a) Sum of probabilities must equal 1:

\( 0.41 + (k – 0.28) + 0.46 + (0.29 – 2k^2) = 1 \)

Combine constants: \( 0.41 – 0.28 + 0.46 + 0.29 + k – 2k^2 = 1 \)

\( 0.88 + k – 2k^2 = 1 \)

Subtract 1: \( 0.88 + k – 2k^2 – 1 = 0 \)

\( -2k^2 + k – 0.12 = 0 \)

Multiply by -1: \( 2k^2 – k + 0.12 = 0 \) [2]

b) Solve \( 2k^2 – k + 0.12 = 0 \):

Quadratic formula: \( k = \frac{-(-1) \pm \sqrt{(-1)^2 – 4 \times 2 \times 0.12}}{2 \times 2} \)

Discriminant: \( 1 – 0.96 = 0.04 \)

\( k = \frac{1 \pm \sqrt{0.04}}{4} = \frac{1 \pm 0.2}{4} \)

\( k = \frac{1.2}{4} = 0.3 \) or \( k = \frac{0.8}{4} = 0.2 \)

Check: For \( k = 0.3 \), \( P(X=1) = 0.3 – 0.28 = 0.02 \geq 0 \), \( P(X=3) = 0.29 – 2 \times 0.3^2 = 0.11 \geq 0 \)

For \( k = 0.2 \), \( P(X=1) = 0.2 – 0.28 = -0.08 < 0 \) (invalid)

Reason: \( k = 0.3 \) is valid as all probabilities are non-negative [2]

c) Use \( k = 0.3 \):

\( P(X=0) = 0.41 \)

\( P(X=1) = 0.3 – 0.28 = 0.02 \)

\( P(X=2) = 0.46 \)

\( P(X=3) = 0.29 – 2 \times 0.3^2 = 0.29 – 0.18 = 0.11 \)

\( E(X) = 0 \times 0.41 + 1 \times 0.02 + 2 \times 0.46 + 3 \times 0.11 \)

\( = 0 + 0.02 + 0.92 + 0.33 = 1.27 \) [2]