Q.If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then (A) $P(B|A) = 1$ (B) $P(A|B) = 1$ (C) $P(B|A) = 0$ (D) $P(A|B) = 0$

Showing the 12 most recent of 30 on this concept.

1. 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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2. 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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3. 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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4. 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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5. CA Foundation 2026Set jan-20261 markMCQ
   Q.If in a class, 50% of the student study mathematics and science and 70% of the student study mathematics, then the probability of a student studying science given that he/she is already studying mathematics is (A) $\frac{3}{7}$ (B) $\frac{6}{7}$ (C) $\frac{4}{7}$ (D) $\frac{5}{7}$

   ›Reveal solutionSolution

   Conditional probability $P(S\mid M) = \dfrac{P(M \cap S)}{P(M)}$.

   Step 1 — identify the probabilities

   50% study both maths and science, so $P(M \cap S) = 0.5$; 70% study maths, so $P(M) = 0.7$.

   Step 2 — apply the conditional-probability formula

   $P(S \mid M) = \frac{P(M \cap S)}{P(M)} = \frac{0.5}{0.7} = \frac{5}{7}.$ …

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6. CA Foundation 2026Set jan-20261 markMCQ
   Q.If two dice are rolled, then the probability of getting a greater number on the first die than the one on the second, given that the sum should be equal to 7 is (A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{6}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   Conditional probability on a reduced sample space: $P(A\mid B) = \dfrac{n(A \cap B)}{n(B)}$.

   Step 1 — list the outcomes with sum 7.

   $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \Rightarrow n(B) = 6.$

   Step 2 — count first die greater than second, among those.

   $(4,3),(5,2),(6,1)$ → 3 outcomes.

   Step 3 — conditional probability.

   $P = \frac{3}{6} = \frac{1}{2}.$ …

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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 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 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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10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.The probability distribution of a random variable X is given below. Then, the standard deviation of X is

    [!FORMULA] $\begin{array}{|c|c|c|c|c|c|} \hline X = x_i & 2 & 3 & 5 & 7 & 12 \\ \hline P(X = x_i) & 3k & k & k & 2k & k \\ \hline \end{array}$

    (A) $11$ (B) $\sqrt{11}$ (C) $5$ (D) $\sqrt{5}$

    ›Reveal solutionSolution

    The key idea is to first find the constant $k$ by ensuring the total probability sums to 1, then compute the mean $\mu$, then the variance $\sigma^2 = E[X^2] - \mu^2$, and finally take the square root to get the standard deviation. The result is $\sqrt{11}$, so the correct option is (B).

    We are given a discrete probability distribution. To find the standard deviation, we need the mean (expected value) and the variance. But first, we must determine the unknown constant $k$ from the condition that probabilities sum to 1.

    Why this approach works:

    Standard deviation measures spread around the mean. For a discrete random variable, we compute the mean $\mu = \sum x_i P(x_i)$, then the variance $\sigma^2 = \sum (x_i - \mu)^2 P(x_i)$ or equivalently $\sigma^2 = E[X^2] - \mu^2$. The square root of variance gives the standard deviation. The constant $k$ is found by normalizing the probabilities.

    ---

    1. Find $k$ using total probability = 1

       The probabilities are: $P(2)=3k$, $P(3)=k$, $P(5)=k$, $P(7)=2k$, $P(12)=k$.

       Sum: $3k + k + k + 2k + k = 8k$.

       Set $8k = 1 \Rightarrow k = \frac{1}{8}$.

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

    $\mu = \sum x_i P(x_i) = 2\cdot\frac{3}{8} + 3\cdot\frac{1}{8} + 5\cdot\frac{1}{8} + 7\cdot\frac{2}{8} + 12\cdot\frac{1}{8}$

    $= \frac{6}{8} + \frac{3}{8} + \frac{5}{8} + \frac{14}{8} + \frac{12}{8} = \frac{40}{8} = 5.$

    So the mean is 5.

    3. Compute $E[X^2]$

    $E[X^2] = \sum x_i^2 P(x_i) = 4\cdot\frac{3}{8} + 9\cdot\frac{1}{8} + 25\cdot\frac{1}{8} + 49\cdot\frac{2}{8} + 144\cdot\frac{1}{8}$

    $$$$

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11. 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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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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