TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2020W/HW01.4

Student athletes

A random sample of 400 college students was asked if college athletes should be payed. The following table gives a two-way classification of the responses.

	Should be paid	Should not be paid
Student athlete	90	10
Student nonathlete	210	90

Problem 4a)

If one student is randomly selected from these 400 students, find the probability that this student

1. is in favor of paying college athletes

2. is an athlete and favors paying student athletes

3. is a nonathlete or is against paying students athletes

Problem 4b)

Are the events student athlete and should be paid mutually exclusive? Justify your answer.

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 13:10:32 CET

Solution for 4ai)

There are ${\textstyle 400}$ students in total, of which ${\textstyle 300}$ are in favor of paying athletes. Now we can divide the desireable outcomes by the number possible ones and get:

$${\frac {300}{400}}=0,75\quad \Box$$

Solution for 4aii)

Only ${\textstyle 90}$ students are athletes and in favor of paying.

$${\frac {90}{400}}=0,225\quad \Box$$

Solution for 4aiii)

To solve this problem we need to apply the Inclusion-Exclusion Principle. Let ${\textstyle A}$ be the set of students that are nonathletes and ${\textstyle B}$ is the set of students that are against paying students.

$${\begin{aligned}\left|A\cup B\right|&=\left|A\right|+\left|B\right|-\left|A\cap B\right|\\&=300+100-90\\&=310\end{aligned}}$$

Now all we have to do is to divide them by the total number of students.

$${\frac {310}{400}}=0,775$$

Solution for 4b)

We describe the set of student athletes as ${\textstyle A}$ and the set of should be paid as ${\textstyle B}$. ${\textstyle A}$ and ${\textstyle B}$ would be mutually exclusive if ${\textstyle A\cap B=\emptyset }$ would be true. However, from the table we know that ${\textstyle \left|A\cap B\right|=90}$ which is contrary to the definition of mutually exclusive. Therefore the two sets are not mutually exclusive.

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