Events $A$ and $B$ are such that $P(A)=0.7$ and $P(B)=0.8$. Show that $P(A \cap B) \ge 0.5$ If $P(A'\ \cup\ B)=0.85$, find $P(A\mid B)$. State, with a reason, whether or not $A$ and $B$ are independent events. What is the relationship between event A and event B if $P(A \cap B)=0.5$ ? Eleven non-parallel lines are drawn in a plane. If no three lines intersect at a common point, show that there are 55 points of intersection between the lines. If three of these points are chosen at random, find the probability that they are all on one of the eleven lines.

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.

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