A \cap B)$.

   ---

   1. Identify the probabilities of each event individually.

      • There are 13 diamonds in a deck of 52, so $P(A) = \frac{13}{52} = \frac{1}{4}$.
      • There are 4 aces, so $P(B) = \frac{4}{52} = \frac{1}{13}$.

   2. Find the probability that both events occur (the intersection).

      • Only one card is both a diamond and an ace: the ace of diamonds.
      • So $P(A \cap B) = \frac{1}{52}$.

   3. Apply the formula for exactly one event.

      • Exactly one of A or B occurs means: $(A \text{ and not } B)$ or $(B \text{ and not } A)$.
      • The probability is:

   $P(\text{exactly one}) = P(A) + P(B) - 2P(A \cap B)$

   • Substitute the values:

   $\frac{13}{52} + \frac{4}{52} - 2 \cdot \frac{1}{52} = \frac{13 + 4 - 2}{52} = \frac{15}{52}$

   4. Check against the options. …

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7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Three persons A, B, C planned to have a running race among themselves. If the probability that A wins the race is thrice that of B and the probability that B wins the race is $\dfrac{3}{2}$ times that of C, then the difference in probabilities of A and C to win the race is (A) $\dfrac{2}{3}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{5}{14}$ (D) $\dfrac{3}{7}$

   ›Reveal solutionSolution

   With $P(A)=\tfrac{9}{14},\ P(C)=\tfrac{2}{14}$, the difference is $P(A)-P(C)=\dfrac12$.

   Let $P(C)=p$. Then $P(B)=\dfrac{3}{2}p$ and $P(A)=3\,P(B)=\dfrac{9}{2}p$.

   The three probabilities sum to $1$ (one of them must win):

   $\frac{9}{2}p+\frac{3}{2}p+p=7p=1\ \Rightarrow\ p=\frac{1}{7}.$

   Hence

   $P(A)=\frac{9}{2}\cdot\frac17=\frac{9}{14},\qquad P(C)=\frac17=\frac{2}{14}.$ …

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8. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.A pair of dice is thrown twice in succession. The probability of getting prime numbers on both the dice in first throw and composite numbers on both the dice in second throw is (A) $\dfrac{1}{216}$ (B) $\dfrac{1}{16}$ (C) $\dfrac{1}{36}$ (D) $\dfrac{1}{9}$

   ›Reveal solutionSolution

   The key idea is to treat the two throws as independent events, multiply their probabilities, and note that each die has 3 prime numbers (2,3,5) and 2 composite numbers (4,6) — 1 is neither. The final probability is $\frac{1}{16}$.

   ---

   We start by recalling what “prime” and “composite” mean for the numbers 1 through 6 on a standard die.

   • Prime numbers on a die: 2, 3, 5 (three numbers).
   • Composite numbers on a die: 4, 6 (two numbers).
   • Neither: 1 (not prime, not composite).

   The problem asks: first throw — both dice show primes; second throw — both dice show composites. The two throws are independent, so we multiply probabilities.

   ---

   1. Probability of both dice showing primes in the first throw For one die, $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$. Since the two dice are independent,

   $P(\text{both prime}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$

   2. Probability of both dice showing composites in the second throw For one die, $P(\text{composite}) = \frac{2}{6} = \frac{1}{3}$. So,

   $P(\text{both composite}) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}.$

   3. Combine the two independent events …

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9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If 4 letters are selected at random from the letters of the word PROBABILITY, then the probability of getting a combination of letters in which atleast one letter is repeated is (A) $\dfrac{43}{170}$ (B) $\dfrac{19}{61}$ (C) $\dfrac{57}{184}$ (D) $\dfrac{29}{155}$

   ›Reveal solutionSolution

   The multiset $\text{PROBABILITY}$ has $9$ distinct letters (with $B$ and $I$ each twice). Total $4$-letter selections $=183$; those with a repeat $=57$, so the probability is $\dfrac{57}{183}=\dfrac{19}{61}$, option (B).

