Q.Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

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1. 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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2. 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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3. 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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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-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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6. 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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7. 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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8. 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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9. 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 $P(I_2 = 1) = \frac{3}{8}.$ …

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10. 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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11. 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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12. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.Two numbers b and c are chosen at random in succession without replacement from the set $\{1,2,3,\ldots,9\}$. Then the probability that $x^2 + bx + c > 0,\ \forall x \in \mathbb{R}$ is (A) $\dfrac{29}{72}$ (B) $\dfrac{32}{81}$ (C) $\dfrac{45}{143}$ (D) $\dfrac{82}{125}$

    ›Reveal solutionSolution

    The condition $x^2+bx+c>0$ for all real $x$ is equivalent to the discriminant $b^2-4c<0$. Counting ordered pairs $(b,c)$ from $\{1,\dots,9\}$ without replacement that satisfy $b^2<4c$ gives $29$ favorable outcomes out of $72$ total, so the probability is $\frac{29}{72}$, which is option (A).

    ---

    Why this approach works

    A quadratic $x^2+bx+c$ that is always positive (for every real $x$) must have no real roots and open upward. Since the coefficient of $x^2$ is $1>0$, the condition reduces to the discriminant being negative: $b^2-4c<0$, i.e. $b^2<4c$.

    We are choosing $b$ and $c$ without replacement from $\{1,\dots,9\}$, so each ordered pair $(b,c)$ with $b\neq c$ is equally likely. The total number of such ordered pairs is $9\times 8 = 72$. We just need to count how many of them satisfy $b^2<4c$.

    ---

    Step-by-step counting

    1. Understand the inequality

    We need $b^2 < 4c$. Since $c$ is an integer from $1$ to $9$, rewrite as $c > \frac{b^2}{4}$. For each $b$, we count the number of $c$ values (different from $b$) that are strictly greater than $b^2/4$.

    2. Compute for each $b$

    • $b=1$: $b^2/4 = 0.25$, so $c > 0.25$ means $c \ge 1$. All $c$ from $1$ to $9$ except $c=1$ (since $b\neq c$) work. That gives $8$ choices.
    • $b=2$: $b^2/4 = 1$, so $c > 1$ means $c \ge 2$. Excluding $c=2$ leaves $\{3,4,5,6,7,8,9\}$ → $7$ choices.
    • $b=3$: $b^2/4 = 2.25$, so $c > 2.25$ means $c \ge 3$. Excluding $c=3$ leaves $\{4,5,6,7,8,9\}$ → $6$ choices.
    • $b=4$: $b^2/4 = 4$, so $c > 4$ means $c \ge 5$. Excluding $c=4$ (which isn't in this set anyway) gives $\{5,6,7,8,9\}$ → $5$ choices.
    • $b=5$: $b^2/4 = 6.25$, so $c > 6.25$ means $c \ge 7$. Excluding $c=5$ (not in set) gives $\{7,8,9\}$ → $3$ choices.
    • $b=6$: $b^2/4 = 9$, so $c > 9$ means $c \ge 10$, but max $c$ is $9$. No $c$ works → $0$ choices. …

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