Question

. The teacher decides that no student can win twice, so she removes the tickets of the first winner before drawing the second winner. What is the probability that she draws Kitzen first and then Ava second? Use a / to represent a fraction bar.

Answer 1

Answer:

There is 8% probability of Kitzen winning first and Ava second.

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

$P_{Kitzen}=\frac{12}{48}=\frac{1}{4}$

Ava has a probability of

$P_{Ava}=\frac{16}{48}=\frac{1}{3}$

Now, the probability of having one event and the other is

$P=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12} \approx 0.08$

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)