   Letters of PROBABILITY: $P,R,O,B,A,B,I,L,I,T,Y$ — $11$ letters, $9$ distinct types, with $B$ and $I$ appearing twice each.

   Step 1 — Total number of $4$-letter selections (order does not matter).

   Count by repetition pattern:

   • All four distinct: $\dbinom{9}{4}=126$.
   • Exactly one repeated pair ($B$ or $I$) plus two other distinct letters: $2\times\dbinom{8}{2}=2\times 28=56$.
   • Two repeated pairs, i.e. $\{B,B,I,I\}$: $1$ way. …

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10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.Two numbers b and c are chosen at random in succession without replacement from the set $\{1,2,3,\ldots,9\}$. Then the probability that $x^2 + bx + c > 0,\ \forall x \in \mathbb{R}$ is (A) $\dfrac{29}{72}$ (B) $\dfrac{32}{81}$ (C) $\dfrac{45}{143}$ (D) $\dfrac{82}{125}$

    ›Reveal solutionSolution

    The condition $x^2+bx+c>0$ for all real $x$ is equivalent to the discriminant $b^2-4c<0$. Counting ordered pairs $(b,c)$ from $\{1,\dots,9\}$ without replacement that satisfy $b^2<4c$ gives $29$ favorable outcomes out of $72$ total, so the probability is $\frac{29}{72}$, which is option (A).

    ---

    Why this approach works

    A quadratic $x^2+bx+c$ that is always positive (for every real $x$) must have no real roots and open upward. Since the coefficient of $x^2$ is $1>0$, the condition reduces to the discriminant being negative: $b^2-4c<0$, i.e. $b^2<4c$.

    We are choosing $b$ and $c$ without replacement from $\{1,\dots,9\}$, so each ordered pair $(b,c)$ with $b\neq c$ is equally likely. The total number of such ordered pairs is $9\times 8 = 72$. We just need to count how many of them satisfy $b^2<4c$.

    ---

    Step-by-step counting

    1. Understand the inequality

    We need $b^2 < 4c$. Since $c$ is an integer from $1$ to $9$, rewrite as $c > \frac{b^2}{4}$. For each $b$, we count the number of $c$ values (different from $b$) that are strictly greater than $b^2/4$.

    2. Compute for each $b$

    • $b=1$: $b^2/4 = 0.25$, so $c > 0.25$ means $c \ge 1$. All $c$ from $1$ to $9$ except $c=1$ (since $b\neq c$) work. That gives $8$ choices.
    • $b=2$: $b^2/4 = 1$, so $c > 1$ means $c \ge 2$. Excluding $c=2$ leaves $\{3,4,5,6,7,8,9\}$ → $7$ choices.
    • $b=3$: $b^2/4 = 2.25$, so $c > 2.25$ means $c \ge 3$. Excluding $c=3$ leaves $\{4,5,6,7,8,9\}$ → $6$ choices.
    • $b=4$: $b^2/4 = 4$, so $c > 4$ means $c \ge 5$. Excluding $c=4$ (which isn't in this set anyway) gives $\{5,6,7,8,9\}$ → $5$ choices.
    • $b=5$: $b^2/4 = 6.25$, so $c > 6.25$ means $c \ge 7$. Excluding $c=5$ (not in set) gives $\{7,8,9\}$ → $3$ choices.
    • $b=6$: $b^2/4 = 9$, so $c > 9$ means $c \ge 10$, but max $c$ is $9$. No $c$ works → $0$ choices. …

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11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.The numbers 2, 3, 5, 7, 11, 13 are written on six distinct paper chits. If 3 of them are chosen at random, then the probability that the sum of the numbers on the obtained chits is divisible by 3, is (A) $\dfrac{7}{20}$ (B) $\dfrac{6}{20}$ (C) $\dfrac{5}{20}$ (D) $\dfrac{1}{5}$

    ›Reveal solutionSolution

    The key idea is to classify each number by its remainder modulo 3, then count only those 3‑card combinations whose remainders sum to a multiple of 3. The probability is $\frac{7}{20}$, which corresponds to option (A).

