Q.Three dice are thrown at the same time. Find the probability of getting three two's, if it is known that the sum of the numbers on the dice was six.

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 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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3. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $X$ is a Poisson variate satisfying the condition $3P(x=2) = P(x=4)$ then $P(x=6) =$ (A) $\dfrac{162}{5e^6}$ (B) $\dfrac{108}{5e^6}$ (C) $\dfrac{324}{5e^6}$ (D) $\dfrac{648}{5e^6}$

   ›Reveal solutionSolution

   The key idea is to use the Poisson probability mass function $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, set up the given condition $3P(X=2) = P(X=4)$, solve for $\lambda$, then compute $P(X=6)$ and match it to one of the options. The final result is $\frac{324}{5e^6}$, which corresponds to option (C).

   We start with the Poisson distribution. A Poisson random variable $X$ with mean $\lambda$ has probability mass function

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

   The problem gives a relationship between $P(X=2)$ and $P(X=4)$. This lets us solve for $\lambda$, the only unknown parameter. Once we know $\lambda$, we can compute $P(X=6)$ directly.

   ---

   1. Write the given condition in terms of $\lambda$. We have $3P(X=2) = P(X=4)$. Substituting the Poisson formula:

   $3 \cdot \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-\lambda} \lambda^4}{4!}$

   2. Cancel the common factor $e^{-\lambda}$ (since $e^{-\lambda} > 0$ for any finite $\lambda$). This gives:

   $3 \cdot \frac{\lambda^2}{2} = \frac{\lambda^4}{24}$

   3. Simplify both sides. Left: $3 \cdot \frac{\lambda^2}{2} = \frac{3\lambda^2}{2}$. Right: $\frac{\lambda^4}{24}$. So:

   $\frac{3\lambda^2}{2} = \frac{\lambda^4}{24}$

   4. Solve for $\lambda$. Multiply both sides by 24:

   $24 \cdot \frac{3\lambda^2}{2} = \lambda^4 \quad \Rightarrow \quad 36\lambda^2 = \lambda^4$

   Rearranging:

   $\lambda^4 - 36\lambda^2 = 0 \quad \Rightarrow \quad \lambda^2(\lambda^2 - 36) = 0$

   Since $\lambda > 0$ for a Poisson distribution (mean cannot be zero if we have nonzero probabilities for $k=2,4$), we take $\lambda^2 = 36$, so $\lambda = 6$. …

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4. 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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5. 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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6. 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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7. 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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8. 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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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 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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11. 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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12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.The variance of a Poisson variate $X$ is $2$. Then $P(X \geq 3) =$ (A) $\dfrac{e^2 - 7}{e^2}$ (B) $\dfrac{e^2 - 3}{e^2}$ (C) $\dfrac{e^2 - 5}{e^2}$ (D) $1 - \dfrac{4}{e^2}$

    ›Reveal solutionSolution

    For a Poisson distribution, variance equals mean ($\lambda$). Given variance $=2$, we have $\lambda=2$. Then $P(X \geq 3) = 1 - P(X \leq 2) = 1 - e^{-2}(1 + 2 + 2) = \frac{e^2 - 5}{e^2}$, which matches option (C).

    The Poisson distribution is defined by a single parameter $\lambda$, which is both its mean and its variance. That’s the key property here — once you know the variance, you know $\lambda$ directly. The question then becomes a straightforward probability sum.

    The probability mass function of a Poisson variate $X$ with parameter $\lambda$ is:

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

    We are told $\operatorname{Var}(X) = 2$. For Poisson, $\operatorname{Var}(X) = \lambda$, so $\lambda = 2$.

    We need $P(X \geq 3)$. It’s often easier to compute the complement: $P(X \geq 3) = 1 - P(X \leq 2)$.

    1. Compute $P(X = 0)$

    $P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2}$

    2. Compute $P(X = 1)$

    $P(X = 1) = \frac{e^{-2} \cdot 2^1}{1!} = 2e^{-2}$

    3. Compute $P(X = 2)$ $P(X = 2) = \frac{e^{-2} \cdot 2^2}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$ …

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