1. **Problem statement:** Given the random variable $X$ with probability mass function $$P(X = k) = \binom{3}{k} \cdot 0.1^k \cdot 0.9^{3-k}$$ (a) List all possible values of $k$. (b) Calculate $P(X \leq 2)$. 2. **Formula and explanation:** This is a binomial distribution with parameters $n=3$ and $p=0.1$. - The possible values of $k$ are integers from $0$ to $n$ inclusive. - The binomial probability formula is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ 3. **Solution for (a):** Since $n=3$, the possible values of $k$ are: $$k = 0, 1, 2, 3$$ 4. **Solution for (b):** Calculate $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$. Calculate each term: - $P(X=0) = \binom{3}{0} 0.1^0 0.9^3 = 1 \cdot 1 \cdot 0.9^3 = 0.729$ - $P(X=1) = \binom{3}{1} 0.1^1 0.9^2 = 3 \cdot 0.1 \cdot 0.81 = 0.243$ - $P(X=2) = \binom{3}{2} 0.1^2 0.9^1 = 3 \cdot 0.01 \cdot 0.9 = 0.027$ Sum these probabilities: $$P(X \leq 2) = 0.729 + 0.243 + 0.027 = 0.999$$ 5. **Final answers:** - (a) $k = 0, 1, 2, 3$ - (b) $P(X \leq 2) = 0.999$