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# PROBABILITY

CONTENT

(a) Addition and multiplication rules of probability: (i) Mutually exclusive events and addition (“or”) rule. (ii) Complimentary events and probability rule. (iii) Independent events and multiplication (“and”) rules.

(b) Solving simple problems on mutually exclusive, Independent and complimentary events.

(c) Experiment with or without replacement.

(d) Practical application of probability in; health, finance, population, etc.

Probability of Event A “OR“ Event B i.e. Pr (A ∪ B) (for Intersecting Sets).

Given any sample space $$W = \{1, 2, 3, …, 8, 9, 10\}$$ and Event $$A = \{2, 3, 5, 7 \}$$ and Event $$B = \{1, 3, 5, 6, 7, 9\}$$

If a number is picked from the sample space W, the probability of picking a number that forms the Set A or B denoted by A ∪ B is explained as follows

From above, Note that

$$A ∪ B = \{1, 2, 3, 5, 6, 7, 9\} \\ n (A ∪ B) = 7$$

∴ Prob.$$(A ∪ B) = \frac{7}{10}………(1)$$

Note that A and B are intersecting, hence suppose

$$Pr(A∪B) = Pr(A) + Pr(B)$$ But $$Pr(A) = \frac{4}{10} \\$$

$$\frac{n(A)}{n(Ω)} + \frac{n(B)}{n(Ω)}$$ But $$Pr(B) = \frac{6}{10} \\$$

$$= \frac{4}{10} + \frac{6}{10} \\ = \frac{10}{10} = 1$$ But $$Pr(A∪B) = \frac{7}{10}$$

From equation (1)

$$∴ Pr(A∪B) ≠ Pr(A) + Pr(B)$$

This is because the Set $$\{3, 5, 7\}$$ was counted twice.

Lesson tags: General Mathematics Lesson Notes, General Mathematics Objective Questions, SS2 General Mathematics, SS2 General Mathematics Evaluation Questions, SS2 General Mathematics Evaluation Questions Third Term, SS2 General Mathematics Objective Questions, SS2 General Mathematics Objective Questions Third Term, SS2 General Mathematics Third Term
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