Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.