Answer :
To answer the questions, let's first understand the context of the problem and solve each event step-by-step.
Event A: Lena is the first prize winner, Soo is second, and Hans is third.
There are 9 people, and the winners need to be selected in a specific order: Lena first, Soo second, and Hans third.
1. The probability that Lena wins first:
- There is only 1 favorable outcome out of 9 possibilities.
[tex]\[ P(\text{Lena first}) = \frac{1}{9} \][/tex]
2. The probability that Soo wins second given that Lena has already been chosen:
- There are now 8 people left.
[tex]\[ P(\text{Soo second} \mid \text{Lena first}) = \frac{1}{8} \][/tex]
3. The probability that Hans wins third given that Lena and Soo have already been chosen:
- There are now 7 people left.
[tex]\[ P(\text{Hans third} \mid \text{Lena first and Soo second}) = \frac{1}{7} \][/tex]
Combining these probabilities together:
[tex]\[ P(A) = \frac{1}{9} \times \frac{1}{8} \times \frac{1}{7} = \frac{1}{504} \][/tex]
Thus,
[tex]\[ P(A) = \frac{1}{504} \][/tex]
Event B: The first three prize winners are Omar, Elsa, and Lena, regardless of order.
1. First, calculate the number of ways to select any 3 winners out of the 9 people. This can be calculated using the combination formula [tex]\( C(n, k) = \frac{n!}{k!(n-k)!} \)[/tex], where [tex]\( n = 9 \)[/tex] and [tex]\( k = 3 \)[/tex].
[tex]\[ C(9, 3) = \frac{9!}{3! \times 6!} = 84 \][/tex]
2. Event B has only one favorable outcome; specifically selecting Omar, Elsa, and Lena in any order is just one specific way.
3. The probability of event B happening:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{84} \][/tex]
Converting this probability to the simplest form:
[tex]\[ P(B) = \frac{1}{84} \][/tex]
Thus, the answer for Event B is:
[tex]$ \begin{array}{l} P(A)=\frac{1}{504} \\ P(B)=\frac{1}{84} \end{array} $[/tex]
Event A: Lena is the first prize winner, Soo is second, and Hans is third.
There are 9 people, and the winners need to be selected in a specific order: Lena first, Soo second, and Hans third.
1. The probability that Lena wins first:
- There is only 1 favorable outcome out of 9 possibilities.
[tex]\[ P(\text{Lena first}) = \frac{1}{9} \][/tex]
2. The probability that Soo wins second given that Lena has already been chosen:
- There are now 8 people left.
[tex]\[ P(\text{Soo second} \mid \text{Lena first}) = \frac{1}{8} \][/tex]
3. The probability that Hans wins third given that Lena and Soo have already been chosen:
- There are now 7 people left.
[tex]\[ P(\text{Hans third} \mid \text{Lena first and Soo second}) = \frac{1}{7} \][/tex]
Combining these probabilities together:
[tex]\[ P(A) = \frac{1}{9} \times \frac{1}{8} \times \frac{1}{7} = \frac{1}{504} \][/tex]
Thus,
[tex]\[ P(A) = \frac{1}{504} \][/tex]
Event B: The first three prize winners are Omar, Elsa, and Lena, regardless of order.
1. First, calculate the number of ways to select any 3 winners out of the 9 people. This can be calculated using the combination formula [tex]\( C(n, k) = \frac{n!}{k!(n-k)!} \)[/tex], where [tex]\( n = 9 \)[/tex] and [tex]\( k = 3 \)[/tex].
[tex]\[ C(9, 3) = \frac{9!}{3! \times 6!} = 84 \][/tex]
2. Event B has only one favorable outcome; specifically selecting Omar, Elsa, and Lena in any order is just one specific way.
3. The probability of event B happening:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{84} \][/tex]
Converting this probability to the simplest form:
[tex]\[ P(B) = \frac{1}{84} \][/tex]
Thus, the answer for Event B is:
[tex]$ \begin{array}{l} P(A)=\frac{1}{504} \\ P(B)=\frac{1}{84} \end{array} $[/tex]
