Q.(b) A random variable $X$ can take all non-negative integral values and the probability that $X$ takes the value $r$ is proportional to $5^{-r}$. Find $P(X < 3)$.

Showing the 12 most recent of 28 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If a random variable X has the following probability distribution, then the mean of X is

   [!FORMULA] $\begin{array}{|c|c|c|c|c|c|} \hline X = x_i & 1 & 3 & 5 & 7 & 9 \\ \hline P(X = x_i) & 3k & 5k & k^2 & 3k^2 + k & 6k^2 \\ \hline \end{array}$

   (A) 9.6 (B) 8.4 (C) 10.2 (D) 3.3

   ›Reveal solutionSolution

   To find the mean of a discrete random variable, first determine the unknown constant $k$ by using the property that the sum of all probabilities must be 1. Then, calculate the mean using the formula $E(X) = \sum x_i P(X=x_i)$. The mean of $X$ is $\boxed{3.3}$.

   The core idea behind this problem is understanding the properties of a probability distribution for a discrete random variable and how to calculate its mean (also known as the expected value).

   A probability distribution lists all possible values a random variable can take and their corresponding probabilities. For any valid probability distribution, two fundamental rules must hold:

   1. Each probability must be non-negative: $P(X=x_i) \ge 0$ for all $x_i$.
   2. The sum of all probabilities must be 1: $\sum P(X=x_i) = 1$.

   The mean, or expected value, of a discrete random variable $X$ is a measure of its central tendency. It represents the average value of $X$ over a very large number of trials. It's calculated as a weighted average, where each possible value $x_i$ is weighted by its probability $P(X=x_i)$.

   The mean (expected value) of a discrete random variable $X$ is given by:

   $E(X) = \sum_{i} x_i P(X=x_i)$

   We will use these concepts to first find the unknown constant $k$, then determine the actual probabilities, and finally calculate the mean.

   1. Determine the value of $k$

      The sum of all probabilities in a probability distribution must be equal to 1.

      From the given table, we have:

      $P(X=1) = 3k$

      $P(X=3) = 5k$

      $P(X=5) = k^2$

      $P(X=7) = 3k^2 + k$

      $P(X=9) = 6k^2$

      Summing these probabilities and setting them equal to 1:

      $3k + 5k + k^2 + (3k^2 + k) + 6k^2 = 1$

      Combine the terms involving $k$ and $k^2$:

      $(3k + 5k + k) + (k^2 + 3k^2 + 6k^2) = 1$

      $9k + 10k^2 = 1$

      Rearrange this into a standard quadratic equation:

      $10k^2 + 9k - 1 = 0$

      We can solve this quadratic equation by factoring. We look for two numbers that multiply to $10 \times (-1) = -10$ and add up to $9$. These numbers are $10$ and $-1$.

      $10k^2 + 10k - k - 1 = 0$

      Factor by grouping:

      $10k(k+1) - 1(k+1) = 0$

      $(10k - 1)(k+1) = 0$

      This gives two possible values for $k$:

      $10k - 1 = 0 \implies k = \frac{1}{10}$

      $k + 1 = 0 \implies k = -1$

      Watch out

      Probabilities must always be non-negative. We must check both values of $k$ to ensure they result in valid probabilities.

      If $k = -1$, then $P(X=1) = 3k = 3(-1) = -3$. A probability cannot be negative, so $k=-1$ is not a valid solution.

      If $k = \frac{1}{10}$, all probability expressions will be non-negative:

      $P(X=1) = 3(\frac{1}{10}) = \frac{3}{10} \ge 0$

      $P(X=3) = 5(\frac{1}{10}) = \frac{5}{10} \ge 0$

      $P(X=5) = (\frac{1}{10})^2 = \frac{1}{100} \ge 0$

      $P(X=7) = 3(\frac{1}{10})^2 + \frac{1}{10} = \frac{3}{100} + \frac{10}{100} = \frac{13}{100} \ge 0$

      $P(X=9) = 6(\frac{1}{10})^2 = \frac{6}{100} \ge 0$

      All probabilities are non-negative, so $k = \frac{1}{10}$ is the correct value.

   2. Calculate the actual probabilities

      Substitute $k = \frac{1}{10}$ (or $0.1$) into each probability expression:

      $P(X=1) = 3k = 3(0.1) = 0.3$

      $P(X=3) = 5k = 5(0.1) = 0.5$

      $P(X=5) = k^2 = (0.1)^2 = 0.01$

      $P(X=7) = 3k^2 + k = 3(0.1)^2 + 0.1 = 3(0.01) + 0.1 = 0.03 + 0.1 = 0.13$

      $P(X=9) = 6k^2 = 6(0.1)^2 = 6(0.01) = 0.06$

      Let's verify the sum of these probabilities: $0.3 + 0.5 + 0.01 + 0.13 + 0.06 = 0.8 + 0.2 = 1.0$. The sum is indeed 1.

