Q.An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.Two balls are drawn at random from a bag containing 5 black balls and 3 white balls. If the random variable X denotes the number of white balls drawn, then the mean of X is (A) $\dfrac{1}{2}$ (B) $\dfrac{5}{8}$ (C) $\dfrac{3}{4}$ (D) $\dfrac{3}{8}$

   ›Reveal solutionSolution

   The mean (expected value) of the number of white balls drawn when picking two balls without replacement from 5 black and 3 white balls is $\frac{3}{4}$. The correct option is (C).

   ---

   We are drawing two balls without replacement from a small finite set. The random variable $X$ counts how many white balls appear. The mean (expected value) is just the average number of whites we’d see if we repeated the draw many times.

   Key insight: Instead of listing all outcomes and probabilities, we can use the linearity of expectation. Each ball drawn is like a “mini-experiment”: define an indicator for whether the first ball is white, and another for the second. The expected number of whites is simply the sum of the probabilities that each individual draw yields a white ball. This works even though the draws are dependent — expectation adds regardless.

   ---

   1. Define indicator variables

      Let $I_1 = 1$ if the first ball is white, $0$ otherwise.

      Let $I_2 = 1$ if the second ball is white, $0$ otherwise.

      Then $X = I_1 + I_2$.

   2. Find the probability the first ball is white

      Initially there are 3 white balls out of 8 total.

   $P(I_1 = 1) = \frac{3}{8}.$

   3. Find the probability the second ball is white By symmetry (or by the law of total probability), the chance the second ball is white is also $\frac{3}{8}$. Why? Because without any information about the first draw, the second ball is equally likely to be any of the 8 original balls. So

   $P(I_2 = 1) = \frac{3}{8}.$

   4. Apply linearity of expectation

   $E[X] = E[I_1 + I_2] = E[I_1] + E[I_2] = \frac{3}{8} + \frac{3}{8} = \frac{6}{8} = \frac{3}{4}.$

   Tip

   A common pitfall is to think the second draw’s probability changes because the first draw removed a ball. But without conditioning on the first result, the second draw still has a $\frac{3}{8}$ chance of being white — symmetry saves us.

   Watch out

   Do not compute the distribution of $X$ from scratch unless you enjoy extra work. The direct method (listing $P(X=0), P(X=1), P(X=2)$) gives the same answer but is slower. Here, linearity makes it a one-liner.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If $P\left(\dfrac{A}{B}\right) = \dfrac{3}{10}$, $P\left(\dfrac{B}{A}\right) = \dfrac{4}{5}$ and $P(A \cup B) = K P(B)$, then $\dfrac{1}{K} =$ (A) $\dfrac{40}{49}$ (B) $\dfrac{40}{43}$ (C) $\dfrac{100}{101}$ (D) $1$

   ›Reveal solutionSolution

   The key idea is to use the definitions of conditional probability to relate $P(A \cap B)$ to $P(A)$ and $P(B)$, then express $P(A \cup B)$ in terms of $P(B)$ alone. The result is $\frac{1}{K} = \frac{40}{43}$.

   We are given two conditional probabilities and a relation involving the union. The goal is to find $\frac{1}{K}$, where $P(A \cup B) = K P(B)$. This is a problem about linking conditional probabilities to the basic probability of events, so we start by writing down what each conditional means.

   Recall: $P(A/B) = \frac{P(A \cap B)}{P(B)}$ and $P(B/A) = \frac{P(A \cap B)}{P(A)}$. These are not symmetric — each gives a different ratio. Our job is to use them to find $P(A)$ and $P(A \cap B)$ in terms of $P(B)$, then compute $P(A \cup B)$.

