10.1 The events A and B are independent - NSC Mathematics - Question 10 - 2016 - Paper 1

For independent events, the probability of A or B is calculated using:

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

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

Therefore, the probability that at least one of the events A or B occurs is 0.7.

Using the complement rule, we have:

$P( ext{Not } A ext{ and Not } B) = 1 - P(A ext{ or } B)$

From the previous step, we know: $P(A ext{ or } B) = 0.7$

Thus, $P( ext{Not } A ext{ and Not } B) = 1 - 0.7 = 0.3$

The probability that neither event A nor event B occurs is 0.3.

To represent this situation, a tree diagram would show the two bags as the first branches (Bag A and Bag B) with respective colors of balls as the subsequent branches:

• Bag A:

  • Pink (P(A) = 0.5)
  • Yellow (P(A) = 0.5)

• Bag B:

  • Pink (P(B) = 0.5)
  • Yellow (P(B) = 0.5)

Each choice has equal probability of 0.5 since it is equally likely to choose either Bag A or Bag B.

Bag A contains 3 pink and 2 yellow balls. The total number of balls in Bag A is:

$3 + 2 = 5$

The probability of selecting a yellow ball from Bag A is:

$P( ext{Yellow from A}) = \frac{2}{5} = 0.4$

So, the probability of choosing a yellow ball from Bag A is 0.4.

For Bag A:

The probability of choosing a pink ball from Bag A is:

$P( ext{Pink from A}) = \frac{3}{5} = 0.6$

For Bag B:

The probability of choosing a pink ball from Bag B is:

$P( ext{Pink from B}) = \frac{5}{9}$ (since Bag B has a total of 9 balls).

The combined probability of selecting a pink ball from either bag is:

$P( ext{Pink}) = P( ext{A}) \times P( ext{Pink from A}) + P( ext{B}) \times P( ext{Pink from B})$

$= 0.5 \times 0.6 + 0.5 \times \frac{5}{9} = 0.3 + 0.2777 = 0.5777$

Thus, the total probability of choosing a pink ball is approximately 0.5777.

Using the information given in the Venn diagram:

• Total learners = 600
• Learners that play rugby = 288
• Learners that play hockey = 372
• Learners that play neither = 56

Let:

• a = learners playing hockey only
• b = learners playing rugby only
• x = learners playing both sports

Thus, we can form the equations:

• For Rugby: $288 = a + x$
• For Hockey: $372 = b + x$
• For neither sport: $a + b + x + 56 = 600$

From here we can express a and b in terms of x: $a = 288 - x$ $b = 372 - x$

Substituting the expressions for a and b in the overall learner equation:

$(288 - x) + (372 - x) + x + 56 = 600$

Simplifying:

$288 + 372 - x + 56 = 600$ $716 - x = 600$ $x = 716 - 600$

Thus, we find that: $x = 116$

No, the events are not mutually exclusive because there are learners who play both sports (indicated by x). If the events were mutually exclusive, it would imply that learners could not participate in both sports, which contradicts the information given that x learners play both rugby and hockey.