Q.If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

Showing the 12 most recent of 30 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If $\dfrac{^{22}P_{r+1} : ^{20}P_{r+2}}{ } = 11 : 52$ then $r =$ (A) 3 (B) 5 (C) 7 (D) 9

   ›Reveal solutionSolution

   This problem involves simplifying a ratio of permutations using the permutation formula and then solving the resulting algebraic equation. We find that $r=7$.

   Concept and Intuition

   Permutations deal with the arrangement of distinct items. The number of permutations of $r$ items chosen from $n$ distinct items is denoted by $^nP_r$ and calculated using the formula:

   $^nP_r = \frac{n!}{(n-r)!}$

   For this formula to be valid, certain conditions must be met:

   1. $n$ must be a non-negative integer.
   2. $r$ must be a non-negative integer.
   3. $n \ge r$. This ensures that the term $(n-r)!$ is well-defined (i.e., we do not have a factorial of a negative number).

   In this problem, we are given a ratio of two permutation expressions. The key to solving such problems is to:

   1. Apply the permutation formula to expand each term.
   2. Simplify the resulting factorial expressions by cancelling common terms. Remember that $k! = k \times (k-1)!$ and generally $k! = k \times (k-1) \times \dots \times (k-m+1) \times (k-m)!$. This property is crucial for simplifying ratios of factorials.
   3. Solve the algebraic equation for $r$.
   4. Finally, check if the obtained value of $r$ satisfies the conditions for permutations for both terms in the original problem.

   Let's apply these ideas to the given problem.

   Step-by-step Solution

   1. Write down the given ratio and apply the permutation formula.

      We are given the ratio $\dfrac{^{22}P_{r+1}}{^{20}P_{r+2}} = \dfrac{11}{52}$.

      Using the formula $^nP_r = \frac{n!}{(n-r)!}$:

      • For $^{22}P_{r+1}$: $n=22$, $r_{perm} = r+1$. So, $^{22}P_{r+1} = \frac{22!}{(22-(r+1))!} = \frac{22!}{(21-r)!}$.
      • For $^{20}P_{r+2}$: $n=20$, $r_{perm} = r+2$. So, $^{20}P_{r+2} = \frac{20!}{(20-(r+2))!} = \frac{20!}{(18-r)!}$.

      Substituting these into the ratio:

   $\frac{\frac{22!}{(21-r)!}}{\frac{20!}{(18-r)!}} = \frac{11}{52}$

   2. Simplify the expression by rearranging and expanding factorials. We can rewrite the left side by inverting the denominator and multiplying:

   $\frac{22!}{(21-r)!} \times \frac{(18-r)!}{20!} = \frac{11}{52}$

   Now, we expand the larger factorials in terms of smaller ones to facilitate cancellation:
   *   $22! = 22 \times 21 \times 20!$
   *   $(21-r)! = (21-r) \times (20-r) \times (19-r) \times (18-r)!$
   
   Substitute these expansions into the equation:
   

   $\frac{22 \times 21 \times 20!}{(21-r)(20-r)(19-r)(18-r)!} \times \frac{(18-r)!}{20!} = \frac{11}{52}$

   3. Cancel common factorial terms. Notice that $20!$ in the numerator and denominator cancel out. Similarly, $(18-r)!$ in the numerator and denominator cancel out:

   $\frac{22 \times 21}{(21-r)(20-r)(19-r)} = \frac{11}{52}$

   Calculate the product in the numerator: $22 \times 21 = 462$.
   

   $\frac{462}{(21-r)(20-r)(19-r)} = \frac{11}{52}$

   4. Solve the algebraic equation for $r$. …

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7. 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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8. 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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9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $X$ is a Poisson variate such that $\frac{5}{3}k = P(X=2) = P(X=3)$, then $P(X=5)=$ (A) $k$ (B) $\frac{1}{4}k$ (C) $\frac{1}{2}k$ (D) $\frac{3}{4}k$

   ›Reveal solutionSolution

   We use the Poisson Probability Mass Function to equate $P(X=2)$ and $P(X=3)$, which allows us to determine the parameter $\lambda$. Once $\lambda$ is known, we can express $P(X=5)$ in terms of $k$. The result is $P(X=5) = \boxed{\frac{3}{4}k}$.

   The Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space, given a constant average rate of occurrence and independence of events. It is characterized by a single parameter, $\lambda$ (lambda), which represents the average number of events in the given interval.

   The core idea here is to use the given equality of probabilities, $P(X=2) = P(X=3)$, to find the value of this parameter $\lambda$. Once $\lambda$ is known, we can calculate any other probability $P(X=x)$ using the Poisson Probability Mass Function (PMF). We are also given a relationship involving $k$, which we will use to express our final answer in terms of $k$.

   1. Recall the Poisson Probability Mass Function (PMF):

      For a Poisson variate $X$ with parameter $\lambda$, the probability of observing exactly $x$ events is given by:

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

      where $x = 0, 1, 2, \dots$ and $\lambda > 0$.

   2. Use the given condition $P(X=2) = P(X=3)$ to find $\lambda$:

      Substitute $x=2$ and $x=3$ into the PMF:

      $P(X=2) = \frac{e^{-\lambda} \lambda^2}{2!}$

      $P(X=3) = \frac{e^{-\lambda} \lambda^3}{3!}$

      Equating these two probabilities:

      $\frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^3}{3!}$

      Since $e^{-\lambda}$ is never zero and $\lambda$ must be positive (as probabilities are non-zero), we can divide both sides by $e^{-\lambda} \lambda^2$:

      $\frac{1}{2!} = \frac{\lambda}{3!}$

      Recall that $2! = 2 \times 1 = 2$ and $3! = 3 \times 2 \times 1 = 6$.

      $\frac{1}{2} = \frac{\lambda}{6}$

      Multiply both sides by 6 to solve for $\lambda$:

      $\lambda = \frac{6}{2} = 3$

      So, the parameter of the Poisson distribution is $\lambda = 3$.

   3. Express $k$ in terms of $\lambda$ (and $e^{-\lambda}$):

      We are given that $\frac{5}{3}k = P(X=2)$.

      We know $P(X=2) = \frac{e^{-\lambda} \lambda^2}{2!}$. Substitute $\lambda=3$:

      $P(X=2) = \frac{e^{-3} 3^2}{2} = \frac{9e^{-3}}{2}$

      Now, equate this to $\frac{5}{3}k$:

      $\frac{5}{3}k = \frac{9e^{-3}}{2}$

      Solve for $k$:

      $k = \frac{3}{5} \times \frac{9e^{-3}}{2} = \frac{27e^{-3}}{10}$

   4. Calculate $P(X=5)$ using $\lambda=3$: …

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10. 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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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 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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