SOLUTION: A sample of students, faculty, and administration from a large university were asked if they were in favor of increasing parking fees to build a new parking lot. Student Fac

Algebra ->  Probability-and-statistics -> SOLUTION: A sample of students, faculty, and administration from a large university were asked if they were in favor of increasing parking fees to build a new parking lot. Student Fac      Log On


   



Question 1185349: A sample of students, faculty, and administration from a large university were asked if they were in favor of increasing parking fees to build a new parking lot.

Student Faculty Admin Total
In Favor 652 665 696 2013
Against 208 586 609 1403
Total 860 1251 1305 3416

Give all of your probabilities as decimals and round to at least two decimal places.

What is the probability that a person selected at random is in favor of increasing the fees?
What is the probability that a person selected at random is both a student and against increasing the fees?
What is the probability that a person selected at random is either administration or against increasing the fees?
What is the probability that a person selected at random is in favor of increasing the fees and is against increasing the fees?
What is the probability that a faculty member selected at random is against increasing the fees?

Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here are the probabilities you requested, calculated from the given table:
* **Total People:** 3416
* **In Favor:** 2013
* **Against:** 1403
* **Students:** 860
* **Faculty:** 1251
* **Administration:** 1305
* **Students In Favor:** 652
* **Students Against:** 208
* **Faculty In Favor:** 665
* **Faculty Against:** 586
* **Admin In Favor:** 696
* **Admin Against:** 609
1. **Probability of being in favor:**
P(In Favor) = (Number In Favor) / (Total People) = 2013 / 3416 ≈ 0.59
2. **Probability of being a student and against:**
P(Student and Against) = (Number of Students Against) / (Total People) = 208 / 3416 ≈ 0.06
3. **Probability of being administration or against:**
P(Admin or Against) = P(Admin) + P(Against) - P(Admin and Against)
P(Admin) = 1305/3416 ≈ 0.38
P(Against) = 1403/3416 ≈ 0.41
P(Admin and Against) = 609/3416 ≈ 0.18
P(Admin or Against) = 0.38 + 0.41 - 0.18 ≈ 0.61
4. **Probability of being in favor and against:**
This is impossible. A person cannot be both in favor and against at the same time. Therefore, the probability is 0.
5. **Probability of a faculty member being against:**
P(Faculty Against | Faculty) = (Number of Faculty Against) / (Number of Faculty) = 586 / 1251 ≈ 0.47