An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. The

classes are open to any of the 100 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 16 in the German class. There are 12 students who are in both Spanish and French, 4 who are in both Spanish and German, and 6 who are in both French and German. In addition, there are 2 students taking all 3 classes.

(a) If a student is chosen randomly, what is the probability that he or she is not in any of the language classes?

(b) If a student is chosen randomly, what is the probability that he or she is taking exactly one language class?

(c) If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

Students in Spanish class$=P\left(A\right)=0.28$

Students in French class$=P\left(B\right)=0.26$

Students in German class$=P\left(C\right)=0.16$

Students who are in both Spanish and French class role="math" localid="1649346042059" $=P(A\cap B)=0.12$

Students who are in both Spanish and German class $=P(A\cap C)=0.04$

Students who are in both French and German class $=P(B\cap C)=0.06$

Students taking all three classes $=P(A\cap B\cap C)=0.02$

The probability that the chosen student is not in any of the language classes $P{\left(A\cup B\cup C\right)}^c$.

role="math" localid="1649346573978" $P{\left(A\cup B\cup C\right)}^c=1-P\left(A\cup B\cup C\right)$

role="math" localid="1649346486028" $$\begin{gathered} P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(A\cap C\right)+P{\left(A\cup B\cup C\right)}^c \\ =0.28+0.26+0.16-0.12-0.04-0.06+0.02 \\ =0.5 \end{gathered}$$

$$\begin{gathered} P{\left(A\cup B\cup C\right)}^c=1-0.5 \\ =0.5 \end{gathered}$$

Therefore, the probability that the chosen student is not in any of the language classes is$0.5$.

The probability that the chosen student is taking exactly one language class $=P\left(A\cap B^c\cap C^c\right)+P\left(A^c\cap B\cap C^c\right)+P\left(A^c\cap B^c\cap C\right)$

$$\begin{gathered} P\left(A\cap B^c\cap C^c\right)=P\left(A\right)-\left[P\left(A\cap B\right)+P\left(A\cap C\right)-P\left(A\cap B\cap C\right)\right] \\ =0.28-\left[0.12+0.04-0.02\right] \\ =0.14 \\ P\left(A^c\cap B\cap C^c\right)=P\left(B\right)-\left[P\left(A\cap B\right)+P\left(B\cap C\right)-P\left(A\cap B\cap C\right)\right] \\ =0.26-\left[0.12+0.06-0.02\right] \\ =0.10 \\ P\left(A^c\cap B^c\cap C\right)=P\left(C\right)-\left[P\left(A\cap C\right)+P\left(B\cap C\right)-P\left(A\cap B\cap C\right)\right] \\ =0.16-\left[0.04+0.06-0.02\right] \\ =0.08 \\ P\left(A\cap B^c\cap C^c\right)+P\left(A^c\cap B\cap C^c\right)+P\left(A^c\cap B^c\cap C\right)=0.32 \end{gathered}$$

Therefore, the probability that the chosen student is taking exactly one language class isrole="math" localid="1649347106423" $0.32$.

The probability that the chosen student is not in any of the language classes is $0.5$.

Total no. of students in class $=100$.

No. of students not in any of the language classes $=100\times 0.5=50$

Probability that the two students chosen randomly are not taking any random class is

$$\begin{gathered} =\frac{\left(\begin{array}{c}50\\ 2\end{array}\right)}{\left(\begin{array}{c}100\\ 2\end{array}\right)} \\ =\frac{50\times 49}{100\times 99} \\ =\frac{49}{198} \end{gathered}$$

Therefore, the probability that at least one of the chosen student is taking a language class is

$$\begin{gathered} =1-\frac{49}{198} \\ =0.7525 \end{gathered}$$