Answer:
Step-by-step explanation:
The given table shows that there are 4 students, so there are 4 possible winner in total, but they have different number of tickets.
The total number of tickets is 48, that's the total possible outcomes. Kitzen has a probability of
[tex]P_{Kitzen}=\frac{12}{48}=\frac{1}{4}[/tex]
Ava has a probability of
[tex]P_{Ava}=\frac{16}{48}=\frac{1}{3}[/tex]
Now, the probability of having one event and the other is
[tex]P=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12} \approx 0.08[/tex]
Therefore, there is 8% probability of Kitzen winning first and Ava second.
(Notice that the probaility is not about Kitzen or Ava winning, it's about winning both, that's why the percentage is low)
