A club with SU college students is doing volunteer work this semester. Each student is volunteering at one of four locations. Here is a summary. Location Number of students Soup kitchen 14 Hospital 12 Pet shelter 15 Library 9 Three students from the club are selected at random, one at a time without replacement. What is the probability that all three of the students volunteer library? Do not round your intermediate computations. Round your final answer to three decimal places. 0 X ?
A club with SU college students is doing volunteer work this semester. Each student is volunteering at one of four locations. Here is a summary. Location Number of students Soup kitchen 14 Hospital 12 Pet shelter 15 Library 9 Three students from the club are selected at random, one at a time without replacement. What is the probability that all three of the students volunteer library? Do not round your intermediate computations. Round your final answer to three decimal places. 0 X ?
A First Course in Probability (10th Edition)
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![**Volunteering Probability Problem**
**Problem Statement:**
A club with 50 college students is doing volunteer work this semester. Each student is volunteering at one of four locations. Here is a summary:
| Location | Number of students |
|---------------|---------------------|
| Soup kitchen | 14 |
| Hospital | 12 |
| Pet shelter | 15 |
| Library | 9 |
Three students from the club are selected at random, one at a time without replacement.
**Question:**
What is the probability that all three of the students volunteer at the library?
*Note: Do not round your intermediate computations. Round your final answer to three decimal places.*
**Solution Steps:**
To solve this, we need to calculate the probability that all three randomly chosen students are from the group of 9 students volunteering at the library, out of the total 50 students.
1. The probability for the first student to be from the library is calculated as:
\[ P(\text{First student from library}) = \frac{9}{50} \]
2. After choosing the first student, there are now 8 students left in the library and a total of 49 students remaining. Hence, the probability that the second student is also from the library is:
\[ P(\text{Second student from library}) = \frac{8}{49} \]
3. After choosing the second student, there are now 7 students left in the library and a total of 48 students remaining. Hence, the probability that the third student is also from the library is:
\[ P(\text{Third student from library}) = \frac{7}{48} \]
Therefore, the overall probability of all three students being from the library is the product of these probabilities:
\[ P(\text{All three from library}) = \left(\frac{9}{50}\right) \times \left(\frac{8}{49}\right) \times \left(\frac{7}{48}\right) \]
**Calculation:**
\[ P(\text{All three from library}) = \frac{9 \times 8 \times 7}{50 \times 49 \times 48} \]
Perform the multiplication:
\[ 9 \times 8 \times 7 = 504 \]
\[ 50 \times 49 \times 48 = 117600 \]
Now, divide the results](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F424746dd-29d3-476a-868c-5a06ada19580%2F7dd8d42e-416b-47e2-9e5f-b3d5306d038f%2Fgprimg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Volunteering Probability Problem**
**Problem Statement:**
A club with 50 college students is doing volunteer work this semester. Each student is volunteering at one of four locations. Here is a summary:
| Location | Number of students |
|---------------|---------------------|
| Soup kitchen | 14 |
| Hospital | 12 |
| Pet shelter | 15 |
| Library | 9 |
Three students from the club are selected at random, one at a time without replacement.
**Question:**
What is the probability that all three of the students volunteer at the library?
*Note: Do not round your intermediate computations. Round your final answer to three decimal places.*
**Solution Steps:**
To solve this, we need to calculate the probability that all three randomly chosen students are from the group of 9 students volunteering at the library, out of the total 50 students.
1. The probability for the first student to be from the library is calculated as:
\[ P(\text{First student from library}) = \frac{9}{50} \]
2. After choosing the first student, there are now 8 students left in the library and a total of 49 students remaining. Hence, the probability that the second student is also from the library is:
\[ P(\text{Second student from library}) = \frac{8}{49} \]
3. After choosing the second student, there are now 7 students left in the library and a total of 48 students remaining. Hence, the probability that the third student is also from the library is:
\[ P(\text{Third student from library}) = \frac{7}{48} \]
Therefore, the overall probability of all three students being from the library is the product of these probabilities:
\[ P(\text{All three from library}) = \left(\frac{9}{50}\right) \times \left(\frac{8}{49}\right) \times \left(\frac{7}{48}\right) \]
**Calculation:**
\[ P(\text{All three from library}) = \frac{9 \times 8 \times 7}{50 \times 49 \times 48} \]
Perform the multiplication:
\[ 9 \times 8 \times 7 = 504 \]
\[ 50 \times 49 \times 48 = 117600 \]
Now, divide the results
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