Given P(A) = 0,48 and P(B) = 0,26. Calculate P(A or B) if: 1.1.1 A and B are mutually exclusive events. 1.1.2 A and B are independent events.

For mutually exclusive events, the probability of A or B occurring is the sum of their individual probabilities.

$$P(A \text{ or } B) = P(A) + P(B)$$ $$P(A \text{ or } B) = 0.48 + 0.26$$ $$P(A \text{ or } B) = \boxed{\textbf{0.74}}$$

For independent events, the probability of A and B occurring is the product of their individual probabilities: $P(A \text{ and } B) = P(A) \times P(B)$.
The general formula for $P(A \text{ or } B)$ is:

$$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$

Substitute the formula for independent events:

$$P(A \text{ or } B) = P(A) + P(B) - (P(A) \times P(B))$$ $$P(A \text{ or } B) = 0.48 + 0.26 - (0.48 \times 0.26)$$ $$P(A \text{ or } B) = 0.74 - 0.1248$$ $$P(A \text{ or } B) = \boxed{\textbf{0.6152}}$$

From the table, the total number of boys is 80. The sum of boys' preferences for Netball, Soccer, and Rugby must equal this total.

$$32 + x + 27 = 80$$ $$59 + x = 80$$ $$x = 80 - 59$$ $$x = \boxed{\textbf{21}}$$

Total learners = 150.
Number of boys = 80.
Number of boys who like Netball = 32.
Number of boys who do not like Netball = Total boys - Boys who like Netball

$$= 80 - 32 = 48$$

The probability is the number of boys who do not like netball divided by the total number of learners.

$$P(\text{boy and not netball}) = \frac{48}{150}$$

Simplify the fraction:

$$P(\text{boy and not netball}) = \frac{24}{75} = \frac{8}{25}$$ $$P(\text{boy and not netball}) = \boxed{\textbf{0.32}}$$

Two events A and B are independent if $P(A \text{ and } B) = P(A) \times P(B)$.
Let A be the event "being a boy".
Let B be the event "liking soccer".

Calculate the individual probabilities:
Total learners = 150.
Number of boys = 80.

$$P(\text{boy}) = \frac{80}{150}$$

Number of learners who like Soccer = 33.

$$P(\text{soccer}) = \frac{33}{150}$$

Number of boys who like Soccer (value of $x$) = 21.

$$P(\text{boy and soccer}) = \frac{21}{150}$$

Check for independence:
Is $P(\text{boy and soccer}) = P(\text{boy}) \times P(\text{soccer})$?

$$\frac{21}{150} = \frac{80}{150} \times \frac{33}{150}$$

Calculate the right side:

$$\frac{80}{150} \times \frac{33}{150} = \frac{8}{15} \times \frac{33}{150} = \frac{264}{2250}$$

Simplify $\frac{264}{2250}$ by dividing by 6:

$$\frac{264 \div 6}{2250 \div 6} = \frac{44}{375}$$

Now compare this to the left side:

$$\frac{21}{150}$$

Simplify $\frac{21}{150}$ by dividing by 3:

$$\frac{21 \div 3}{150 \div 3} = \frac{7}{50}$$

Since $\frac{7}{50} \neq \frac{44}{375}$ (or $0.14 \neq 0.1173$), the events are not independent.

$$\boxed{\textbf{The events "being a boy" and "liking soccer" are NOT independent.}}$$