   3. Calculate the mean of $X$

      Using the formula $E(X) = \sum x_i P(X=x_i)$:

      $E(X) = (1 \times P(X=1)) + (3 \times P(X=3)) + (5 \times P(X=5)) + (7 \times P(X=7)) + (9 \times P(X=9))$

      $E(X) = (1 \times 0.3) + (3 \times 0.5) + (5 \times 0.01) + (7 \times 0.13) + (9 \times 0.06)$

      $E(X) = 0.3 + 1.5 + 0.05 + 0.91 + 0.54$

      Now, sum these values:

      $E(X) = 0.30 + 1.50 + 0.05 + 0.91 + 0.54 = 3.30$

   4. Compare with options

      The calculated mean is $3.3$. This matches option (D).

   ✓Final answer

   The mean of $X$ is $\boxed{3.3}$.

2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Let $p$ be the probability of getting a success in one trial and $0 < p < 1$. If $X$ is the random variable which represents the number of trials until first success occurs is given by

   [!FORMULA] $P(X = k) = \lambda (1 - p)^{k-1},\ k = 1, 2, 3, \ldots, \infty$

   then $\lambda =$ (A) $1 - p$ (B) $p$ (C) $p^2$ (D) $\dfrac{1}{p}$

   ›Reveal solutionSolution

   The key idea is that the probabilities for a geometric distribution must sum to 1 over all possible outcomes. Using the infinite geometric series formula, we find that $\lambda = p$, so the correct option is (B).

   We are told that $X$ counts the number of trials until the first success, and the probability mass function is given as

   $P(X = k) = \lambda (1 - p)^{k-1}, \quad k = 1, 2, 3, \dots$

   This is almost the standard geometric distribution — except the constant $\lambda$ is unknown. Our job is to find $\lambda$ so that this is a valid probability distribution.

   ---

   Why this approach works

   A probability distribution must satisfy that the sum of probabilities over all possible values equals 1. Here, $k$ runs from 1 to infinity, so we need

   $\sum_{k=1}^\infty P(X = k) = 1.$

   Substituting the given form gives an infinite geometric series. The sum of such a series has a simple closed form, which lets us solve for $\lambda$.

   ---

   Step-by-step reasoning

   1. Write the sum condition Since $P(X = k)$ must be a valid probability mass function, we have

   $\sum_{k=1}^\infty \lambda (1-p)^{k-1} = 1.$

   2. Factor out the constant $\lambda$ does not depend on $k$, so

   $\lambda \sum_{k=1}^\infty (1-p)^{k-1} = 1.$

   3. Recognize the geometric series Let $r = 1-p$. Then the sum is

   $\sum_{k=1}^\infty r^{k-1} = 1 + r + r^2 + r^3 + \cdots$

   This is an infinite geometric series with first term $1$ and common ratio $r$. Since $0 < p < 1$, we have $0 < r < 1$, so the series converges.

   4. Apply the geometric series formula For $|r| < 1$,

   $\sum_{n=0}^\infty r^n = \frac{1}{1-r}.$

   Here $n = k-1$, so

   $\sum_{k=1}^\infty r^{k-1} = \frac{1}{1-r} = \frac{1}{1 - (1-p)} = \frac{1}{p}.$

   5. Solve for $\lambda$ Substituting back:

   $\lambda \cdot \frac{1}{p} = 1 \quad \Rightarrow \quad \lambda = p.$

   ---

   Tip

   A quick check: For the standard geometric distribution, $P(X = k) = p(1-p)^{k-1}$. So $\lambda = p$ makes the given formula exactly that — confirming our result.

   Watch out

   A common mistake is to forget that the sum starts at $k=1$, not $k=0$. If you mistakenly sum from $k=0$, you'd get $\frac{1}{1-(1-p)} = \frac{1}{p}$ but then the exponent would be off, leading to a wrong $\lambda$.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the probability function of a random variable $X$ is $P(X=x) = a^x k$, $x = 0,1,2,\ldots \infty$, then $k =$ (A) $1-a$ when $0<a<1$ (B) $1-a$ for all positive $a$ (C) $a-1$ when $a>1$ (D) $1+a$ for all real $a$

   ›Reveal solutionSolution

   This tests recognizing a geometric probability distribution and using $\sum P(X=x)=1$ together with the convergence condition on $a$. Answer: $k=1-a$, valid only for $0<a<1$.

