(a) [2 marks]

Probability distribution: \( P(X=1) = \frac{1}{6} \), \( P(X=2) = \frac{2}{6} = \frac{1}{3} \), \( P(X=3) = \frac{3}{6} = \frac{1}{2} \) (M1).
Expected value: \( E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{2}{6} + 3 \cdot \frac{3}{6} = \frac{1}{6} + \frac{4}{6} + \frac{9}{6} = \frac{14}{6} = \frac{7}{3} \approx 2.33 \) (A1).

(b)(i) [3 marks]

Total of three numbers is 5: cases are (1,1,3) and (1,2,2) with arrangements (M1).
\( P(1,1,3) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{3}{6} = \frac{3}{216} \), with 3 arrangements: \( 3 \cdot \frac{3}{216} = \frac{9}{216} \).
\( P(1,2,2) = \frac{1}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} = \frac{4}{216} \), with 3 arrangements: \( 3 \cdot \frac{4}{216} = \frac{12}{216} \).
Total probability: \( \frac{9}{216} + \frac{12}{216} = \frac{21}{216} = \frac{7}{72} \approx 0.0972 \) (A1A1).

(b)(ii) [3 marks]

Median is 1: cases are (1,1,1), (1,1,2), (1,1,3) (M1).
\( P(1,1,1) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{216} \).
\( P(1,1,2) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{2}{6} = \frac{2}{216} \), with 3 arrangements: \( 3 \cdot \frac{2}{216} = \frac{6}{216} \).
\( P(1,1,3) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{3}{6} = \frac{3}{216} \), with 3 arrangements: \( 3 \cdot \frac{3}{216} = \frac{9}{216} \).
Total probability: \( \frac{1}{216} + \frac{6}{216} + \frac{9}{216} = \frac{16}{216} = \frac{2}{27} \approx 0.0741 \) (A1A1).

(c) [3 marks]

Number of twos: \( X \sim B\left(10, \frac{1}{3}\right) \) (M1).
\( P(X < 4) = P(X \leq 3) = \sum_{k=0}^{3} \binom{10}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{10-k} \approx 0.559 \) (M1A1).

(d) [3 marks]

Probability of at least one 2: \( P(X \geq 1) = 1 – P(X = 0) > 0.95 \) (M1).
\( P(X = 0) = \left(\frac{2}{3}\right)^n < 0.05 \).
Solve: \( \left(\frac{2}{3}\right)^n < 0.05 \), so \( n \ln\left(\frac{2}{3}\right) < \ln(0.05) \).
\( n > \frac{\ln(0.05)}{\ln\left(\frac{2}{3}\right)} \approx 7.39 \), thus \( n = 8 \) (M1A1).

(e) [8 marks]

Expected number of 1’s: \( 8p_1 = 4.8 \Rightarrow p_1 = \frac{4.8}{8} = \frac{3}{5} \) (M1A1).
Variance of number of 2’s: \( 8p_2(1 – p_2) = 1.5 \) (M1).
Solve: \( 8p_2 – 8p_2^2 = 1.5 \Rightarrow p_2^2 – p_2 + \frac{1.5}{8} = 0 \Rightarrow p_2^2 – p_2 + 0.1875 = 0 \) (M1).
\( \Delta = 1 – 4 \cdot 0.1875 = 0.25 \), \( p_2 = \frac{1 \pm \sqrt{0.25}}{2} = \frac{1 \pm 0.5}{2} \).
\( p_2 = \frac{1}{4} \) or \( p_2 = \frac{3}{4} \) (A1).
\( p_1 + p_2 + p_3 = 1 \). For \( p_2 = \frac{1}{4} \): \( p_3 = 1 – \frac{3}{5} – \frac{1}{4} = \frac{3}{20} \). For \( p_2 = \frac{3}{4} \): \( p_3 = \frac{-2}{20} \) (invalid) (A1).
Probabilities: \( p_1 = \frac{12}{20} \), \( p_2 = \frac{5}{20} \), \( p_3 = \frac{3}{20} \). Least common multiple is 20, so minimum total balls is 20, with 3 balls numbered 3 (M1A1).