Q.A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

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 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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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 the probability that an individual will suffer a bad reaction from an injection is 0.001, then the probability that out of 2000 individuals, exactly 3 individuals suffer a bad reaction is (A) $\dfrac{4}{3e^2}$ (B) $\dfrac{2}{e^2}$ (C) $\dfrac{2}{3e^2}$ (D) $\dfrac{4}{5e^2}$

   ›Reveal solutionSolution

   This is a rare-event problem where $n=2000$ is large and $p=0.001$ is small, so the Poisson approximation to the binomial distribution applies. The probability that exactly 3 individuals suffer a reaction is $\boxed{\dfrac{4}{3e^2}}$, which corresponds to option (A).

   The key insight here is that we are dealing with a binomial experiment: 2000 independent trials, each with a constant probability of "success" (bad reaction) $p = 0.001$. The exact probability of exactly 3 successes would be $\binom{2000}{3} (0.001)^3 (0.999)^{1997}$. That expression is perfectly correct but computationally messy — and more importantly, it misses the conceptual point.

   When $n$ is large and $p$ is very small, the binomial distribution is well approximated by the Poisson distribution with parameter $\lambda = np$. This is the classic "rare event" scenario. The Poisson distribution gives us a clean, closed-form answer that matches one of the given options exactly.

   Let's work through it step by step.

   1. Identify the exact binomial probability. For $n = 2000$, $p = 0.001$, the probability of exactly $k = 3$ reactions is

   $P(X = 3) = \binom{2000}{3} (0.001)^3 (0.999)^{1997}.$

   This is the truth, but we won't evaluate it directly — the numbers are unwieldy and the exam expects the Poisson approximation.

   2. Compute the Poisson parameter $\lambda$.

   $\lambda = np = 2000 \times 0.001 = 2.$

   This $\lambda = 2$ is the average number of bad reactions expected in 2000 individuals.

   3. Apply the Poisson approximation. For a Poisson random variable $Y$ with mean $\lambda$,

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

   Here $\lambda = 2$ and $k = 3$, so …

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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 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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7. 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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8. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.In a Poisson distribution with parameter $\lambda$, if $5P(X=3) = P(X=5)$, then $P(X=2) =$ (A) $\dfrac{25}{e^5}$ (B) $\dfrac{50}{e^{10}}$ (C) $\dfrac{30}{e^6}$ (D) $\dfrac{40}{e^8}$

   ›Reveal solutionSolution

   The key idea is to use the Poisson probability formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ and the given relation $5P(X=3) = P(X=5)$ to solve for $\lambda$, then compute $P(X=2)$. The final answer is $\frac{50}{e^{10}}$, which corresponds to option (B).

   The Poisson distribution models the number of events in a fixed interval when events occur independently at a constant average rate $\lambda$. The probability mass function is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$ for $k = 0, 1, 2, \dots$. Here, the condition $5P(X=3) = P(X=5)$ gives a direct equation in $\lambda$ because the $e^{-\lambda}$ factor cancels, leaving a simple algebraic relation. Once $\lambda$ is found, plugging into $P(X=2)$ yields the answer.

   1. Write the given condition using the Poisson formula:

   $5 \cdot \frac{e^{-\lambda} \lambda^3}{3!} = \frac{e^{-\lambda} \lambda^5}{5!}$

   The factor $e^{-\lambda}$ cancels on both sides (since $\lambda$ is finite), giving:

   $5 \cdot \frac{\lambda^3}{6} = \frac{\lambda^5}{120}$

   2. Simplify the equation. Multiply both sides by 120 to clear denominators:

   $5 \cdot \frac{\lambda^3}{6} \cdot 120 = \lambda^5$

   Compute $120/6 = 20$, so $5 \cdot 20 \cdot \lambda^3 = \lambda^5$, i.e., $100 \lambda^3 = \lambda^5$.

