已知事件 $A$ 和 $B$ 满足 $P(A)=0.7$ 且 $P(B)=0.8$。证明 $P(A \cap B) \ge 0.5$ 若 $P(A'\ \cup\ B)=0.85$，求 $P(A\mid B)$。说明并解释 $A$ 和 $B$ 是否为独立事件。 若 $P(A \cap B)=0.5$，事件 A 与事件 B 之间有什么关系？ 在一平面上画出 11 条互不平行的直线。若没有任意三条直线交于同一点，证明这些直线之间有 55 个交点。 若随机选择其中三个交点，求它们都在这 11 条直线中的某一条上的概率。

1. Find $P(A \cap B)$ using the addition rule.

$P(A' \cup B) = P(A') + P(B) - P(A' \cap B)$
$0.85 = (1 - 0.7) + 0.8 - P(A' \cap B)$
$0.85 = 0.3 + 0.8 - P(A' \cap B) \implies P(A' \cap B) = 0.25$
Since $B$ consists of the mutually exclusive events $(A \cap B)$ and $(A' \cap B)$:
$P(B) = P(A \cap B) + P(A' \cap B)$
$0.8 = P(A \cap B) + 0.25 \implies P(A \cap B) = 0.55$

2. Calculate the conditional probability $P(A \mid B)$.

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$
$P(A \mid B) = \frac{0.55}{0.8} = \frac{11}{16} = 0.6875$

3. Test for independence of events A and B.

For $A$ and $B$ to be independent events, $P(A \cap B)$ must equal $P(A)P(B)$.
$P(A)P(B) = 0.7 \times 0.8 = 0.56$
Since $P(A \cap B) = 0.55 \neq 0.56$, the events $A$ and $B$ are not independent.

1. Calculate the probability of the union of events $A$ and $B$.

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.7 + 0.8 - 0.5 = 1$

2. State the relationship between the two events based on the result.

Since $P(A \cup B) = 1$, the union of events $A$ and $B$ covers the entire sample space. Therefore, $A$ and $B$ are exhaustive events.

1. Show that the total number of intersection points is 55.

Since the $11$ lines are non-parallel and no three lines intersect at a common point, each pair of lines intersects at exactly one unique point.
$\text{Total number of intersection points} = {}^{11}C_2$
$= \frac{11 \times 10}{2} = 55 \text{ (shown)}$

2. Determine the total number of ways to choose 3 points.

$\text{Total ways to choose } 3 ext{ points from } 55 = {}^{55}C_3$
$= \frac{55 \times 54 \times 53}{3 \times 2 \times 1} = 26235$

3. Determine the number of favourable outcomes (choosing 3 points that all lie on one of the 11 lines).

Each of the $11$ lines intersects the other $10$ lines, so there are exactly $10$ intersection points on each line.
For a specific line, the number of ways to choose $3$ points on it is:
${}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
Since two distinct lines share at most $1$ point, no $3$ points can be collinear on more than one line simultaneously. Thus, the subsets of $3$ points on each of the $11$ lines are mutually exclusive.
$\text{Total favourable outcomes} = 11 \times {}^{10}C_3$
$= 11 \times 120 = 1320$

4. Calculate the final probability.

$P(\text{all } 3 \text{ points are on one of the } 11 \text{ lines}) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}}$
$= \frac{1320}{26235}$
$= \frac{8}{159}$