1. **Problem Statement:** Given the probability distribution for $X$ with values $0,1,2,3$ and probabilities $0.30,0.40,0.20,$ and an unknown $P(X=3)$, find the missing probability and compute requested probabilities. 2. **Formula and Rules:** The sum of all probabilities in a distribution must equal 1: $$\sum P(X=x) = 1$$ Probabilities must be between 0 and 1 inclusive. 3. **Find $P(X=3)$:** $$P(X=3) = 1 - (0.30 + 0.40 + 0.20) = 1 - 0.90 = 0.10$$ 4. **Compute requested probabilities:** - a. $P(X=3) = 0.10$ - b. $P(X=2) = 0.20$ - c. $P(X \leq 1) = P(X=0) + P(X=1) = 0.30 + 0.40 = 0.70$ - d. $P(X \geq 2) = P(X=2) + P(X=3) = 0.20 + 0.10 = 0.30$ - e. $P(1 < X \leq 3) = P(X=2) + P(X=3) = 0.20 + 0.10 = 0.30$ **Final answers:** $$P(X=3) = 0.10, P(X=2) = 0.20, P(X \leq 1) = 0.70, P(X \geq 2) = 0.30, P(1 < X \leq 3) = 0.30$$