   3. Assuming $\lambda > 0$ (since it's a rate parameter), divide both sides by $\lambda^3$:

   $100 = \lambda^2$

   Hence $\lambda = 10$ (we take the positive root because $\lambda$ is a mean, always positive). …

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9. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.A non-zero integer $x$ is selected randomly from the set of integers $\{x \in \mathbb{Z} / -25 \leq x \leq 25,\, x \neq 0\}$. The probability that $x + 6 \leq \dfrac{135}{x}$ is (A) $\dfrac{12}{25}$ (B) $\dfrac{2}{5}$ (C) $\dfrac{3}{5}$ (D) $\dfrac{14}{25}$

   ›Reveal solutionSolution

   We need to find the probability that a non-zero integer $x$ from the set $\{-25, \dots, 25\}$ satisfies the inequality $x + 6 \leq \frac{135}{x}$. We first determine the total number of possible integers (the sample space), which is 50. Then, we solve the inequality to find the integers that satisfy it within the given range (the event space), which are 20 integers. The probability is $\boxed{\frac{2}{5}}$.

   The problem asks for a probability, which is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. Our strategy will be to first identify the complete set of possible integers $x$ (the sample space) and count them. Then, we will solve the given inequality to find which of these integers satisfy the condition (the event space) and count those. Finally, we will compute the ratio.

   1. Determine the Sample Space:

      The problem states that $x$ is a non-zero integer selected from the set $\{x \in \mathbb{Z} / -25 \leq x \leq 25,\, x \neq 0\}$.

      This means $x$ can be any integer from $-25$ to $25$, but $x$ cannot be $0$.

      The integers in this range are $\{-25, -24, \dots, -1, 0, 1, \dots, 24, 25\}$.

      The total count of integers from $-25$ to $25$ (inclusive) is $25 - (-25) + 1 = 51$.

      Since $x \neq 0$, we must exclude $0$ from this count.

      Therefore, the total number of possible outcomes (the size of the sample space) is $51 - 1 = 50$.

   2. Solve the Inequality:

      We need to find the integers $x$ that satisfy the inequality $x + 6 \leq \dfrac{135}{x}$.

      To solve rational inequalities, the most reliable method is to move all terms to one side and combine them into a single fraction. This avoids potential errors that arise from multiplying by a variable whose sign is unknown.

      $x + 6 - \dfrac{135}{x} \leq 0$

      To combine these terms, we find a common denominator, which is $x$:

      $\dfrac{x \cdot x}{x} + \dfrac{6 \cdot x}{x} - \dfrac{135}{x} \leq 0$

      $\dfrac{x^2 + 6x - 135}{x} \leq 0$

   3. Factor the Numerator:

      Now, we need to find the roots of the quadratic expression in the numerator, $x^2 + 6x - 135 = 0$. We can use the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

      $x = \dfrac{-6 \pm \sqrt{6^2 - 4(1)(-135)}}{2(1)}$

      $x = \dfrac{-6 \pm \sqrt{36 + 540}}{2}$

      $x = \dfrac{-6 \pm \sqrt{576}}{2}$

      Recognizing that $24^2 = 576$, we have:

      $x = \dfrac{-6 \pm 24}{2}$

      This gives two roots:

      $x_1 = \dfrac{-6 - 24}{2} = \dfrac{-30}{2} = -15$

      $x_2 = \dfrac{-6 + 24}{2} = \dfrac{18}{2} = 9$

      So, the numerator can be factored as $(x - (-15))(x - 9) = (x+15)(x-9)$.

      The inequality now becomes $\dfrac{(x+15)(x-9)}{x} \leq 0$.

      Watch out

      A common mistake is to multiply both sides of the inequality by $x$. This is incorrect because the sign of $x$ is unknown. If $x$ is negative, multiplying by $x$ would reverse the inequality sign. If $x$ is positive, it would not. Handling these two cases separately is cumbersome and prone to error. The method of moving all terms to one side and analyzing critical points is more robust.

   4. Determine Intervals Satisfying the Inequality:

      The critical points are the values of $x$ where the numerator or the denominator is zero. These are $x = -15$, $x = 0$, and $x = 9$. These points divide the number line into four intervals. We will test a value from each interval to determine the sign of the expression $\dfrac{(x+15)(x-9)}{x}$.

