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In a school in Canberra, there are 18 students in Year 9 and 12 in Year 10. Two students are selected at random without replacement to form a leadership team.
Use a table or list to determine all possible outcomes for the year levels of the two selected students. What is the exact probability that the first student selected is from Year 10 and the second is from Year 9?
Express your answer as a simplified fraction.
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Worked solution
- There are 18 Year 9 students and 12 Year 10 students, so the total number of students is 18 + 12 = 30.
- Two students are selected at random without replacement, so the selection is sequential and the first choice affects the second.
- We need to find the probability that the first student is from Year 10 and the second is from Year 9.
- List all possible outcomes for the year levels of the two selected students using a table of combinations: (Year 9, Year 9), (Year 9, Year 10), (Year 10, Year 9), (Year 10, Year 10).
- Calculate the probability of the first student being from Year 10: 12/30.
- After selecting one Year 10 student, 29 students remain, with 18 still in Year 9. So the probability the second student is from Year 9 is 18/29.
- Multiply the two probabilities: (12/30) × (18/29).
- Simplify 12/30 to 2/5. So the expression becomes (2/5) × (18/29).
- Multiply numerators: 2 × 18 = 36. Multiply denominators: 5 × 29 = 145. So the result is 36/145.
- Check if 36/145 can be simplified further. The greatest common divisor of 36 and 145 is 1, so the fraction is already in simplest form.
- The exact probability that the first student is from Year 10 and the second is from Year 9 is 36/145.
- Verify by checking total outcomes: number of ways to choose one Year 10 and one Year 9 in order is 12 × 18 = 216. Total possible ordered pairs: 30 × 29 = 870. 216/870 simplifies to 36/145. Correct.
Reference answer: 36/145
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