\(\quad\text{Required probability}\\ = \dfrac35 \cdot \dfrac24 \cdot \dfrac23 + \dfrac25 \cdot \dfrac34 \cdot \dfrac13\\ =\dfrac3{10}\)

Suppose the friends are A, B, C, Dhruv (D), E. Then denote the number of chocolate gotten by each person as \(x_A, x_B, \cdots, x_E\).

The problem becomes to count the nonnegative integer solutions of \(x_A+x_B+x_C+x_D+x_E=8\) with \(x_D \geq 6\).

We make the substitution \(y_A=x_A, y_B=x_B, y_C=x_C, y_D=x_D-6,y_E=x_E\). Then the equation becomes \(y_A+y_B+y_C+y_D+y_E=2\). We will count the nonnegative integer solutions of this equation to get the answer. The answer is \(\displaystyle \binom{2 + 5 - 1}{5 - 1} = 15\).

The pair \((x_2, x_5)\) can be (0, 1), (0, 3), ..., (0, 9), (2, 1), (2, 3), ..., (2, 7), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (8, 1).

From here, 5 pairs add up to 9, 4 pairs add up to 7, 3 pairs add up to 5, 2 pairs add up to 3, and 1 pair add up to 1.

If \(x_2 + x_5 = k\), then \(x_1 + x_3 + x_4 + x_6 + x_7 = 10 - k\). The number of nonnegative integer solutions of \(x_1 + x_3 + x_4 + x_6 + x_7 = N\) is \(\displaystyle \binom{N+5-1}{5-1} = \binom{N+4}4\). Then, we simply calculate the required number to be \(\displaystyle 5\binom{1 + 4}{4} + 4\binom{3 + 4}{4} + 3\binom{5 + 4}{4} + 2\binom{7+ 4}{4} + 1\binom{9+4}{4} = 1918\).

\(\begin{cases} x + y = 10\\ x^2 + y^2 = 62 + 2xy \end{cases}\\ (x - y)^2 = 62\\ x - y = \pm \sqrt{62}\\ \)

When x - y = \(\sqrt{62}\), \((x, y) = \left(\dfrac{10 + \sqrt{62}}2, \dfrac{10 - \sqrt{62}}2\right)\).

Otherwise, \((x, y) = \left(\dfrac{10 -\sqrt{62}}2, \dfrac{10 + \sqrt{62}}2\right)\).

These 2 pairs are all the solutions.

Suppose the answer is x%.

\(60\% \cdot 30\% + (1 - 60\%) x\% = 50\%\\ 0.18 + 0.004x = 0.5\\ x = 80\)

80% of the remainder of the games must be won.

The polynomial is just x^2 - 24x + c. When it is the square of a binomial,

\(24^2 - 4(1)(c) = 0\\ c = 144\)

The polynomial is just x^2 + sx + 4. For the polynomial to be the square of a binomial, the discriminant is equal to 0.

\(s^2 - 4(1)(4) = 0\\ s^2 = 16\\ s = \pm 4\)

\(\displaystyle \frac{x}{x - 5} = \frac{4}{x - 4} + \frac{3}{x} - 2\\ \dfrac x{x-5} = \dfrac{7x-12-2x(x-4)}{x(x-4)} = \dfrac{-2x^2+15x-12}{x^2 - 4x}\\ x^3 - 4x^2 = (x - 5)(-2x^2 + 15x - 12)\\ x^3 - 4x^2 = -2x^3 + 25x^2 - 87x + 60\\ 3x^3 - 29x^2 + 87x - 60 = 0 \)

Solving numerically (or using Cardano's formula or Tartaglia's formula) we have \(x \approx 0.97404\text{ or }x \approx 4.3463 \pm 1.2816i\)

By Vieta's formulae we have b = -(5 + 3i + (8-2i)) = -13 - i and c = (5 + 3i)(8 - 2i) = 46 + 14i so b + c = 33 + 13i.

Case 1: \(c \leq -5\)

Then 

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

But then c is not less than or equal to -5. This root is rejected.

Case 2: \(-5 < c \leq 0\)

Then

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This root is in the range -5 < c <= 0. So we keep it.

Case 3: \(0 < c \leq 4\)

Then

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This root is in range 0 < c <= 4. So we keep it.

Case 4: \(c > 4\)

Then

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

But then c is not greater than 4. This root is rejected.

Hence, the solutions are: \(c = \dfrac{13}6\), \(c = -\dfrac{13}6\).