   Concept and Intuition

   $P(X=x)=a^x k$ is the geometric distribution's probability mass function (up to the constant $k$). Any valid pmf must sum to exactly $1$ over all its support, and here the support is infinite ($x=0,1,2,\dots$), so we need the infinite geometric series $\sum_{x=0}^\infty a^x$ to actually converge — which only happens for $|a|<1$.

   Step-by-Step Solution

   1. Require $\displaystyle\sum_{x=0}^{\infty} P(X=x) = 1$: $\displaystyle k\sum_{x=0}^{\infty}a^x = 1$.
   2. The infinite geometric series $\sum_{x=0}^\infty a^x = \dfrac{1}{1-a}$ converges only when $|a|<1$.
   3. Since $a^xk \geq 0$ must hold for every $x$, and $a$ must be positive for the series to make sense as probabilities (a negative $a$ gives sign-alternating terms), we need $0<a<1$.
   4. Then $k\cdot\dfrac{1}{1-a}=1 \Rightarrow k=1-a$.

   Common Mistakes

   • Forgetting the convergence restriction $0<a<1$ and claiming $k=1-a$ "for all positive $a$" (fails for $a\geq 1$, where the series diverges and no valid $k$ exists).
   • Sign errors turning $k=1-a$ into $k=a-1$ (which would be negative for $0<a<1$, invalid as a probability constant).

   ✓Final answer

   The correct option is (A) — $1-a$ when $0<a<1$.

   ANSWER: A

4. COMEDK 2026Set 2026-A1 markMCQ
   Q.A company is migrating its database, and two software engineers, Ishaan and Kavya, take turns running a data-sync script that has a constant success rate of $\frac{3}{8}$ per attempt. If Ishaan initiates the first attempt and they persist until the migration is successful, what is the probability that Kavya is the one who initiates the successful sync? (A) $\frac{5}{13}$ (B) $\frac{8}{11}$ (C) $\frac{8}{13}$ (D) $\frac{3}{11}$

   ›Reveal solutionSolution

   The probability that Kavya succeeds first is the sum of an infinite geometric series where she wins on her 1st, 2nd, 3rd, … turn. The result is $\frac{5}{13}$, which corresponds to option (A).

   Concept & Intuition

   This is a classic “alternating turns with constant success probability” problem. Because the attempts are independent and the success rate is fixed, the game can continue indefinitely. The key is to notice that Kavya only gets a turn if Ishaan fails on his first attempt. Then, on her first turn, she might succeed immediately, or if she fails, the whole pattern repeats from the start. This recursive structure lets us set up a simple equation.

   Step-by-step reasoning

   1. Define the probability we want.

      Let $P$ be the probability that Kavya eventually initiates the successful sync. Since Ishaan goes first, the only way Kavya ever gets to attempt is if Ishaan fails on his first try.

   2. Condition on the first two attempts.

      • Ishaan’s first attempt: success probability $p = \frac{3}{8}$, failure probability $q = 1-p = \frac{5}{8}$.
      • If Ishaan succeeds immediately (probability $\frac{3}{8}$), Kavya never gets a turn — so that branch contributes $0$ to $P$.
      • If Ishaan fails (probability $\frac{5}{8}$), then it’s Kavya’s turn. On her first attempt:
        • She succeeds with probability $\frac{3}{8}$ → game ends, and she wins. This contributes $\frac{5}{8} \cdot \frac{3}{8}$ to $P$.
        • She fails with probability $\frac{5}{8}$ → then the game resets exactly to the starting state (Ishaan’s turn again), because both have failed once and the next attempt is Ishaan’s. From that point, the probability that Kavya eventually wins is again $P$.

   3. Write the equation.

      Putting the branches together:

   $P = 0 \cdot \frac{3}{8} \;+\; \frac{5}{8}\left( \frac{3}{8} + \frac{5}{8} \cdot P \right)$

   Simplify:

   $P = \frac{5}{8} \cdot \frac{3}{8} + \frac{5}{8} \cdot \frac{5}{8} \cdot P$

   $P = \frac{15}{64} + \frac{25}{64} P$

   4. Solve for $P$. Subtract $\frac{25}{64}P$ from both sides:

   $P - \frac{25}{64}P = \frac{15}{64}$

   $\frac{39}{64}P = \frac{15}{64}$

   Multiply both sides by $\frac{64}{39}$:

   $P = \frac{15}{39} = \frac{5}{13}$

   Tip

   You can also sum the infinite geometric series directly: Kavya wins on her 1st turn (Ishaan fails, she succeeds): $\left(\frac{5}{8}\right)^1 \cdot \frac{3}{8}$; on her 2nd turn (both fail once, then she succeeds): $\left(\frac{5}{8}\right)^3 \cdot \frac{3}{8}$; on her $k$th turn: $\left(\frac{5}{8}\right)^{2k-1} \cdot \frac{3}{8}$. Summing gives $\frac{3}{8} \cdot \frac{5/8}{1 - (25/64)} = \frac{5}{13}$.

