A group of people participated in a trial to test a new headache pill - NSC Mathematics - Question 10 - 2023 - Paper 1

To find the probability that a person is NOT cured, we sum the probabilities of NOT being cured from both branches:

• P(Not Cured from Headache Pill) = 0.2
• P(Not Cured from Sugar Pill) = 0.35

Thus, the total probability that a person chosen at random will NOT be cured is:

$P(Not Cured) = 0.2 + 0.35 = 0.55$.

To determine if events A and B are mutually exclusive, we check if they can occur simultaneously. If P(A and B) = 0, then they are mutually exclusive.

From the problem we know: P(A) = \frac{2}{5} P(B) = \frac{1}{4} P(A \text{ or } B) = \frac{13}{20}

Since P(A or B) = P(A) + P(B) - P(A and B) and given P(A or B) is greater than the sum of P(A) and P(B), then:

$P(A and B) > 0$

Thus, A and B are not mutually exclusive.

To find P(only C), we need to use the information given:

1. We know that:

   • P(A or C) = P(A) + P(C) - P(A and C)
   • Hence, $P(A or C) = \frac{7}{10} = \frac{2}{5} + P(C) - \frac{2}{5}$ This simplifies to $P(C) = \frac{7}{10} - \frac{2}{5} = \frac{1}{10}.$

2. Given that 2P(B and C) = P(A and C), we can determine the sequential probabilities accordingly.

To find the probability that events A, B or C do NOT take place, we first calculate the combined probabilities of events A, B, and C:

$P(A) + P(B) + P(C) - P(A \text{ and } B ) - P(B \text{ and } C) - P(A \text{ and } C) + P(A \text{ and } B \text{ and } C).$

Then, we subtract this value from 1 to find the probability of none of the events occurring:

$P(None) = 1 - (P(A) + P(B) + P(C))$

To solve this, we can treat the 3 girls as one single unit (block). Therefore, we have 4 units to arrange: 1 block of girls + 4 boys.

The total arrangements of these 5 units is: $5! = 120.$

Within the block, the 3 girls can be arranged among themselves: $3! = 6.$

Thus, the total arrangements is: $5! \times 3! = 120 \times 6 = 720.$

To determine the probability that Selwyn and Lindiwe do NOT stand next to each other:

1. First calculate the total arrangements of the 7 friends: $7! = 5040.$

2. Then calculate the arrangements if Selwyn and Lindiwe are treated as a single unit: $6! \times 2! = 720 \times 2 = 1440.$

3. Then, subtract this from the total arrangements to find the arrangements where they do NOT stay together: $5040 - 1440 = 3600.$

4. Finally, the probability that they do NOT stand next to each other is: $P(Not next) = \frac{3600}{5040} = \frac{12}{21} \approx 0.571.$