   1. From $P(A/B)$ we get $P(A \cap B)$ in terms of $P(B)$.

   $P(A/B) = \frac{3}{10} \quad\Rightarrow\quad \frac{P(A \cap B)}{P(B)} = \frac{3}{10}$

   So

   $P(A \cap B) = \frac{3}{10} P(B).$

   2. From $P(B/A)$ we get $P(A)$ in terms of $P(A \cap B)$.

   $P(B/A) = \frac{4}{5} \quad\Rightarrow\quad \frac{P(A \cap B)}{P(A)} = \frac{4}{5}$

   Hence

   $P(A) = \frac{5}{4} P(A \cap B).$

   3. Substitute the expression for $P(A \cap B)$ from step 1 into step 2.

   $P(A) = \frac{5}{4} \cdot \frac{3}{10} P(B) = \frac{15}{40} P(B) = \frac{3}{8} P(B).$

   So $P(A)$ is $\frac{3}{8}$ of $P(B)$.

   4. Now write $P(A \cup B)$ using the inclusion-exclusion formula.

   $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

   Substitute the expressions we have:

   $P(A \cup B) = \frac{3}{8} P(B) + P(B) - \frac{3}{10} P(B).$

   5. Combine the terms over a common denominator. The denominators are 8, 1, and 10. The LCM is 40. So:

   $\frac{3}{8} = \frac{15}{40},\quad 1 = \frac{40}{40},\quad \frac{3}{10} = \frac{12}{40}.$

   Therefore:

   $P(A \cup B) = \left( \frac{15}{40} + \frac{40}{40} - \frac{12}{40} \right) P(B) = \frac{43}{40} P(B).$

   6. Compare with the given relation $P(A \cup B) = K P(B)$. We have $P(A \cup B) = \frac{43}{40} P(B)$, so $K = \frac{43}{40}$. Hence

   $\frac{1}{K} = \frac{40}{43}.$

   Watch out

   A common mistake is to treat $P(A/B)$ and $P(B/A)$ as if they were the same, or to forget that $P(A \cap B)$ appears in both but must be expressed consistently. Always write the definition first.

   ✓Final answer

   The value is $\boxed{\frac{40}{43}}$, which corresponds to option (B).

3. 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:

   $P(B_2 \mid A) = \frac{P(B_2) \, P(A \mid B_2)}{P(A)}$

   Plug in the numbers:

   $P(B_2 \mid A) = \frac{0.30 \times 0.04}{0.0380} = \frac{0.0120}{0.0380}$

   4. Simplify the fraction Divide numerator and denominator by 0.002 (or multiply by 1000 to clear decimals):

   $\frac{0.0120}{0.0380} = \frac{12}{38} = \frac{6}{19}$

   So $P(B_2 \mid A) = \frac{6}{19}$.

   Tip

   A common pitfall is forgetting to compute $P(A)$ correctly — students sometimes use only the numerator. Always check that the denominator is the total probability of A, not just one term.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. 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.
      • (1,1,1): Only 2 numbers with remainder 1, so can’t pick 3. Count = 0.
      • (2,2,2): We have 3 numbers with remainder 2. Number of ways = $\binom{3}{3} = 1$.
      • (0,1,2): Pick 1 from remainder 0 (1 way), 1 from remainder 1 (2 ways), 1 from remainder 2 (3 ways). Total = $1 \times 2 \times 3 = 6$.

   4. Total favorable outcomes

      $1 + 6 = 7$.

   5. Probability

   $\frac{7}{20}$

   Tip

   A common mistake is to forget that (0,0,0) and (1,1,1) are impossible here because there aren’t enough numbers in those classes. Always check availability before counting.

   Watch out

   Do not just sum remainders of the original numbers — that would be meaningless. The modulo approach is what makes the problem simple.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. 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 $P(B)$:

   $P(B) - \frac{3}{20} = \frac{3}{5} P(B)$

   Bring terms together:

   $P(B) - \frac{3}{5}P(B) = \frac{3}{20}$

   $\frac{2}{5}P(B) = \frac{3}{20}$

   So:

   $P(B) = \frac{3}{20} \cdot \frac{5}{2} = \frac{15}{40} = \frac{3}{8}$

   3. Compute the required sum

   $P(A \cap B) + P(B) = \frac{3}{20} + \frac{3}{8} = \frac{6}{40} + \frac{15}{40} = \frac{21}{40}$

   Tip

   Notice that $P(A) = P(B) = \frac{3}{8}$ here — a nice symmetry that emerges from the given numbers.