   Watch out

   A common mistake is to forget that after both fail once, the game resets exactly — the probability for Kavya from that point is the same $P$, not a different value. Also, do not confuse “Kavya initiates the successful sync” with “Kavya’s turn is the successful one” — that is exactly what we computed.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. COMEDK 2026Set 2026-M1 markMCQ
   Q.An engineering team is testing a new prototype drone. The drone has constant success rate of $\frac{\mathbf{2}}{\mathbf{7}}$ for every autonomous landing attempt. Two engineers, Sarah and Swarna, take turns initiating the landing sequence, with Swarna going first. If they continue the process until a landing is successful, what is the probability that Sarah is the one who initiates the successful landing? (A) $\frac{5}{12}$ (B) $\frac{30}{37}$ (C) $\frac{7}{37}$ (D) $\frac{7}{12}$

   ›Reveal solutionSolution

   This is a geometric probability problem with alternating turns. The probability that Sarah (the second player) makes the first successful landing is the sum of an infinite geometric series: $\frac{5}{7} \cdot \frac{2}{7} + \left(\frac{5}{7}\right)^3 \cdot \frac{2}{7} + \cdots = \frac{10}{49} \cdot \frac{1}{1 - \frac{25}{49}} = \frac{10}{24} = \frac{5}{12}$. The correct option is (A).

   ---

   Concept and intuition:

   When two players take turns and each attempt has the same independent probability of success, the game is a "first success" contest. The key is to realize that Sarah can only win if Swarna fails on her first turn, then Sarah succeeds; or if both fail twice, then Sarah succeeds; and so on. This creates an infinite series where each term corresponds to a round of two failures followed by Sarah's success. Because the attempts are independent, we multiply probabilities along each path and sum over all possible paths.

   ---

   1. Define the probabilities.

      Let $p = \frac{2}{7}$ be the probability of a successful landing on any single attempt.

      Let $q = 1 - p = \frac{5}{7}$ be the probability of failure.

   2. Identify when Sarah can win.

      Swarna goes first. So Sarah’s first possible turn is attempt #2.

      Sarah wins if:

      • Swarna fails (attempt 1), then Sarah succeeds (attempt 2).
      • Or Swarna fails (1), Sarah fails (2), Swarna fails (3), Sarah succeeds (4).
      • Or in general, after $k$ full rounds of both failing, Sarah succeeds on her next turn.

   3. Write the probability for the first scenario.

      Probability Sarah wins on her first turn:

   $q \cdot p = \frac{5}{7} \cdot \frac{2}{7} = \frac{10}{49}.$

   4. Write the probability for the second scenario. Both fail in the first round, then Swarna fails again, then Sarah succeeds:

   $q \cdot q \cdot q \cdot p = q^3 p = \left(\frac{5}{7}\right)^3 \cdot \frac{2}{7} = \frac{125}{343} \cdot \frac{2}{7} = \frac{250}{2401}.$

   5. General pattern. For Sarah to win on her $n$-th turn (where $n \ge 1$), there must be $n-1$ full rounds of both failing (each round: $q^2$), then Swarna fails once more, then Sarah succeeds. So the probability for the $n$-th turn is:

   $(q^2)^{n-1} \cdot q \cdot p = q^{2n-1} p.$

   6. Sum the infinite series.

   $P(\text{Sarah wins}) = \sum_{n=1}^{\infty} q^{2n-1} p = p q \sum_{n=1}^{\infty} (q^2)^{n-1}.$

   This is a geometric series with first term $p q$ and common ratio $q^2$:

   $= p q \cdot \frac{1}{1 - q^2}.$

   7. Plug in the numbers. $p = \frac{2}{7}$, $q = \frac{5}{7}$, so $q^2 = \frac{25}{49}$.

   $P = \frac{2}{7} \cdot \frac{5}{7} \cdot \frac{1}{1 - \frac{25}{49}} = \frac{10}{49} \cdot \frac{1}{\frac{24}{49}} = \frac{10}{49} \cdot \frac{49}{24} = \frac{10}{24} = \frac{5}{12}.$

   Tip

   A faster way: The ratio of Sarah’s winning probability to Swarna’s is $q : 1$ (since Sarah only gets a turn if Swarna fails first). So $P(\text{Sarah}) = \frac{q}{1+q} = \frac{5/7}{1 + 5/7} = \frac{5/7}{12/7} = \frac{5}{12}$. This works because the game is memoryless and symmetric after the first turn.

