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 ?

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter12: Probability
Section12.4: Discrete Random Variables; Applications To Decision Making
Problem 30E
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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
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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