1. **Problem statement:** We consider a random variable $X$ with the given distribution: $$\begin{array}{c|c|c|c|c|c|c} x & 1 & 2 & 4 & 5 & 8 & 9 \\ \hline P(X=x) & 0.25 & 0.10 & 0.15 & 0.20 & 0.25 & 0.05 \\ \end{array}$$ We need to: a) Justify that this table defines a probability distribution. b) Calculate the following probabilities: ① $P(X=8)$ ② $P(X>5)$ ③ $P(X<3)$ ④ $P(X \geq 5)$ ⑤ $P(2 \leq X \leq 6)$ ⑥ $P(|X - 5| \leq 3)$ --- 2. **Step a) Justification of probability distribution:** 1. The probabilities must be between 0 and 1. 2. The sum of all probabilities must be 1. Check: $$0.25 + 0.10 + 0.15 + 0.20 + 0.25 + 0.05 = 1.00$$ All probabilities are between 0 and 1, and the sum is exactly 1, so this is a valid probability distribution. --- 3. **Step b) Calculate requested probabilities:** ① $P(X=8) = 0.25$ ② $P(X>5) = P(X=8) + P(X=9) = 0.25 + 0.05 = 0.30$ ③ $P(X<3) = P(X=1) + P(X=2) = 0.25 + 0.10 = 0.35$ ④ $P(X \geq 5) = P(X=5) + P(X=8) + P(X=9) = 0.20 + 0.25 + 0.05 = 0.50$ ⑤ $P(2 \leq X \leq 6) = P(X=2) + P(X=4) + P(X=5) = 0.10 + 0.15 + 0.20 = 0.45$ ⑥ $P(|X - 5| \leq 3)$ means $X$ values with distance at most 3 from 5, i.e. $2 \leq X \leq 8$. So, $$P(|X - 5| \leq 3) = P(2) + P(4) + P(5) + P(8) = 0.10 + 0.15 + 0.20 + 0.25 = 0.70$$ --- **Final answers:** ① 0.25 ② 0.30 ③ 0.35 ④ 0.50 ⑤ 0.45 ⑥ 0.70