Let the 11 consecutive natural numbers be $1, 2, 3, \dots, 11.$ Total ways to choose any 3 numbers = $\displaystyle \binom{11}{3} = 165.$
Now, we need to count the number of 3-number selections that can form an arithmetic progression (A.P.) with positive common difference.
For an A.P., let the middle term be $a$ and common difference be $d>0$. Then the three terms are: $(a-d,\ a,\ a+d)$ These must all lie between $1$ and $11$.
That means $1 \le a-d$ and $a+d \le 11$ ⟹ $d \le \min(a-1,\ 11-a)$
Now we count possible values of $d$ for each $a$:

$a$	$\min(a-1,\ 11-a)$	Possible $d$ values
1	0	–
2	1	1
3	2	1,2
4	3	1,2,3
5	4	1,2,3,4
6	5	1,2,3,4,5
7	4	1,2,3,4
8	3	1,2,3
9	2	1,2
10	1	1
11	0	–

Total = $1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25.$
Hence, number of favorable triplets = $25.$
Therefore, $\displaystyle P = \frac{25}{165} = \frac{5}{33}.$
Final Answer: $\boxed{\dfrac{5}{33}}$

Let there be two families with 3 members each (say Family A and Family B), and one family with 4 members (Family C).

Step 1: Since members of the same family must sit together, treat each family as a single block.
Thus, there are 3 blocks: A, B, and C.
They can be arranged in  $3! = 6$ ways.

Step 2: Now arrange members within each family:
Family A (3 members): $3!$ ways
Family B (3 members): $3!$ ways
Family C (4 members): $4!$ ways

Step 3: Total number of arrangements = $3! \times 3! \times 3! \times 4!$

Step 4: Simplify:
$3! = 6$ and $4! = 24$
$\Rightarrow (3!)^3\times4!$

∴ Total number of ways = $\boxed{5184}$

Given:
Box I → cards numbered 1 to 30 (30 cards)
Box II → cards numbered 31 to 50 (20 cards)
A box is selected at random → probability of each box = $\dfrac{1}{2}$

Non-prime numbers in each box:
Box I (1–30): Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 → 10 primes.
Non-prime numbers = 30 − 10 = 20
(Including 1 as non-prime)

Box II (31–50): Prime numbers are 31, 37, 41, 43, 47 → 5 primes.
Non-prime numbers = 20 − 5 = 15

Let
A = “card drawn from Box I”
B = “card drawn from Box II”
N = “number on the card is non-prime”

We need $P(A|N)$.

Using Bayes’ theorem:
P(A|N) = \frac{P(A)P(N|A)}{P(A)P(N|A) + P(B)P(N|B)} \]
Now substitute:
\[ P(A) = P(B) = \frac{1}{2}, \quad \] \[ P(N|A) \frac{20}{30} = \frac{2}{3}, \quad \] \[ P(N|B) = \frac{15}{20} = \frac{3}{4} \]
\[ P(A|N) = \frac{\frac{1}{2}\cdot\frac{2}{3}}{\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{3}{4}} \] \[= \frac{\frac{1}{3}}{\frac{1}{3} + \frac{3}{8}} \] \[= \frac{\frac{1}{3}}{\frac{17}{24}}\] \[= \frac{8}{17} \]

Final Answer: $\boxed{\dfrac{8}{17}}$

Let the G.P. be \(a,\ ar,\ ar^2\).
Given:
$$S = a(1 + r + r^2)$$ and $$a^3 r^3 = 27 \Rightarrow (ar)^3 = 27 \Rightarrow ar = 3.$$
Hence, $$a = \frac{3}{r}.$$
Substitute in \(S\): $$ S = \frac{3}{r}(1 + r + r^2) = 3\left(r + \frac{1}{r} + 1\right) $$
For real \(r \ne 0\):
If \(r > 0,\) then $(r + \frac{1}{r} \ge 2 \Rightarrow S \ge 3(2 + 1) = 9)$
If $(r < 0)$ then $(r + \frac{1}{r} \le -2 \Rightarrow S \le 3(-2 + 1) = -3)$

Hence all possible values of \(S\) lie in the intervals:
$$ \boxed{S \in (-\infty,\ -3]\ \cup\ [9,\ \infty)} $$

Let the cubic polynomial be $$p(x) = a x^3 + b x^2 + c x + d$$ where \(a, b, c, d\) are real constants.