   Watch out

   A common mistake is to forget that Swarna goes first, so the first term in Sarah’s series is $q p$, not $p$. Also, be careful not to double-count the case where Swarna succeeds immediately — that’s not part of Sarah’s probability.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If three dice are thrown, then the mean of the sum of the numbers appearing on them is (A) 58.5 (B) 76.66 (C) 71.75 (D) 10.5

   ›Reveal solutionSolution

   The mean of the sum of three dice is just 3 times the mean of a single die. Since a fair die has mean 3.5, the answer is $3 \times 3.5 = 10.5$, which corresponds to option (D).

   Concept & Intuition

   When you roll three dice, the total is the sum of three independent random variables, each representing the number on one die. A key principle in probability is linearity of expectation: the expected value of a sum is always the sum of the expected values, regardless of whether the variables are independent. So instead of listing all 216 possible outcomes and averaging them, we can simply find the mean of one die and multiply by 3. This saves enormous time and avoids error.

   Step-by-step reasoning

   1. Mean of a single fair die A standard die has faces 1, 2, 3, 4, 5, 6, each equally likely. The expected value is

   $E[X] = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5.$

   2. Apply linearity of expectation Let $X_1, X_2, X_3$ be the numbers on the three dice. Then the total sum is $S = X_1 + X_2 + X_3$. By linearity,

   $E[S] = E[X_1] + E[X_2] + E[X_3] = 3.5 + 3.5 + 3.5 = 10.5.$

   3. Interpret the result The mean sum is 10.5, which is exactly halfway between 3 (all ones) and 18 (all sixes). This makes intuitive sense: the average of many rolls will cluster around the center of the possible range.

   Watch out

   A common mistake is to think the mean of three dice is the mean of the possible sums (3 to 18), which is $(3+18)/2 = 10.5$ — that happens to give the same number here, but that shortcut only works because the distribution is symmetric. For non-symmetric cases, always use expectation.

   Tip

   Linearity of expectation works even if the dice were loaded or had different numbers of faces — just compute each die’s mean separately and add them.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the probability distribution of a discrete random variable X is given by $P(X=k) = \dfrac{2^{-k}(3k+1)}{2^c}, k=0,1,2,\ldots\infty$ then $P(X\le c) =$ (A) $\dfrac{c}{5}$ (B) $\dfrac{c}{4}$ (C) $\dfrac{c+2}{5}$ (D) $\dfrac{c-2}{7}$

   ›Reveal solutionSolution

   Normalizing the distribution pins down $c=3$, and then $P(X\le c)=P(X\le3)=\dfrac34=\dfrac{c}{4}$.

   Concept and Intuition

   First use $\sum_k P(X=k)=1$ to determine the unknown constant $2^c$ in the denominator (splitting the series into an arithmetico-geometric sum), then directly sum the finite number of terms up to $k=c$.

   Step-by-Step Solution

   1. Require $\displaystyle\sum_{k=0}^\infty \frac{2^{-k}(3k+1)}{2^c}=1$, i.e. $\displaystyle\sum_{k=0}^\infty (3k+1)\left(\frac12\right)^k = 2^c$.
   2. With $x=\frac12$: $\sum kx^k=\dfrac{x}{(1-x)^2}=\dfrac{1/2}{1/4}=2$, and $\sum x^k=\dfrac{1}{1-x}=2$.
   3. So the sum $=3(2)+2=8$, giving $2^c=8\Rightarrow c=3$.
   4. Now compute $P(X\le3)=\dfrac{1}{8}\sum_{k=0}^3 2^{-k}(3k+1)$:
      • $k=0$: $1\cdot1=1$
      • $k=1$: $\frac12\cdot4=2$
      • $k=2$: $\frac14\cdot7=\frac74$
      • $k=3$: $\frac18\cdot10=\frac54$
      • Sum $=1+2+\frac74+\frac54=3+3=6$.
   5. $P(X\le3)=\dfrac68=\dfrac34$.
   6. Since $c=3$: $\dfrac{c}{4}=\dfrac34$ matches — option (B).

   Common Mistakes

   • Forgetting to first solve for $c$ from the normalization condition before evaluating $P(X\le c)$ — the answer choices are expressed symbolically in $c$, so both steps are needed.
   • Errors in summing the arithmetico-geometric series $\sum k x^k$ (misremembering the formula $\frac{x}{(1-x)^2}$).

   ✓Final answer

   The correct option is (B) — $\dfrac{c}{4}$.

   ANSWER: B

8. COMEDK 2025Set 2025-E1 markMCQ
   Q.In a game, a man wins ₹ 1000 if he gets an even number greater than or equal to 4 on a fair dice and loses ₹ 200 for getting any other number on the dice. If he decides to throw the dice until he wins or maximum of three times, then his expected gain/loss in (₹) is ----------- (A) $\frac{2200}{9}$ loss (B) $\frac{3800}{9}$ gain (C) $\frac{2200}{9}$ gain (D) $\frac{3800}{9}$ loss

   ›Reveal solutionSolution

   Winning (a 4 or 6) has probability $\tfrac13$ each throw; totalling the payoffs over up to three throws gives an expected value of $\tfrac{3800}{9}$ gain — option (B).