   Watch out

   A common mistake is to treat $P(\overline{A} \mid B)$ as $1 - P(A \mid B)$ incorrectly — that works only if you adjust carefully. Always go back to the definition $P(\overline{A} \mid B) = \frac{P(\overline{A} \cap B)}{P(B)}$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

6. 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 The first throw and second throw are independent (the dice don’t remember previous rolls). Therefore,

   $P(\text{first both prime AND second both composite}) = \frac{1}{4} \times \frac{1}{9} = \frac{1}{36}.$

   Watch out

   A common mistake is to count 1 as composite — it is neither prime nor composite. Also, some forget that “both dice” means squaring the single-die probability.

   Tip

   Notice that $\frac{1}{4} \times \frac{1}{9} = \frac{1}{36}$ matches option (C). Always check whether the problem treats the two throws as independent — here they are, so multiplication is correct.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. 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).

   Watch out

   A common mistake is to forget that $\lambda$ must be positive. Also, do not cancel $\lambda^3$ if $\lambda = 0$ — but $\lambda = 0$ would make all probabilities zero, which contradicts the given relation (since $P(X=3)$ and $P(X=5)$ would both be zero, making $5 \cdot 0 = 0$ trivially true, but that degenerate case is not intended in such problems). Always check that the solution makes physical sense.

   4. Now compute $P(X=2)$ with $\lambda = 10$:

   $P(X=2) = \frac{e^{-10} \cdot 10^2}{2!} = \frac{e^{-10} \cdot 100}{2} = 50 e^{-10}$

   5. Write $e^{-10}$ as $\frac{1}{e^{10}}$, so:

   $P(X=2) = \frac{50}{e^{10}}$

   Tip

   Notice that the answer choices are all of the form $\frac{\text{integer}}{e^{\text{integer}}}$. Once you find $\lambda = 10$, the denominator $e^{10}$ immediately points to option (B). This can be a quick sanity check.

   ✓Final answer

   The value is $\boxed{\frac{50}{e^{10}}}$, which corresponds to option (B).

8. 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.
   • $b=7$: $b^2/4 = 12.25$, so $c > 12.25$ impossible → $0$ choices.
   • $b=8$: $b^2/4 = 16$, impossible → $0$ choices.
   • $b=9$: $b^2/4 = 20.25$, impossible → $0$ choices.

   3. Sum the favorable outcomes

   Total favorable ordered pairs = $8 + 7 + 6 + 5 + 3 = 29$.

   4. Compute probability

   Total ordered pairs without replacement = $9 \times 8 = 72$.

   Probability = $\frac{29}{72}$.

   ---

   Watch out

   A common mistake is to treat the selection as with replacement (giving $9^2=81$ total outcomes) or to forget that $b$ and $c$ must be different. The problem explicitly says "without replacement," so ordered pairs with $b=c$ are not allowed.

   Tip

   Notice that for $b \ge 6$, $b^2/4 \ge 9$, so no $c$ in $\{1,\dots,9\}$ can satisfy $c > b^2/4$. This immediately cuts the work to $b=1$ through $5$.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.If $A$ and $B$ are two events of a random experiment such that $P(\overline{A}) = \dfrac{2}{3}$, $P(B) = \dfrac{4}{15}$ and $P(A \cap \overline{B}) = \dfrac{1}{5}$, then $\sqrt{195 \left[ P\left(B \mid (A \cup \overline{B})\right) + P(A \cup B) \right]} =$ (A) 9 (B) 11 (C) 13 (D) 15

   ›Reveal solutionSolution

   This problem requires us to calculate probabilities of various event combinations (complement, intersection, union) and a conditional probability using fundamental set theory identities. We then substitute these values into the given expression to find the final numerical result. The final value is $\boxed{11}$.

