Question

there are 9 people taking part in a raffle.
ann, bob, elsa, hans, jim, kira, lena, omar, and ravi.
suppose that prize winners are randomly selected from the 9 people.
compute the probability of each of the following events.
event a: the first four prize winners are hans, ann, ravi, and omar, regardless of order.
event b: lena is the first prize winner, ravi is second, ann is third, and elsa is fourth.
write your answers as fractions in simplest form.
p(a) =
p(b) =

Explanation:

For Event A:

Step1: Find total permutations for 4 winners.

The number of ways to choose and arrange 4 people out of 9 for the first four prize winners is given by the permutation formula \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 9 \) and \( r=4 \). So \( P(9, 4)=\frac{9!}{(9 - 4)!}=\frac{9!}{5!}=9\times8\times7\times6 = 3024 \).

Step2: Find favorable permutations for Event A.

Event A requires the first four winners to be Hans, Ann, Ravi, and Omar (regardless of order). The number of ways to arrange 4 specific people is \( 4! = 24 \) (since there are 4 people and we can arrange them in any order).

Step3: Calculate probability for Event A.

The probability \( P(A) \) is the number of favorable outcomes divided by the number of total outcomes. So \( P(A)=\frac{4!}{P(9, 4)}=\frac{24}{3024}=\frac{1}{126} \).

For Event B:

Step1: Find total permutations for 4 winners (same as before).

We already know that the total number of ways to arrange 4 people out of 9 is \( P(9, 4)=9\times8\times7\times6 = 3024 \).

Step2: Find favorable permutations for Event B.

Event B has a specific order: Lena (1st), Ravi (2nd), Ann (3rd), Elsa (4th). There's only 1 way to have this specific arrangement.

Step3: Calculate probability for Event B.

The probability \( P(B) \) is the number of favorable outcomes divided by the number of total outcomes. So \( P(B)=\frac{1}{P(9, 4)}=\frac{1}{3024} \).

Answer:

s:
\( P(A)=\boxed{\dfrac{1}{126}} \)

\( P(B)=\boxed{\dfrac{1}{3024}} \)