There are 8  people taking part in a raffle.
Ann, Elsa, Jim, Kira, Lena, Omar, Ravi, and Soo.
Suppose that prize winners are randomly selected from the 8  people.
Compute the probability of each of the following events.

Event A: Lena is the first prize winner, Omar is second, Ravi is third, and Kira is fourth.

Event B: The first four prize winners are Omar, Ann, Soo, and Elsa, regardless of order.

Write your answers as fractions in simplest form.

### Event A: Lena is the first prize winner, Omar is second, Ravi is third, and Kira is fourth.

In this event, the selection order matters, and there is only **one specific way** for Lena, Omar, Ravi, and Kira to be the first four winners in exactly that order.

The total number of possible ways to assign 8 people to 8 positions (since each person wins only once) is given by the **factorial of 8**:

\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320
\]

However, for Event A, only one specific arrangement (Lena, Omar, Ravi, Kira as the first four in that order) satisfies the condition, so:

\[
\text{Probability of Event A} = \frac{1}{8!} = \frac{1}{40,320}
\]

### Event B: The first four prize winners are Omar, Ann, Soo, and Elsa, regardless of order.

For this event, we are only concerned with the group of four people selected as the first four winners, regardless of their specific order. To calculate this:

1. The number of ways to choose any 4 people from the group of 8 can be done in:

\[
\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \quad \text{ways}
\]

2. Once those 4 people are chosen, the number of ways they can be arranged among the first four positions (since order does not matter) is:

\[
4! = 4 \times 3 \times 2 \times 1 = 24 \quad \text{ways}
\]

Therefore, the total number of possible outcomes is:

\[
8! = 40,320
\]

The number of favorable outcomes where the first four winners are Omar, Ann, Soo, and Elsa (in any order) is:

\[
4! = 24
\]

Thus, the probability for Event B is:

\[
\text{Probability of Event B} = \frac{24}{40,320} = \frac{1}{1,680}
\]

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### Summary of Answers:

- **Event A**: The probability that Lena is first, Omar is second, Ravi is third, and Kira is fourth is:

\[
\frac{1}{40,320}
\]

- **Event B**: The probability that the first four prize winners are Omar, Ann, Soo, and Elsa in any order is:

\[
\frac{1}{1,680}
\]

---

Would you like a further breakdown of these steps, or do you have any questions?

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### Related Questions:
1. What is the probability that Lena is the first prize winner, regardless of the order of the remaining winners?
2. How many different ways can the first 3 prize winners be selected?
3. What is the probability that Omar, Kira, and Ravi are among the first four winners, regardless of the fourth person?
4. How many different ways can all 8 people be arranged in prize-winning order?
5. What is the probability that the first prize winner is a female (Ann, Elsa, Kira, Lena, or Soo)?

---

### Tip:
When calculating probabilities with permutations, always check if the order of events matters. If order matters, use permutations; if not, use combinations.

There are 8 people taking part in a raffle. Ann, Elsa, Jim, Kira, Lena, Omar, Ravi, and Soo. Suppose that prize winners are randomly selected from the 8 people. Compute the probability of each of the following events. Event A: Lena is the first prize winner, Omar is second, Ravi is third, and Kira is fourth. Event B: The first four prize winners are Omar, Ann, Soo, and Elsa, regardless of order.

Explore the probability of selecting raffle winners from a group of 8 people. Event A calculates the exact order of winners using permutations, while Event B considers winners in any order with combinations. Perfect for students learning advanced probability concepts.

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