   To solve this problem, we need to systematically break down the given expression and calculate each probability term using the fundamental rules of probability and set theory. The key is to correctly apply the formulas for complements, unions, intersections, and conditional probabilities, often using set identities to simplify complex event descriptions.

   Here's a step-by-step approach:

   1. Determine $P(A)$ from $P(\overline{A})$:

      The probability of an event $A$ and its complement $\overline{A}$ always sum to 1.

      $P(A) + P(\overline{A}) = 1$

      Given $P(\overline{A}) = \frac{2}{3}$, we can find $P(A)$:

      $P(A) = 1 - P(\overline{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.

   2. Determine $P(A \cap B)$ using $P(A \cap \overline{B})$:

      The event $A$ can be partitioned into two mutually exclusive events: $A \cap B$ (A and B both occur) and $A \cap \overline{B}$ (A occurs, but B does not).

      $P(A) = P(A \cap B) + P(A \cap \overline{B})$

      We are given $P(A \cap \overline{B}) = \frac{1}{5}$ and we found $P(A) = \frac{1}{3}$.

      So, $P(A \cap B) = P(A) - P(A \cap \overline{B}) = \frac{1}{3} - \frac{1}{5}$.

      To subtract these fractions, we find a common denominator, which is 15:

      $P(A \cap B) = \frac{5}{15} - \frac{3}{15} = \frac{2}{15}$.

   3. Calculate $P(A \cup B)$:

      The probability of the union of two events $A$ and $B$ is given by the addition rule.

      $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

      We have $P(A) = \frac{1}{3}$, $P(B) = \frac{4}{15}$ (given), and $P(A \cap B) = \frac{2}{15}$ (from Step 2).

      $P(A \cup B) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15}$.

      Using a common denominator of 15:

      $P(A \cup B) = \frac{5}{15} + \frac{4}{15} - \frac{2}{15} = \frac{5+4-2}{15} = \frac{7}{15}$.

      This is the first part of the sum inside the square root.

   4. Prepare for $P(B \mid (A \cup \overline{B}))$: Identify the intersection term:

      The conditional probability $P(X \mid Y)$ is defined as $\frac{P(X \cap Y)}{P(Y)}$. Here, $X=B$ and $Y=(A \cup \overline{B})$.

      So, we need to find $P(B \cap (A \cup \overline{B}))$.

      Using the distributive property of set intersection over union:

      $B \cap (A \cup \overline{B}) = (B \cap A) \cup (B \cap \overline{B})$.

      The event $B \cap \overline{B}$ means that event $B$ occurs AND event $B$ does NOT occur, which is impossible. Thus, $B \cap \overline{B} = \emptyset$.

      So, $B \cap (A \cup \overline{B}) = (B \cap A) \cup \emptyset = B \cap A$.

      Therefore, $P(B \cap (A \cup \overline{B})) = P(A \cap B)$.

      From Step 2, we know $P(A \cap B) = \frac{2}{15}$. This is the numerator for our conditional probability.

   5. Prepare for $P(B \mid (A \cup \overline{B}))$: Calculate $P(\overline{B})$:

      Similar to Step 1, we use the complement rule for event $B$.

      $P(\overline{B}) = 1 - P(B)$.

      Given $P(B) = \frac{4}{15}$:

      $P(\overline{B}) = 1 - \frac{4}{15} = \frac{15-4}{15} = \frac{11}{15}$.

   6. Prepare for $P(B \mid (A \cup \overline{B}))$: Calculate $P(A \cup \overline{B})$:

      This is the denominator for our conditional probability. We use the addition rule for $A$ and $\overline{B}$.

      $P(A \cup \overline{B}) = P(A) + P(\overline{B}) - P(A \cap \overline{B})$

      We have $P(A) = \frac{1}{3}$ (from Step 1), $P(\overline{B}) = \frac{11}{15}$ (from Step 5), and $P(A \cap \overline{B}) = \frac{1}{5}$ (given).