Given:
Local maximum at \(x = 1 \Rightarrow p'(1) = 0,\ p(1) = 8\)
Local minimum at \(x = 2 \Rightarrow p'(2) = 0,\ p(2) = 4\)

Derivative:
$$p'(x) = 3a x^2 + 2b x + c$$ Apply the stationary point conditions:
\[ \begin{cases} 3a(1)^2 + 2b(1) + c = 0 \\[4pt] 3a(2)^2 + 2b(2) + c = 0 \end{cases} \] Simplify: \[ \begin{cases} 3a + 2b + c = 0 \\[4pt] 12a + 4b + c = 0 \end{cases} \] Subtracting gives: \[ 9a + 2b = 0 \Rightarrow b = -\frac{9a}{2}. \] Substitute into \(3a + 2b + c = 0\): \[ 3a + 2\left(-\frac{9a}{2}\right) + c = 0 \Rightarrow 3a - 9a + c = 0 \Rightarrow c = 6a. \]
Using the value conditions:
\[ \begin{cases} p(1) = a + b + c + d = 8 \\[4pt] p(2) = 8a + 4b + 2c + d = 4 \end{cases} \] Substitute \(b = -\frac{9a}{2},\ c = 6a\):
\[ a - \frac{9a}{2} + 6a + d = 8 \Rightarrow \frac{5a}{2} + d = 8 \Rightarrow d = 8 - \frac{5a}{2}. \] and \[ 8a + 4\left(-\frac{9a}{2}\right) + 2(6a) + d = 4 \Rightarrow 2a + d = 4. \] Substitute \(d = 8 - \frac{5a}{2}\): \[ 2a + 8 - \frac{5a}{2} = 4 \Rightarrow -\frac{a}{2} = -4 \Rightarrow a = 8. \]
Now, \[ b = -\frac{9a}{2} = -36, \quad c = 48, \quad d = 8 - \frac{5(8)}{2} = -12. \]
Therefore, $$p(0) = d = -12.$$
Final Answer: $$\boxed{p(0) = -12}$$

Let the quadratic polynomial be \( f(x) = a x^2 + b x + c \).

Given that $$ f(-1) + f(2) = 0 $$ and one of the roots of \( f(x) = 0 \) is \(3\).

Let the other root be \(\alpha\). Hence, \( f(x) = k(x - 3)(x - \alpha) \), where \(k \ne 0\).

Substitute the condition \( f(-1) + f(2) = 0 \): $$ k(-1 - 3)(-1 - \alpha) + k(2 - 3)(2 - \alpha) = 0 $$ Simplify: $$ (-4)(-1 - \alpha) + (-1)(2 - \alpha) = 0 $$ $$ 4(1 + \alpha) - 2 + \alpha = 0 $$ $$ 2 + 5\alpha = 0 \Rightarrow \alpha = -\frac{2}{5}. $$
Therefore, the other root is \(-\dfrac{2}{5}\), which lies in $$\boxed{(-1,\ 0)}.$$

There are \(n\) stations on a circle. Each pair is connected by a straight track.

Blue lines connect nearest neighbours, so the number of blue lines is 

Blue $ = n. $ 

Total lines is $ \binom{n}{2} = \frac{n(n-1)}{2}. $ 

Hence red lines are $ \text{Red} = \binom{n}{2} - n. $ 

Given Red =$ 99 \times \text{Blue} $ 

$ \binom{n}{2} - n = 99n $

$ \;\;\Longrightarrow\;\; \frac{n(n-1)}{2} - n = 99n$

$ \;\;\Longrightarrow\;\; \frac{n(n-1)}{2} = 100n$

$ \;\;\Longrightarrow\;\; n-1 = 200$

$ \;\;\Longrightarrow\;\; n = 201.$

 Final Answer: \(\boxed{201}\)