Question 10

The die has 4 green faces and 2 red faces. The probability of rolling green on any roll is:

The probability of rolling red is $q = 1 - p = \frac{1}{3}$.

For $n = 7$ independent rolls, let $X$ denote the number of green outcomes. Then $X \sim \text{Binomial}\left(7, \frac{2}{3}\right)$. The probability of getting $k$ green faces is:

Evaluate $\binom{7}{k} 2^k$ for each option:

• $k = 3$ (3 green, 4 red): $\binom{7}{3} \times 2^3 = 35 \times 8 = 280$
• $k = 4$ (4 green, 3 red): $\binom{7}{4} \times 2^4 = 35 \times 16 = 560$
• $k = 5$ (5 green, 2 red): $\binom{7}{5} \times 2^5 = 21 \times 32 = 672$
• $k = 6$ (6 green, 1 red): $\binom{7}{6} \times 2^6 = 7 \times 64 = 448$

The maximum value is 672, occurring at $k = 5$. Thus, 5 green faces and 2 red faces is the most likely outcome.