How many of these learners do not play any of the sports on the survey (that is netball, tennis or hockey)?

Let H be Hockey, N be Netball, T be Tennis.

• $N(H \cap N \cap T) = 2$ (This is the center region, already labeled 2).
• $N(H \cap T) = 7$. From the diagram, $N(H \cap T) = a + 2$. So, $a + 2 = 7 \Rightarrow a = 5$.
• $N(H \cap N) = 5$. From the diagram, $N(H \cap N) = 3 + 2 = 5$. (This matches the diagram).
• $N(T \cap N) = 6$. From the diagram, $N(T \cap N) = 4 + 2 = 6$. (This matches the diagram).
• $N(\text{Netball ONLY}) = b = 4$.
• $N(H) = 18$. From the diagram, $N(H) = c + a + 2 + 3$. Substitute $a=5$: $c + 5 + 2 + 3 = 18 \Rightarrow c + 10 = 18 \Rightarrow c = 8$.
• $N(T) = 12$. From the diagram, $N(T) = d + a + 2 + 4$. Substitute $a=5$: $d + 5 + 2 + 4 = 12 \Rightarrow d + 11 = 12 \Rightarrow d = 1$.

The total number of learners surveyed is 33.
The sum of all learners in the circles is:
$c + a + d + 3 + 2 + 4 + b = 8 + 5 + 1 + 3 + 2 + 4 + 4 = 27$.
The number of learners who do not play any sport ($e$) is the total learners minus those who play at least one sport.
$e = 33 - 27 = 6$.

The values are: $\boxed{a=5, b=4, c=8, d=1, e=6}$.

5.1.2 How many of these learners do not play any of the sports on the survey (that is netball, tennis or hockey)?

This is the value $e$ calculated in 5.1.1.

The number of learners who do not play any of the sports is $\boxed{6}$.

5.1.3 Write down the probability that a learner selected at random from this sample plays netball ONLY.

From 5.1.1, $b = 4$ learners play netball ONLY.
The total number of learners is 33.

The probability that a learner plays netball ONLY is $\boxed{\frac{4}{33}}$.

5.1.4 Determine the probability that a learner selected at random from this sample plays hockey or netball.

This includes all regions within the Hockey (H) circle or the Netball (N) circle.
$N(H \cup N) = c + a + 2 + 3 + b + 4$
$N(H \cup N) = 8 + 5 + 2 + 3 + 4 + 4 = 26$.
The total number of learners is 33.

The probability that a learner plays hockey or netball is $\boxed{\frac{26}{33}}$.

Question 5.2:
In all South African schools, EVERY learner must choose to do either Mathematics or Mathematical Literacy.
At a certain South African school, it is known that 60% of the learners are girls.
The probability that a randomly chosen girl at the school does Mathematical Literacy is 55%.
The probability that a randomly chosen boy at the school does Mathematical Literacy is 65%.

Determine the probability that a learner selected at random from this school does Mathematics.

Let G be the event a learner is a girl, B be the event a learner is a boy.
Let ML be the event a learner does Mathematical Literacy, M be the event a learner does Mathematics.
$P(G) = 0.60$
$P(B) = 1 - P(G) = 1 - 0.60 = 0.40$
$P(ML | G) = 0.55$
$P(ML | B) = 0.65$

Since learners must choose either Mathematics or Mathematical Literacy:
$P(M | G) = 1 - P(ML | G) = 1 - 0.55 = 0.45$
$P(M | B) = 1 - P(ML | B) = 1 - 0.65 = 0.35$

The probability that a learner selected at random does Mathematics is $\boxed{0.41}$.

Question 5.3:
William writes a Mathematics examination and an Accounting examination.
He estimates that he has a 40% chance of passing the Mathematics examination.
He estimates that he has a 60% chance of passing the Accounting examination.
He estimates that he has a 30% chance of passing both.

Determine the probability that William will fail Mathematics and Accounting.

Let M be the event William passes Mathematics, A be the event William passes Accounting.
$P(M) = 0.40$
$P(A) = 0.60$
$P(M \cap A) = 0.30$ (Probability of passing both)

Using the Addition Rule for probabilities:

This is the complement of passing Mathematics OR Accounting, i.e., $P((M \cup A)')$.

The probability that William will fail Mathematics and Accounting is $\boxed{0.30}$.