    ---

    We have six numbers: 2, 3, 5, 7, 11, 13.

    We pick 3 at random. The total number of ways is $\binom{6}{3} = 20$.

    We want the probability that the sum of the three chosen numbers is divisible by 3.

    Why classify by remainder?

    A number’s remainder modulo 3 determines whether it contributes 0, 1, or 2 to the total sum mod 3. The sum of three numbers is divisible by 3 exactly when the sum of their remainders is 0 mod 3. This turns a problem about specific numbers into a simple counting problem about remainder classes.

    ---

    Step-by-step

    1. Find each number’s remainder mod 3

       • $2 \equiv 2$
       • $3 \equiv 0$
       • $5 \equiv 2$
       • $7 \equiv 1$
       • $11 \equiv 2$
       • $13 \equiv 1$

       So we have:

       • Remainder 0: {3} → 1 number
       • Remainder 1: {7, 13} → 2 numbers
       • Remainder 2: {2, 5, 11} → 3 numbers

    2. Which remainder combinations sum to 0 mod 3?

       Let (r1, r2, r3) be the remainders of the three chosen numbers. We need $r_1 + r_2 + r_3 \equiv 0 \pmod{3}$.

       The possible triples (order doesn’t matter) are:

       • (0,0,0) — all three have remainder 0
       • (1,1,1) — all three have remainder 1
       • (2,2,2) — all three have remainder 2
       • (0,1,2) — one of each remainder

       No other triple works (e.g., (0,0,1) sums to 1, etc.).

    3. Count the number of 3‑card combinations for each case

       • (0,0,0): Only 1 number with remainder 0, so impossible. Count = 0. …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.A typist claims that he prepares a typed page with typo errors of 1 per 10 pages. In a typing assignment of 40 pages, if the probability that the typo errors are at most 2 is $p$, then $e^{2}p=$ (A) $5$ (B) $13$ (C) $13e^{-2}$ (D) $5e^{-2}$

    ›Reveal solutionSolution

    The problem models rare typos with a Poisson distribution (mean = 4 typos in 40 pages). The probability of at most 2 typos is $p = e^{-4}(1 + 4 + 8) = 13e^{-4}$, so $e^{2}p = 13e^{-2}$, matching option (C).

    We have a typist who averages 1 typo per 10 pages. That’s a small rate for a rare event over a fixed “area” (pages). When events are rare and independent, the Poisson distribution is the natural choice — it counts the number of occurrences in a fixed interval when the average rate is known. Here, the “interval” is 40 pages.

    Why Poisson?

    • Each page has a small chance of a typo.
    • Pages are independent.
    • We care about the count of typos, not their arrangement. The Poisson distribution with parameter $\lambda$ (the mean number of events in the interval) fits perfectly.

    ---

    1. Find the average number of typos in 40 pages. The rate is 1 typo per 10 pages, so in 40 pages:

    $\lambda = \frac{1 \text{ typo}}{10 \text{ pages}} \times 40 \text{ pages} = 4.$

    2. Set up the Poisson probability formula. For a Poisson random variable $X$ with mean $\lambda$:

    $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}.$

    We need $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.

    3. Compute each term.

       • $P(X=0) = \frac{e^{-4} \cdot 4^0}{0!} = e^{-4}$.
       • $P(X=1) = \frac{e^{-4} \cdot 4^1}{1!} = 4e^{-4}$.
       • $P(X=2) = \frac{e^{-4} \cdot 4^2}{2!} = \frac{16e^{-4}}{2} = 8e^{-4}$.

    4. Sum them to get $p$.

       $p = e^{-4} + 4e^{-4} + 8e^{-4} = 13e^{-4}.$ …

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