A fair dice is rolled twice. Let $$X$$ denote the number of times a composite number showed up. If $$\mu$$ and $$\sigma^2$$ represent the mean and variance of $$X$$ respectively, then what is the value of $$9(\mu+\sigma^2)$$?

The sample space when a fair die is rolled is $$\{1, 2, 3, 4, 5, 6\}$$.

• Composite numbers on a die: The numbers with more than two factors are $$4$$ and $$6$$. Thus, there are $$2$$ favourable outcomes.
  
• Probability of success ($$p$$): Getting a composite number in a single roll is:
  $$p = \frac{2}{6} = \frac{1}{3}$$
• Probability of failure ($$q$$):
  $$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$$

Since the die is rolled twice independently ($$n = 2$$), the random variable $$X$$ representing the number of times a composite number appears follows a Binomial distribution:

$$X \sim B\left(n = 2, p = \frac{1}{3}\right)$$

Calculate the mean ($$\mu$$) and variance ($$\sigma^2$$) using standard binomial formulas:

• Mean ($$\mu$$):
  $$\mu = np = 2 \times \frac{1}{3} = \frac{2}{3}$$
• Variance ($$\sigma^2$$):
  $$\sigma^2 = npq = 2 \times \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}$$

Find the sum of the mean and variance:

$$\mu + \sigma^2 = \frac{2}{3} + \frac{4}{9} = \frac{6 + 4}{9} = \frac{10}{9}$$

Evaluate the required expression $$9(\mu + \sigma^2)$$:

$$9(\mu + \sigma^2) = 9 \times \frac{10}{9} = 10$$