(a) Constant Probability Model, \( p = 0.1 \):

(i) Probability first boost on third action:

Recognize geometric distribution, \( P(X = x) = (1 – p)^{x-1} p \) (M1).

For \( x = 3 \), \( p = 0.1 \):

\[ P(X = 3) = (0.9)^2 \times 0.1 = 0.81 \times 0.1 = 0.081 \quad (A1) \]

[2 marks]

(ii) Probability at least one boost in first six actions:

Use complement: \( P(\text{at least one}) = 1 – P(\text{no boosts}) \) (M1).

\[ P(\text{no boosts}) = 0.9^6 = 0.531441 \]

\[ P(\text{at least one}) = 1 – 0.531441 = 0.468559 \approx 0.469 \quad (A1) \]

[2 marks]

(b) General Constant Probability Model, \( 0 < p < 1 \):

(i) Explain \( P(X = x) = p (1 – p)^{x-1} \):

For first boost on \( x \)-th action:

\( x – 1 \) failures (\( 1 – p \)), then one success (\( p \)) (M1).

Since actions are independent:

\[ P(X = x) = (1 – p)^{x-1} \times p \quad (A1) \]

[2 marks]

(ii) Expression for \( E(X) \):

\[ E(X) = \sum_{x=1}^\infty x p (1 – p)^{x-1} \quad (A1) \]

[1 mark]

(c) Geometric Series and Expected Value:

(i) Find \( \sum_{n=1}^\infty n a r^{n-1} \):

Given: \( S = a + ar + ar^2 + \cdots = \frac{a}{1 – r} \), \( |r| < 1 \) (M1).

Differentiate with respect to \( r \):

\[ \frac{d}{dr} (a + ar + ar^2 + \cdots) = a + 2ar + 3ar^2 + \cdots = \sum_{n=1}^\infty n a r^{n-1} \quad (A1) \]

\[ \frac{d}{dr} \left( \frac{a}{1 – r} \right) = \frac{a}{(1 – r)^2} \quad (A1) \]

\[ \sum_{n=1}^\infty n a r^{n-1} = \frac{a}{(1 – r)^2} \quad (A1) \]

[3 marks]

(ii) Show \( E(X) = \frac{1}{p} \):

From (b)(ii): \( E(X) = \sum_{x=1}^\infty x p (1 – p)^{x-1} = p \sum_{x=1}^\infty x (1 – p)^{x-1} \) (M1).

Using (c)(i), set \( a = 1 \), \( r = 1 – p \):

\[ \sum_{x=1}^\infty x (1 – p)^{x-1} = \frac{1}{(1 – (1 – p))^2} = \frac{1}{p^2} \quad (A1) \]

\[ E(X) = p \cdot \frac{1}{p^2} = \frac{1}{p} \quad (A1) \]

[2 marks]

(d) Expected Value and Variance for \( p = 0.1 \):

Using \( E(X) = \frac{1}{p} \), \( \text{Var}(X) = \frac{1 – p}{p^2} \):

\[ E(X) = \frac{1}{0.1} = 10 \quad (A1) \]

\[ \text{Var}(X) = \frac{1 – 0.1}{(0.1)^2} = \frac{0.9}{0.01} = 90 \quad (A1) \]

[2 marks]

(e) Second Model, Increasing Probability:

Show probability of first boost on third action is 0.288:

Probability sequence: 0.2, 0.4, 0.6, 0.8, 1.0 (M1).

\[ P(Y = 3) = P(\text{no boost 1}) \times P(\text{no boost 2}) \times P(\text{boost 3}) \]

\[ = 0.8 \times 0.6 \times 0.6 = 0.288 \quad (A1) \]

[2 marks]

(f) Explain \( Y \leq 5 \):

By action 5, \( P(\text{boost}) = 1.0 \), ensuring a boost if none occurred earlier (R1).

Thus, \( Y \leq 5 \).

[1 mark]

(g) Probability Distribution of \( Y \):

(i) Find \( m \) and \( n \):

Sum of probabilities: \( 0.2 + m + 0.288 + n + 0.0384 = 1 \) (M1).

\[ m + n = 1 – 0.5264 = 0.4736 \]

Calculate: \( P(Y = 2) = 0.8 \times 0.4 = 0.32 \), so \( m = 0.32 \).

\[ P(Y = 4) = 0.8 \times 0.6 \times 0.4 \times 0.8 = 0.1536 \], so \( n = 0.1536 \quad (A1) \)

[2 marks]

(ii) Show \( E(Y) = 2.5104 \):

\[ E(Y) = 1 \cdot 0.2 + 2 \cdot 0.32 + 3 \cdot 0.288 + 4 \cdot 0.1536 + 5 \cdot 0.0384 \quad (M1) \]

\[ = 0.2 + 0.64 + 0.864 + 0.6144 + 0.192 = 2.5104 \quad (A1) \]

[2 marks]

(iii) Find \( \text{Var}(Y) \):

\[ E(Y^2) = 1^2 \cdot 0.2 + 2^2 \cdot 0.32 + 3^2 \cdot 0.288 + 4^2 \cdot 0.1536 + 5^2 \cdot 0.0384 \quad (M1) \]

\[ = 0.2 + 1.28 + 2.592 + 2.4576 + 0.96 = 7.4896 \quad (A1) \]

\[ \text{Var}(Y) = 7.4896 – (2.5104)^2 = 7.4896 – 6.3026816 = 1.1869184 \approx 1.19 \quad (A1) \]

[3 marks]

(h) Comparing the Two Models:

(i) Find \( p \) where \( E(X) = E(Y) \):

\[ E(X) = \frac{1}{p} \], \( E(Y) = 2.5104 \quad (M1) \)

\[ \frac{1}{p} = 2.5104 \implies p = \frac{1}{2.5104} \approx 0.398 \quad (A1) \]

[2 marks]

(ii) Find \( \text{Var}(X) \) for \( p \approx 0.3982 \):

\[ \text{Var}(X) = \frac{1 – p}{p^2} \quad (M1) \]

\[ p \approx 0.3982, \quad 1 – p \approx 0.6018, \quad p^2 \approx 0.1586 \]

\[ \text{Var}(X) = \frac{0.6018}{0.1586} \approx 3.7933 \approx 3.79 \quad (A1) \]

[2 marks]

(iii) Determine which model is more consistent:

\[ \text{Var}(X) \approx 3.7933 \], \( \text{Var}(Y) \approx 1.1869 \quad (M1) \)

Lower variance indicates more consistent timing of boosts. Since \( \text{Var}(Y) < \text{Var}(X) \), the second model is more consistent (R1).

[2 marks]