B部分：概率与统计 [60 分] 离散随机变量 $X$ 具有以下概率分布：
$\mathrm{P}(X = x) = k(x^2 + x) \text{ for } x = 1, \ 2, \ 3, \ 4, \ 5, \text{ where } k \text{ is a constant.}$

1. State the condition that the sum of all probabilities in a probability distribution equals 1.

$\sum_{x=1}^{5} \mathrm{P}(X=x) = 1$

2. Substitute the given function for each value of $x$ and solve for $k$.

$k(1^2 + 1) + k(2^2 + 2) + k(3^2 + 3) + k(4^2 + 4) + k(5^2 + 5) = 1$
$k(2 + 6 + 12 + 20 + 30) = 1$
$70k = 1$
$k = \frac{1}{70} \text{ (shown)}$

1. Calculate the expected value $\mathrm{E}(X)$ using the definition $\sum x \mathrm{P}(X=x)$.

$\begin{aligned} \mathrm{E}(X) &= \sum_{x=1}^{5} x \cdot \mathrm{P}(X=x) \\ &= \frac{1}{70} \left( 1(1^2+1) + 2(2^2+2) + 3(3^2+3) + 4(4^2+4) + 5(5^2+5) \right) \\ &= \frac{1}{70} (1(2) + 2(6) + 3(12) + 4(20) + 5(30)) \\ &= \frac{1}{70} (2 + 12 + 36 + 80 + 150) \\ &= \frac{280}{70} \\ &= 4 \end{aligned}$

2. Calculate $\mathrm{E}(X^2)$ as an intermediate step for the variance.

$\begin{aligned} \mathrm{E}(X^2) &= \sum_{x=1}^{5} x^2 \cdot \mathrm{P}(X=x) \\ &= \frac{1}{70} \left( 1^2(2) + 2^2(6) + 3^2(12) + 4^2(20) + 5^2(30) \right) \\ &= \frac{1}{70} (2 + 24 + 108 + 320 + 750) \\ &= \frac{1204}{70} \\ &= \frac{86}{5} \quad \text{or } 17.2 \end{aligned}$

3. Calculate the variance $\operatorname{Var}(X)$ using the formula $\operatorname{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2$.

$\begin{aligned} \operatorname{Var}(X) &= \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 \\ &= \frac{86}{5} - 4^2 \\ &= 17.2 - 16 \\ &= 1.2 \quad \text{or } \frac{6}{5} \end{aligned}$

1. Calculate the relevant individual probabilities for $X$.

Using $\mathrm{P}(X = x) = \frac{1}{70}(x^2 + x)$:
$\mathrm{P}(X=1) = \frac{2}{70}$
$\mathrm{P}(X=2) = \frac{6}{70}$
$\mathrm{P}(X=4) = \frac{20}{70}$
$\mathrm{P}(X=5) = \frac{30}{70}$

2. Identify the possible pairs $(x_1, x_2)$ where the difference is at least 3.

For $|X_1 - X_2| \ge 3$, the possible combinations $(X_1, X_2)$ are:

• Difference of 3: $(1, 4), (4, 1), (2, 5), (5, 2)$
• Difference of 4: $(1, 5), (5, 1)$

3. Calculate the total probability by adding the probabilities of these mutually exclusive pairs.

Since the observations are independent, $\mathrm{P}(X_1=x_1, X_2=x_2) = \mathrm{P}(X=x_1) \times \mathrm{P}(X=x_2)$.

$\begin{aligned}\mathrm{P}(|X_1 - X_2| \ge 3) &= 2\, \mathrm{P}(1, 4) + 2\, \mathrm{P}(2, 5) + 2\, \mathrm{P}(1, 5) \\ &= 2 \left( \frac{2}{70} \times \frac{20}{70} \right) + 2 \left( \frac{6}{70} \times \frac{30}{70} \right) + 2 \left( \frac{2}{70} \times \frac{30}{70} \right) \\ &= \frac{80}{4900} + \frac{360}{4900} + \frac{120}{4900} \\ &= \frac{560}{4900} \\ &= \frac{4}{35} \end{aligned}$