Question

there are 6 people taking part in a raffle. bob, elsa, hans, jim, omar, and soo. suppose that prize winners are randomly selected from the 6 people. compute the probability of each of the following events. event a: the first four prize winners are elsa, bob, hans, and jim, regardless of order. event b: soo is the first prize winner, omar is second, hans is third, and elsa is fourth. write your answers as fractions in simplest form. p(a) = p(b) =

Explanation:

For Event A:

Step 1: Find total permutations of 4 winners from 6

The number of permutations of choosing 4 people out of 6 for the first four prize winners is given by the permutation formula \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 6 \) and \( r=4 \). So \( P(6,4)=\frac{6!}{(6 - 4)!}=\frac{6!}{2!}=\frac{6\times5\times4\times3\times2!}{2!}=6\times5\times4\times3 = 360 \).

Step 2: Find permutations of the 4 specific people

The number of permutations of Elsa, Bob, Hans, and Jim (4 people) among themselves is \( P(4,4)=\frac{4!}{(4 - 4)!}=4!=4\times3\times2\times1 = 24 \).

Step 3: Calculate probability for Event A

The probability \( P(A) \) is the number of favorable outcomes (permutations of the 4 people) divided by the total number of possible outcomes (permutations of 4 from 6). So \( P(A)=\frac{24}{360}=\frac{1}{15} \). (Alternatively, using combinations: The number of ways to choose 4 people out of 6 is \( C(6,4)=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15 \), and there's 1 favorable combination of the 4 people, so probability is \( \frac{1}{15} \) since each combination is equally likely when order doesn't matter in selection, but here we can also think in terms of combinations as the problem says "regardless of order" for the first four winners. So the number of favorable combinations is 1 (choosing Elsa, Bob, Hans, Jim) and total combinations of 4 from 6 is 15, so \( P(A)=\frac{1}{15} \))

For Event B:

Step 1: Analyze the number of favorable and total outcomes

For Event B, we are looking for a specific order: Soo first, Omar second, Hans third, Elsa fourth. The total number of possible permutations for the first four prize winners is still \( P(6,4) = 360 \) (as calculated above). The number of favorable outcomes is 1 (only one specific order: Soo, Omar, Hans, Elsa in that order for the first four).

Step 2: Calculate probability for Event B

We can also calculate it step - by - step using the multiplication rule of probability. The probability that Soo is first is \( \frac{1}{6} \), then Omar is second (given Soo is first) is \( \frac{1}{5} \), then Hans is third (given Soo and Omar are first and second) is \( \frac{1}{4} \), then Elsa is fourth (given the first three) is \( \frac{1}{3} \). So \( P(B)=\frac{1}{6}\times\frac{1}{5}\times\frac{1}{4}\times\frac{1}{3}=\frac{1}{360} \).

Answer:

\( P(A)=\frac{1}{15} \)
\( P(B)=\frac{1}{360} \)