      $P(A \cup \overline{B}) = \frac{1}{3} + \frac{11}{15} - \frac{1}{5}$.

      Using a common denominator of 15:

      $P(A \cup \overline{B}) = \frac{5}{15} + \frac{11}{15} - \frac{3}{15} = \frac{5+11-3}{15} = \frac{13}{15}$.

   7. Calculate $P(B \mid (A \cup \overline{B}))$:

      Now we have both the numerator and the denominator for the conditional probability.

      $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$

      $P(B \mid (A \cup \overline{B})) = \frac{P(B \cap (A \cup \overline{B}))}{P(A \cup \overline{B})} = \frac{P(A \cap B)}{P(A \cup \overline{B})}$.

      Substituting the values from Step 4 and Step 6:

      $P(B \mid (A \cup \overline{B})) = \frac{2/15}{13/15} = \frac{2}{13}$.

      This is the second part of the sum inside the square root.

   8. Calculate the sum inside the square root:

      We need to find $P\left(B \mid (A \cup \overline{B})\right) + P(A \cup B)$.

      From Step 7, $P\left(B \mid (A \cup \overline{B})\right) = \frac{2}{13}$.

      From Step 3, $P(A \cup B) = \frac{7}{15}$.

      Sum $= \frac{2}{13} + \frac{7}{15}$.

      To add these fractions, find a common denominator, which is $13 \times 15 = 195$.

      Sum $= \frac{2 \times 15}{13 \times 15} + \frac{7 \times 13}{15 \times 13} = \frac{30}{195} + \frac{91}{195} = \frac{30+91}{195} = \frac{121}{195}$.

   9. Calculate the final expression:

      The expression we need to evaluate is $\sqrt{195 \left[ P\left(B \mid (A \cup \overline{B})\right) + P(A \cup B) \right]}$.

      Substitute the sum we just calculated:

      $\sqrt{195 \times \frac{121}{195}}$.

      The $195$ in the numerator and denominator cancel out:

      $\sqrt{121}$.

      $\sqrt{121} = 11$.

   The final value is 11. Comparing this with the given options, it matches option (B).

   ✓Final answer

   The value of the expression is $\boxed{11}$.

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.

    $\text{Total}=126+56+1=183.$

    Step 2 — Selections with all distinct letters.

    $\dbinom{9}{4}=126.$

    Step 3 — Selections with at least one repeated letter.

    $183-126=57.$

    Step 4 — Probability.

    $P=\frac{57}{183}=\frac{19}{61}.$

    ✓Final answer

    The required probability is $\dfrac{19}{61}$ — option (B).

    ANSWER: B

11. 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

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

    $P(X=2) = 28\cdot\frac{3^{2}}{5^{2}}\cdot\frac{2^{6}}{5^{6}} = \frac{28\cdot 3^{2}\cdot 2^{6}}{5^{8}}.$

    Write $28 = 2^{2}\times 7$:

    $= \frac{2^{2}\times 7\times 3^{2}\times 2^{6}}{5^{8}} = \frac{2^{8}\times 3^{2}\times 7}{5^{8}}.$

    ✓Final answer

    $P(\text{exactly two}) = \dfrac{2^{8}\times 3^{2}\times 7}{5^{8}}$. The correct option is (D).

    ANSWER: D

12. 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.
       • $\frac{15}{52}$ matches option (A).
       • For completeness: $\frac{4}{13} = \frac{16}{52}$, $\frac{17}{52}$, and $\frac{5}{13} = \frac{20}{52}$ are all different.

    Tip

    A quick sanity check: The probability of exactly one is also $P(A) + P(B) - 2P(A \cap B)$. Many students mistakenly use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, which gives "at least one" instead of "exactly one."

    Watch out

    A common pitfall is forgetting to subtract the intersection twice. If you only subtract it once, you get $\frac{16}{52} = \frac{4}{13}$, which is option (B) — the probability of at least one, not exactly one.

    ✓Final answer

    The correct option is (A).

    ANSWER: A