Exercise 1.3: Fictional University Somewhere

Fictional University Somewhere

From the adjacent graph you can read some information about $\rm FUS$ ("Fictional University Somewhere"). The whole square represents the universal set $G$ of $960$ students. Of these

$25\%$ female (German: "weiblich") (set $W$, purple rectangle),
$75\%$ male (German: "männlich") (set $M$, yellow rectangle).

At the university there are the faculties of

Theology (set $T$,nbsp; black triangle),
Information Technology (set $I$,nbsp; blue triangle),
Business Administration (set $B$,nbsp; green rectangle).

Each student must be assigned to at least one of these faculties,nbsp; but can belong to two or three faculties at the same time.

Hints:

The exercise belongs to the chapter Set Theory Basics.
The topic of this chapter is illustrated with examples in the (German language) learning video

Mengentheoretische Begriffe und Gesetzmäßigkeiten $\Rightarrow$ "Set Theoretical Concepts and Laws".

The areas in the above diagram are to scale,&nbsp, so you can easily give the (percentage) occupancy figures using the numerical values given and simple geometric considerations.

Questions

$N_{\rm T} \ = \ $

	$I$ is a subset of $M$.
	$W$ is a subset of $B$.
	$W$ and $M$ together form a "complete system".
	$B$, $I$ and $T$ together form a "complete system".
	$W$ and $T$ are disjoint sets.
	The union of $B$, $I$ and $T$ gives the universal set $G$.
	The intersection of $B$, $I$ and $T$ gives the empty set $\phi$.

$\text{Pr}\big[\text{female IT student}\big] \ = \ $

$\ \%$

$\text{Pr}\big[\text{one field of study}\big] \ = \ $

$\ \%$

$\text{Pr}\big[\text{three fields of study}\big] \ = \ $

$\ \%$

$\text{Pr}\big[\text{two fields of study}\big] \ = \ $

$\ \%$

Solution

(1) From simple geometric considerations, we arrive at the results:

$${\rm Pr}(B) = 3/4 \cdot 1 = 3/4\hspace{0.3cm}(\text{absolute:}\ 720),$$

$${\rm Pr}(I) = {1}/{2}\cdot 1\cdot 1 = 1/2\hspace{0.3cm}(\text{absolute:} \ 480),$$

$${\rm Pr}(T) = {1}/{2} \cdot {3}/{4} \cdot {3}/{4} = {9}/{32} \hspace{0.3cm}(\text{absolute:}\ 270)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}N_{\rm T} \;\underline{= 270}.$$

(2) Proposed solutions 2, 3, 5 and 6 are correct ⇒ proposed solutions 1, 4, 7 are consequently incorrect:

There are also female IT students, although very few.
The union of $B$, $I$ and $T$ gives the universal set, but not a complete system (not all combinations of $B$, $I$ and $T$ are disjoint).
For the same reason, the intersection of $B$, $I$ and $T$ does not yield the empty set.

Geometric solution of a probability problem

(3) In set theory, an IT student is the intersection of $I$ and $W$
(shown as a shaded area in the upper left of the graph):

$$\text{Pr[female IT student] = Pr}(I \cap W) = {1}/{2}\cdot {1}/{4} \cdot {1}/{4} = {1}/{32} \hspace{0.15cm}\underline { \thickapprox 3.13 \%}.$$

In words, there are $30$ female IT students among the $960$ students.

(4) The probability can be calculated as the sum of three individual probabilities:

$$ \text{Pr[one field of study] = Pr}( \overline{B} \cap \overline{I} \cap T) + {\rm Pr}( \overline{B} \cap I \cap \overline{T}) + {\rm Pr}( \it B \cap \overline{I} \cap \overline{T}).$$

Each individual probability corresponds to an area in the Venn diagram and can be determined by addition or subtraction of triangles or rectangles (see graph):

$$p_1 = {\rm Pr}( \overline{B} \cap \overline{I} \cap T) = {\rm Triangle\ (ABC)}= \frac{1}{2}\hspace{0.02cm} \cdot \hspace{0.02cm}\frac{1}{4}\hspace{0.02cm}\cdot \hspace{0.02cm}\frac{1}{4}= \frac{1}{32}\hspace{0.1cm}\underline{\approx 0.0313},$$

$$p_2 ={\rm Pr}( \overline{B} \cap I \cap \overline{T}) = {\rm Rectangle\hspace{0.1cm}(DEFG)}= \frac{1}{4}\hspace{0.02cm}\cdot \hspace{0.02cm} \frac{1}{4}\hspace{0.02cm}+ \hspace{0.02cm}\frac{1}{2}\hspace{0.02cm}\cdot \hspace{0.02cm} \frac{1}{4}\hspace{0.02cm}\cdot \hspace{0.02cm} \frac{1}{4} = \frac{3}{32}\hspace{0.1cm}\underline{\approx 0.0938},$$

$$p_3 = {\rm Pr}( B \cap \overline{I} \cap \overline{T}) ={\rm Rectangle\hspace{0.1cm}(HIJK)}= {\rm Triangle\hspace{0.1cm}(HLK)}- {\rm Triangle\hspace{0.1cm}(ILJ)} = \frac{1}{2}\hspace{0.02cm} \cdot \hspace{0.02cm} \frac{5}{4}\hspace{0.02cm} \cdot \hspace{0.02cm} \frac{5}{8}\hspace{0.02cm} - \hspace{0.02cm}\frac{1}{2}\hspace{0.02cm} \cdot \hspace{0.02cm} \frac{1}{4} \cdot \frac{1}{4} = \frac{23}{64}\hspace{0.1cm}\underline{\approx 0.3594}.$$

$\text{Or:}\hspace{0.3cm}$

$$p_3 = {\rm Triangle\hspace{0.1cm}(HIC)}- {\rm Triangle\hspace{0.1cm}(KJC)} ={1}/{2}\hspace{0.1cm} \cdot \hspace{0.1cm} 1 \hspace{0.1cm} \cdot \hspace{0.1cm} 1 \hspace{0.1cm} - \hspace{0.1cm}{1}/{2}\hspace{0.05cm} \cdot \hspace{0.1cm} {3}/{4} \cdot {3}/{8} = {23}/{64}.$$

The sum of these three probabilities leads to the final result $ \text{Pr[one field of study] } = 31/64 \;\underline {\approx 48.43 \%}$.

(5) This probability is expressed by the triangle $\text{Triangle(AGK)}$ . This has the area

$$\rm Pr[three fields of study] = {1}/{2}\cdot {1}/{4}\cdot {1}/{8} = {1}/{64}\hspace{0.15cm}\underline{\approx 1.56 \%}.$$

(6) The three events

"only one field of study",
"two fields of study" and
"three fields of study"

form a complete system. Thus, using the results of the last subtasks, we obtain:

$$\rm Pr[two fields of study] = 1- \text{Pr[one field of study] } - \rm Pr[three fields of study]= 1- {31}/{64} - {1}/{64} \hspace{0.15cm}\underline{= 50\%}.$$

One would arrive at exactly the same result – but with considerably more effort – in the direct way accordingly:

$${\rm Pr[two fields of study] = Pr}(B\cap I \cap\overline{T}) + {\rm Pr}(B\cap\overline{I}\cap{T}) + {\rm Pr}(\overline{B}\cap I \cap T).$$