In a school of 70 Grade 10 learners, 32 learners take Economics (E), 43 take Mathematics (M) and 15 learners take neither of these two subjects. Let the number of learners who take both subjects be x.

The probability of A or B is $\boxed{\text{0.65}}$.

3.1.3 Determine $P(\text{not } A \text{ and } B)$

The probability of not A and B is $\boxed{\text{0.25}}$.

Question 3.2
Given: Total learners = 70.
Learners taking Economics (E) = 32.
Learners taking Mathematics (M) = 43.
Learners taking neither subject = 15.
Learners taking both subjects = $x$.

3.2.1 Show that $x = 20$.

The total number of learners is the sum of those taking only E, only M, both, and neither.

Thus, it is shown that $\boxed{x = 20}$.

3.2.2 Draw a Venn diagram to illustrate the above information.

• Only Economics: $32 - x = 32 - 20 = 12$
• Only Mathematics: $43 - x = 43 - 20 = 23$
• Both Economics and Mathematics: $x = 20$
• Neither: 15

(Imagine a rectangle representing the universal set of 70 learners. Inside, draw two overlapping circles. Label one circle 'E' for Economics and the other 'M' for Mathematics.)

• In the section of circle E that does not overlap with M, write '12'.
• In the section of circle M that does not overlap with E, write '23'.
• In the overlapping section (intersection), write '20'.
• Outside both circles but inside the rectangle, write '15'. (Sum check: $12 + 23 + 20 + 15 = 70$)

3.2.3 Determine the probability of a learner taking:
Total learners = 70.

a) Economics and Mathematics.

The probability of taking Economics and Mathematics is $\boxed{\frac{2}{7}}$.

b) at least ONE of the two subjects.

This is the total learners minus those taking neither.

The probability of taking at least one subject is $\boxed{\frac{11}{14}}$.

c) taking EXACTLY ONE subject.

The probability of taking exactly one subject is $\boxed{\frac{1}{2}}$.

3.2.4 Determine the number of learners that take Physical Sciences.
Let P be the event of taking Physical Sciences.
Given: "The probability that a Mathematics learner is also a Physical Sciences learner is 79%."
This is a conditional probability: $P(P | M) = 0.79$.
We know $N(M) = 43$.

Note: The question asks for "the number of learners that take Physical Sciences" ($N(P)$). From the given information, we can only directly calculate the number of learners who take both Physical Sciences and Mathematics ($N(P \cap M)$). To determine $N(P)$ uniquely, we would need additional information (e.g., the number of learners taking only Physical Sciences). Assuming this common interpretation, we provide the number of learners taking both.
The number of learners that take Physical Sciences (and Mathematics) is $\boxed{\text{34}}$.