Show that \( p = 0.4 \), \( q = 0.2 \).

Sum of probabilities: \( p + 0.3 + q + 0.1 = 1 \), so \( p + q = 0.6 \) (1).

Given \( E(X) = 2 \): \( p \cdot 1 + 0.3 \cdot 2 + q \cdot 3 + 0.1 \cdot 4 = p + 0.6 + 3q + 0.4 = 2 \), so \( p + 3q = 1.0 \) (2).

Solve: From (1), \( p = 0.6 – q \). Substitute into (2): \( 0.6 – q + 3q = 1.0 \), \( 2q = 0.4 \), \( q = 0.2 \). Then \( p = 0.6 – 0.2 = 0.4 \).

Answer: \( p = 0.4 \), \( q = 0.2 \).

Find \( P(X > 2) \).

\[ P(X > 2) = P(X = 3) + P(X = 4) = 0.2 + 0.1 = 0.3 \]

Answer: \( 0.3 \).

Probability Nicky wins (score \(\geq 10\)) after scoring 4 in three rolls.

Need sum \(\geq 6\) in two rolls. Let \( Y \) be the sum of two rolls.

\[ P(Y \geq 6) \]:

• \( Y = 6 \): (2,4), (3,3), (4,2). \[ P = (0.3 \cdot 0.1) + (0.2 \cdot 0.2) + (0.1 \cdot 0.3) = 0.1 \]
• \( Y = 7 \): (3,4), (4,3). \[ P = (0.2 \cdot 0.1) + (0.1 \cdot 0.2) = 0.04 \]
• \( Y = 8 \): (4,4). \[ P = 0.1 \cdot 0.1 = 0.01 \]

\[ P(Y \geq 6) = 0.1 + 0.04 + 0.01 = 0.15 \]

Answer: 0.15.

Determine \( b \).

Yellow pair max sum is 8. Red pair: First die (1,2,2,3), second die (1,\( a \),\( a \),\( b \)). Max sum: \( 3 + b = 8 \), so \( b = 5 \).

Verify: (3,5), \( P = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \), matches \( P(S=8) \).

Answer: \( b = 5 \).

Find \( a \), with evidence.

Red pair: First die (1,2,2,3), second die (1,\( a \),\( a \),5). Match yellow pair’s distribution.

Method 1:

Yellow pair: \( P(S = 5) = \frac{4}{16} \).

Red pair: \( S = a + 2 \): (2,\( a \)), \( P = \frac{2}{4} \cdot \frac{2}{4} = \frac{4}{16} \).

\[ a + 2 = 5 \]

\[ a = 3 \]

Method 2:

Yellow pair: \( P(S = 6) = \frac{3}{16} \).

Red pair: \( S = a + 3 \): (3,\( a \)), \( P = \frac{1}{4} \cdot \frac{2}{4} = \frac{2}{16} \); \( S = 6 \): (1,5), (3,5), \( P = \frac{1}{4} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{4} = \frac{2}{16} \).

\[ a + 3 = 6 \]

\[ a = 3 \]

Method 3:

Yellow pair: \( P(S = 4) = \frac{3}{16} \).

Red pair: \( S = a + 1 \): (1,\( a \)), \( P = \frac{1}{4} \cdot \frac{2}{4} = \frac{2}{16} \); \( S = 4 \): (1,3), \( P = \frac{1}{4} \cdot \frac{2}{4} = \frac{2}{16} \).

\[ a + 1 = 4 \]

\[ a = 3 \]

Verify with \( a = 3 \):

• \( S = 2 \): (1,1). \[ P = \frac{1}{16} \]
• \( S = 3 \): (2,1). \[ P = \frac{2}{16} \]
• \( S = 4 \): (1,3), (3,1). \[ P = \frac{2}{16} + \frac{1}{16} = \frac{3}{16} \]
• \( S = 5 \): (2,3). \[ P = \frac{4}{16} \]
• \( S = 6 \): (1,5), (3,3). \[ P = \frac{1}{16} + \frac{2}{16} = \frac{3}{16} \]
• \( S = 7 \): (2,5). \[ P = \frac{2}{16} \]
• \( S = 8 \): (3,5). \[ P = \frac{1}{16} \]

Answer: \( a = 3 \).