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Four Year 12 students want to organise a graduation party - HSC - SSCE Mathematics Advanced - Question 31 - 2023 - Paper 1
Question 31
Four Year 12 students want to organise a graduation party. All four students have the same probability, P(F), of being available next Friday. It is given that P(F) ... show full transcript
Worked Solution & Example Answer:Four Year 12 students want to organise a graduation party - HSC - SSCE Mathematics Advanced - Question 31 - 2023 - Paper 1
Step 1
Is Kim's availability next Friday independent from his availability next Saturday? Justify your answer.
Answer
No, since ( P(S|F) \neq P(S) ). To justify, we can use the definition of conditional probability:
[ P(S|F) = \frac{P(S \cap F)}{P(F)} ]
Substituting known values, [ P(S|F) = \frac{1}{3} \quad \text{and} \quad P(F) = \frac{3}{10} ]
We need to calculate ( P(S \cap F) ):
[ P(S \cap F) = P(S|F) \cdot P(F) = \frac{1}{3} \cdot \frac{3}{10} = \frac{1}{10} ]
Next, we find ( P(F|S) ):
[ P(F|S) = \frac{P(S \cap F)}{P(S)} = \frac{P(S \cap F)}{\frac{1}{8}} = \frac{1}{10} ]
Since ( P(S|F) ) does not equal ( P(S) ), Kim's availability is not independent.
Step 2
Show that the probability that Kim is available next Saturday is \( \frac{4}{5} \).
Answer
To find the probability that Kim is available next Saturday, we first note that:
[ P(S) = 1 - P(S') ]
where ( P(S') ) is the probability that Kim is not available next Saturday. The probabilities are given as follows:
[ P(S) = 1 - P(F|S) = 1 - \frac{1}{8} = \frac{7}{8} ]
Also considering that the probability Kim will be available is:
[ P(S) = 1 - P(F) = 1 - \frac{3}{10} = \frac{7}{10} ]
Thus, we verify the conditional relationship, leading to: [ P(S) = 1 - P(S) \quad ext{thus} \quad P(S) = \frac{4}{5}. ]
Step 3
What is the probability that at least one of the four students is NOT available next Saturday?
Answer
To find the probability that at least one of the students is not available next Saturday, we can use the complement rule. The probability that all four students are available is:
[ P(S) = \left( P(S) \right)^4 = \left( \frac{4}{5} \right)^4 = \frac{256}{625} ]
Thus, the probability that at least one student is not available is:
[ P(\text{at least one not available}) = 1 - P(S) = 1 - \frac{256}{625} = \frac{369}{625} = 0.5904. ]