   Set up the probabilities. A "win" means an even number $\ge 4$, i.e. $\{4,6\}$:

   $P(\text{win})=\tfrac{2}{6}=\tfrac13,\qquad P(\text{lose})=\tfrac23,\ \text{payoff } -200 \text{ per losing throw}.$

   He keeps throwing until he wins or has thrown three times. Enumerate the outcomes (a losing throw costs $200$ and he continues):

   Outcome	Probability	Net amount (₹)
   Win on throw 1	$\tfrac13$	$+1000$
   Lose, Win on throw 2	$\tfrac23\cdot\tfrac13=\tfrac29$	$-200+1000=+800$
   Lose, Lose, Win on throw 3	$\tfrac23\cdot\tfrac23\cdot\tfrac13=\tfrac{4}{27}$	$-400+1000=+600$
   Lose all three	$\left(\tfrac23\right)^3=\tfrac{8}{27}$	$-600$

   Expected value:

   $E=\tfrac13(1000)+\tfrac29(800)+\tfrac{4}{27}(600)+\tfrac{8}{27}(-600).$

   Over a common denominator of $27$:

   $E=\frac{9000+4800+2400-4800}{27}=\frac{11400}{27}=\frac{3800}{9}.$

   Since $E>0$, it is a gain of $\dfrac{3800}{9}$.

   ✓Final answer

   Expected gain of $\dfrac{3800}{9}$ ₹ — (B) $\dfrac{3800}{9}$ gain

9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If on an average 4 customers visit a shop in an hour, then the probability that more than 2 customers visit the shop in a specific hour is (A) $\dfrac{e^4 - 13}{e^4}$ (B) $\dfrac{8}{e^4}$ (C) $\dfrac{4}{e^4}$ (D) $\dfrac{e^4 - 21}{e^4}$

   ›Reveal solutionSolution

   This is a Poisson probability problem with mean λ = 4. We need P(X > 2) = 1 − P(X ≤ 2). The correct probability simplifies to (e⁴ − 13)/e⁴, which corresponds to option (A).

   Concept & Intuition

   When events occur independently at a constant average rate (here, 4 customers per hour), the number of events in a fixed time interval follows a Poisson distribution. The key is that “more than 2” means 3, 4, 5, … — an infinite sum. Instead of summing infinitely, we use the complement: subtract the probabilities of 0, 1, and 2 from 1. The Poisson formula is simple:

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

   with λ = 4.

   Step-by-step solution

   1. Identify the distribution and parameter

      Average rate λ = 4 customers per hour. Let X = number of customers in a given hour. Then X ~ Poisson(λ = 4).

   2. Express the desired probability

      We want P(X > 2). Since the total probability is 1,

   $P(X > 2) = 1 - P(X \le 2) = 1 - \big[P(X=0) + P(X=1) + P(X=2)\big].$

   3. Compute each term using the Poisson formula
      • For k = 0:

   $P(X=0) = \frac{e^{-4} \cdot 4^0}{0!} = e^{-4}.$

   • For k = 1:

   $P(X=1) = \frac{e^{-4} \cdot 4^1}{1!} = 4e^{-4}.$

   • For k = 2:

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

   4. Sum the probabilities for k = 0, 1, 2

   $P(X \le 2) = e^{-4} + 4e^{-4} + 8e^{-4} = (1 + 4 + 8)e^{-4} = 13e^{-4}.$

   5. Subtract from 1 to get P(X > 2)

   $P(X > 2) = 1 - 13e^{-4} = \frac{e^4}{e^4} - \frac{13}{e^4} = \frac{e^4 - 13}{e^4}.$

   Watch out

   A common mistake is to forget that “more than 2” excludes 2 itself. Also, don’t confuse “more than 2” with “at least 2” — that would include 2, giving a different result.

   Tip

   Notice that the denominator e⁴ appears in all options. Once you compute the numerator as e⁴ − 13, you can match it directly to option (A) without further simplification.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.The variance of the first 10 natural numbers which are multiples of 3 is (A) 53 (B) 73 (C) 52.5 (D) 74.25

    ›Reveal solutionSolution

    The variance of the first 10 multiples of 3 (3, 6, …, 30) is 74.25, which corresponds to option (D). The key is to treat the data as an arithmetic progression and use the variance formula for such a sequence.