      Interval	Test Value ($x$)	Sign of $(x+15)$	Sign of $(x-9)$	Sign of $x$	Sign of $\dfrac{(x+15)(x-9)}{x}$	Condition $\leq 0$
      $x < -15$	$-20$	Negative	Negative	Negative	$\dfrac{(-)(-) }{(-)} = (-)$	True
      $-15 < x < 0$	$-1$	Positive	Negative	Negative	$\dfrac{(+)(-) }{(-)} = (+)$	False

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.In a Poisson distribution, if $\dfrac{P(X=5)}{P(X=2)} = \dfrac{1}{7500}$ and $\dfrac{P(X=5)}{P(X=3)} = \dfrac{1}{500}$, then the mean of the distribution is (A) $\dfrac{1}{15}$ (B) $\dfrac{1}{5}$ (C) $\dfrac{1}{25}$ (D) $\dfrac{1}{3}$

    ›Reveal solutionSolution

    The key idea is to use the ratio formulas for Poisson probabilities to eliminate the common factor and solve for the mean λ. The mean is found to be 1/5, so option (B) is correct.

    The Poisson distribution has probability mass function

    $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0,1,2,\dots$

    where $\lambda > 0$ is the mean. When we take ratios of probabilities, the factor $e^{-\lambda}$ cancels, leaving only powers of $\lambda$ and factorials. This makes ratios a clean way to solve for $\lambda$ without needing the actual probabilities.

    1. Write the given ratios in terms of $\lambda$.

    $\frac{P(X=5)}{P(X=2)} = \frac{\frac{e^{-\lambda}\lambda^5}{5!}}{\frac{e^{-\lambda}\lambda^2}{2!}} = \frac{\lambda^5 / 120}{\lambda^2 / 2} = \frac{\lambda^3}{60}$

    The problem states this equals $\frac{1}{7500}$. So:

    $\frac{\lambda^3}{60} = \frac{1}{7500}$

    2. Solve for $\lambda^3$ from the first ratio. Multiply both sides by 60:

    $\lambda^3 = \frac{60}{7500} = \frac{1}{125}$

    Hence $\lambda = \sqrt[3]{\frac{1}{125}} = \frac{1}{5}$.

    3. Check consistency with the second ratio.

    $\frac{P(X=5)}{P(X=3)} = \frac{\lambda^5 / 120}{\lambda^3 / 6} = \frac{\lambda^2}{20}$

    Plug $\lambda = 1/5$: …

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11. 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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12. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.If the probability that a student selected at random from a particular college is good at mathematics is 0.6, then the probability of having two students who are good at mathematics in a group of 8 students of that college standing in front of the college is (A) $\dfrac{2^6 \times 3^2 \times 7}{5^8}$ (B) $\dfrac{2^6 \times 3^2 \times 7}{5^6}$ (C) $\dfrac{2^8 \times 3^2 \times 7}{5^6}$ (D) $\dfrac{2^8 \times 3^2 \times 7}{5^8}$

    ›Reveal solutionSolution

    This is a binomial trial with $n=8$, $p=0.6$. $P(X=2)=\binom{8}{2}(0.6)^2(0.4)^6 = \dfrac{2^{8}\times 3^{2}\times 7}{5^{8}}$, option (D).

    Binomial model

    Each student is independently good at mathematics with probability $p = 0.6 = \dfrac{3}{5}$, so $q = 0.4 = \dfrac{2}{5}$. For $n=8$ students, the number good at mathematics is binomial, and we want exactly two:

    $P(X=2) = \binom{8}{2}\,p^{2}\,q^{6} = \binom{8}{2}\left(\frac{3}{5}\right)^{2}\left(\frac{2}{5}\right)^{6}.$

    Simplify

    $\binom{8}{2}=28,\qquad \left(\frac{3}{5}\right)^{2}=\frac{3^{2}}{5^{2}},\qquad \left(\frac{2}{5}\right)^{6}=\frac{2^{6}}{5^{6}}.$ …

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