10.1 Die gebeurtenisse A en B is onafhanklik - NSC Mathematics - Question 10 - 2016 - Paper 1

To find the probability of either event A or event B occurring, we use the formula:

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

Substituting the values: $P(A \text{ or } B) = 0.4 + 0.5 - 0.2 = 0.7$

The probability of neither event A nor event B occurring can be found by:

$P(nie A \text{ and } nie B) = 1 - P(A \text{ or } B)$

Using the previously calculated probability: $P(nie A \text{ and } nie B) = 1 - 0.7 = 0.3$

To illustrate the probabilities with a tree diagram, we can follow these logic branches:

1. Choose a Bag:

   • Probability of choosing Bag A: 0.5
   • Probability of choosing Bag B: 0.5

2. Choosing a Ball from Bag A: (3 pink, 2 yellow)

   • Probability of pink from A: ( \frac{3}{5} = 0.6 )
   • Probability of yellow from A: ( \frac{2}{5} = 0.4 )

3. Choosing a Ball from Bag B: (5 pink, 4 yellow)

   • Probability of pink from B: ( \frac{5}{9} )
   • Probability of yellow from B: ( \frac{4}{9} )

This gives us a complete tree representation of all outcomes.

The probability of selecting a yellow ball from Bag A can be calculated using: $P(yellow \text{ from } A) = P(choosing A) \times P(yellow | A)$ Substituting the values: $P(yellow \text{ from } A) = 0.5 \times 0.4 = 0.2$

The probability of selecting a pink ball can be calculated as follows:

1. From Bag A: $P(pink \text{ from } A) = P(choosing A) \times P(pink | A) = 0.5 \times 0.6 = 0.3$

2. From Bag B: $P(pink \text{ from } B) = P(choosing B) \times P(pink | B) = 0.5 \times \frac{5}{9} = \frac{5}{18} \approx 0.278$

Adding these probabilities gives: $P(pink) = P(pink \text{ from } A) + P(pink \text{ from } B) = 0.3 + 0.278 \approx 0.578$