    Concept and Intuition

    The first 10 natural numbers that are multiples of 3 form an arithmetic progression: 3, 6, 9, …, 30. Variance measures how spread out numbers are from their mean. For an arithmetic progression, the variance depends only on the common difference and the number of terms, not on the starting point. Here, the common difference is 3, so the variance will be $3^2$ times the variance of the first 10 natural numbers (1, 2, …, 10). This scaling property saves us from computing each term individually.

    Step-by-step solution

    1. Identify the data set The first 10 multiples of 3 are:

    $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$

    This is an arithmetic progression with first term $a = 3$, common difference $d = 3$, and number of terms $n = 10$.

    2. Find the mean The mean of an arithmetic progression is the average of the first and last terms:

    $\bar{x} = \frac{a + l}{2} = \frac{3 + 30}{2} = 16.5$

    Alternatively, sum of terms = $\frac{n}{2}(a + l) = 5 \times 33 = 165$, so mean = $165/10 = 16.5$.

    3. Use the variance formula for grouped data

       Variance $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2$. We could compute each deviation, but a smarter way is to use the property of arithmetic progressions.

    4. Apply the scaling trick

       Let $y_i = 1, 2, 3, \dots, 10$. Then $x_i = 3y_i$. Variance scales with the square of the multiplier:

    $\text{Var}(x) = 3^2 \cdot \text{Var}(y)$

    So we first find the variance of the first 10 natural numbers.

    5. Variance of first 10 natural numbers For numbers $1, 2, \dots, n$, the variance is $\frac{n^2 - 1}{12}$. For $n = 10$:

    $\text{Var}(y) = \frac{10^2 - 1}{12} = \frac{100 - 1}{12} = \frac{99}{12} = 8.25$

    Tip

    The formula $\frac{n^2-1}{12}$ comes from $\frac{1}{n}\sum y_i^2 - (\bar{y})^2$. For $n=10$, $\bar{y}=5.5$, $\sum y_i^2 = 385$, so $\frac{385}{10} - 30.25 = 38.5 - 30.25 = 8.25$. It matches.

    6. Scale back to the multiples of 3

    $\text{Var}(x) = 9 \times 8.25 = 74.25$

    This is the population variance (since we consider the entire set of first 10 multiples).

    7. Check against options The value 74.25 appears as option (D). Options (A) 53, (B) 73, and (C) 52.5 are incorrect.

    Watch out

    A common mistake is to use the sample variance formula (dividing by $n-1$) instead of population variance. The problem asks for "the variance" of the given numbers, which typically means population variance (divide by $n$). If you mistakenly divided by 9, you’d get 82.5, which isn’t even an option — so the intended interpretation is clear.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.If a random variable $X$ has the following probability distribution, then its variance is

    [!FORMULA] $\begin{array}{c|cccc} X = x & 1 & 3 & 5 & 2 \\ \hline P(X = x) & 3K^2 & K & K^2 & 2K \\ \end{array}$

    (A) $\dfrac{9}{4}$ (B) $\dfrac{25}{8}$ (C) $\dfrac{27}{16}$ (D) $\dfrac{15}{16}$

    ›Reveal solutionSolution

    The key idea is to first find the constant $K$ using the total probability condition, then compute the mean $\mu$ and variance $\sigma^2 = E[X^2] - \mu^2$. The variance is $\frac{27}{16}$, so the correct option is (C).

    We are given a discrete probability distribution with values $x = 1, 3, 5, 2$ and corresponding probabilities $3K^2, K, K^2, 2K$. The first step is always to find $K$ so that the probabilities sum to 1. Then we compute the mean and variance.

    ---

    1. Find $K$ using the total probability condition

    The sum of all probabilities must equal 1:

    $3K^2 + K + K^2 + 2K = 1$

    Simplify:

    $4K^2 + 3K - 1 = 0$

    Solve the quadratic:

    $K = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm 5}{8}$

    So $K = \frac{2}{8} = \frac{1}{4}$ or $K = \frac{-8}{8} = -1$. Since probabilities cannot be negative, we take:

    $K = \frac{1}{4}$

    Watch out

    A common mistake is to forget that $K$ must be positive. The negative root is extraneous here.

    ---

    2. Compute the mean $\mu = E[X]$

    $\mu = \sum x \cdot P(X=x) = 1 \cdot 3K^2 + 3 \cdot K + 5 \cdot K^2 + 2 \cdot 2K$

    Substitute $K = \frac{1}{4}$:

    • $3K^2 = 3 \cdot \frac{1}{16} = \frac{3}{16}$
    • $K = \frac{1}{4} = \frac{4}{16}$
    • $K^2 = \frac{1}{16}$
    • $2K = \frac{2}{4} = \frac{1}{2} = \frac{8}{16}$

    Now compute:

    $\mu = 1 \cdot \frac{3}{16} + 3 \cdot \frac{4}{16} + 5 \cdot \frac{1}{16} + 2 \cdot \frac{8}{16}$

    $\mu = \frac{3}{16} + \frac{12}{16} + \frac{5}{16} + \frac{16}{16} = \frac{36}{16} = \frac{9}{4}$

    So the mean is $\mu = \frac{9}{4}$.

