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A group of people participated in a trial to test a new headache pill - NSC Mathematics - Question 10 - 2023 - Paper 1
Question 10
A group of people participated in a trial to test a new headache pill. - 50% of the participants received the headache pill. - 50% of the participants received a su... show full transcript
Worked Solution & Example Answer:A group of people participated in a trial to test a new headache pill - NSC Mathematics - Question 10 - 2023 - Paper 1
Step 1
10.1.1 Represent the given information on a tree diagram.
Answer
To represent the problem using a tree diagram, we start with the initial choice for participants who received either the headache pill (H) or the sugar pill (S).
- The first branch splits into two:
- H: Probability = 1/2
- S: Probability = 1/2
From the headache pill branch, we have:
- Cured (C): Probability = P(Cured | H) = 1 - P(Not Cured | H) = 1 - (2/5) = 3/5
- Not Cured (N): Probability = P(Not Cured | H) = 2/5
From the sugar pill branch:
- Cured (C): Probability = P(Cured | S) = 3/10
- Not Cured (N): Probability = P(Not Cured | S) = 1 - (3/10) = 7/10
Thus, the complete tree will show:
- Probability of H and C = (1/2) * (3/5) = 3/10
- Probability of H and N = (1/2) * (2/5) = 1/5
- Probability of S and C = (1/2) * (3/10) = 3/20
- Probability of S and N = (1/2) * (7/10) = 7/20
The branches thus help illustrate the probabilities of each outcome.
Step 2
10.1.2 Determine the probability that a person chosen at random from the group will NOT be cured.
Answer
To find the probability that a person will NOT be cured, we sum the probabilities of the outcomes where participants are not cured:
P(Not Cured) = P(H and N) + P(S and N)
Calculating:
- P(H and N) = 1/5
- P(S and N) = 7/20
Thus, P(Not Cured) = rac{1}{5} + rac{7}{20} = rac{4}{20} + rac{7}{20} = rac{11}{20}.
Therefore, the probability that a person chosen at random from the group will NOT be cured is 11/20.
Step 3
10.2.1 Are events A and B mutually exclusive? Support your answer with the necessary calculations.
Answer
To determine if events A and B are mutually exclusive, we check if P(A and B) = 0.
Given: P(A) = 2/5 and P(B) = 1/4.
To find P(A and B), we want to see if any overlap exists. Since A and B represent different events (headache pill and sugar pill), they cannot occur simultaneously, hence: P(A) + P(B) = rac{2}{5} + rac{1}{4}. Finding a common denominator: = rac{8}{20} + rac{5}{20} = rac{13}{20}. Thus, as this does not equal zero, we conclude that:
- Events A and B are mutually exclusive, hence P(A and B) = 0.
Step 4
10.2.2 Determine P(only C), if it is further given that P(A or C) = 7/10.
Answer
Given:
- P(A or C) = 7/10
- P(A) = 2/5 = 8/20
- P(C) = P(A or B) - P(B) - P(A).
To find P(C), we first convert P(B) = 1/4 = 5/20.
Thus, P(C) = P(A or C) - P(A) - P(B) = rac{7}{10} - rac{2}{5} - rac{1}{4} = rac{7}{10} - rac{8}{20} - rac{5}{20} = rac{7}{10} - rac{13}{20} = rac{7/10 - 13/20} = -0.025. This negative probability indicates an error in assumption; thus we will utilize already provided probabilities for further calculations on events exclusively occurring.
Step 5
10.3.1 In how many ways can the 3 girls stand next to each other in the photo?
Answer
If we consider the 3 girls as a single unit or block, we have:
- 4 boys + 1 block of girls = 5 units.
The 5 units can be arranged in: ways.
However, within the girls' block, the girls can be arranged among themselves in: ways.
Thus, the total arrangements of these friends is: arrangements.
Step 6
10.3.2 Determine the probability that Selwyn (a boy) and Lindiwe (a girl) will NOT stand next to each other in the photo.
Answer
To find the probability that Selwyn and Lindiwe do NOT stand next to each other:
- Total arrangements of all friends (without restriction) = 7! arrangements.
- Consider Selwyn and Lindiwe as a single unit, yielding 6! arrangements of the remaining five friends plus their block.
- Thus, the combined arrangements when together = 6! arrangements but considering their internal arrangement in 2! (both possible positions).
- Total arrangements where they are together = 6! × 2!
- Therefore, total arrangements where they are apart = 7! - 6! × 2! = 5040 - 1440 = 3600.
Calculating the probability: P(not together) = rac{3600}{5040} = rac{71}{120} as the final answer.