    ---

    3. Compute $E[X^2]$

    $E[X^2] = \sum x^2 \cdot P(X=x) = 1^2 \cdot 3K^2 + 3^2 \cdot K + 5^2 \cdot K^2 + 2^2 \cdot 2K$

    Substitute the same probabilities:

    $E[X^2] = 1 \cdot \frac{3}{16} + 9 \cdot \frac{4}{16} + 25 \cdot \frac{1}{16} + 4 \cdot \frac{8}{16}$

    $E[X^2] = \frac{3}{16} + \frac{36}{16} + \frac{25}{16} + \frac{32}{16} = \frac{96}{16} = 6$

    ---

    4. Compute the variance

    $\text{Var}(X) = E[X^2] - \mu^2 = 6 - \left(\frac{9}{4}\right)^2$

    $\mu^2 = \frac{81}{16}$

    So:

    $\text{Var}(X) = 6 - \frac{81}{16} = \frac{96}{16} - \frac{81}{16} = \frac{15}{16}$

    Tip

    Notice that $\frac{15}{16}$ is option (D), but wait — we must double-check the order of the $x$ values in the table. The table lists $x = 1, 3, 5, 2$ — not in ascending order. That’s fine; we used each value correctly. However, the computed variance is $\frac{15}{16}$, which matches option (D). But let’s verify the probabilities again: $3K^2 = 3/16$, $K = 4/16$, $K^2 = 1/16$, $2K = 8/16$. Sum = $3+4+1+8 = 16/16 = 1$. Correct.

    So the variance is $\frac{15}{16}$.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.Two persons A and B throw a pair of dice alternately until one of them gets the sum of the numbers appeared on the dice as 4 and the person who gets this result first is declared as the winner. If A starts the game, then the probability that B wins the game is (A) $\dfrac{11}{23}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{5}{11}$ (D) $\dfrac{8}{17}$

    ›Reveal solutionSolution

    With $p=1/12$ (probability of rolling a sum of 4), the alternating-turns geometric-series argument gives B's winning probability as $\dfrac{q}{1+q}=\dfrac{11}{23}$.

    Concept and Intuition

    When two players alternate independent trials until one succeeds, and the first player (A) starts, the second player's (B's) win probability is a geometric series: B can only win on their 1st turn (2nd throw overall, requiring A to fail first), or their 2nd turn (4th throw overall, requiring both to fail twice), and so on.

    Step-by-Step Solution

    1. Ways to get sum 4 with two dice: $(1,3),(2,2),(3,1)$ — 3 favourable outcomes out of 36. So $p = P(\text{sum}=4) = \dfrac{3}{36}=\dfrac1{12}$, and $q=1-p=\dfrac{11}{12}$ (probability of NOT getting sum 4 on a given throw).
    2. B wins on their first opportunity (the 2nd throw overall) if A fails (prob $q$) then B succeeds (prob $p$): probability $qp$.
    3. B wins on their second opportunity (4th throw overall) if A fails, B fails, A fails again, then B succeeds: probability $q\cdot q\cdot q\cdot p = q^3p$.
    4. In general, B wins on their $k$-th opportunity with probability $q^{2k-1}p$ (since $2k-1$ prior throws must all fail before B's $k$-th throw succeeds).
    5. Total: $P(B\text{ wins}) = \sum_{k=1}^\infty q^{2k-1}p = qp(1+q^2+q^4+\cdots) = \dfrac{qp}{1-q^2} = \dfrac{qp}{(1-q)(1+q)} = \dfrac{qp}{p(1+q)} = \dfrac{q}{1+q}$.
    6. Substitute $q=\dfrac{11}{12}$: $P(B\text{ wins}) = \dfrac{11/12}{1+11/12} = \dfrac{11/12}{23/12} = \dfrac{11}{23}$.

    Common Mistakes

    • Forgetting that A goes first, which asymmetrically favours A — B's winning probability should come out less than $1/2$ (indeed $11/23 < 1/2$), a good sanity check.
    • Mis-setting the geometric series ratio — the common ratio between consecutive "B wins on the $k$-th vs $(k+1)$-th opportunity" terms is $q^2$ (two extra failed throws, one by each player), not $q$.

    ✓Final answer

    The correct option is (A) — $\dfrac{11}{23}$.

    ANSWER: A