Respuesta :
Answer:
0.5 = 50% probability that he or she is not in any of the language classes.
Step-by-step explanation:
We treat the number of students in each class as Venn sets.
I am going to say that:
Set A: Spanish class
Set B: French class
Set C: German class
We start building these sets from the intersection of the three.
In addition, there are 2 students taking all 3 classes.
This means that:
[tex](A \cap B \cap C) = 2[/tex]
6 that are in both French and German
This means that:
[tex](B \cap C) + (A \cap B \cap C) = 6[/tex]
So
[tex](B \cap C) = 4[/tex]
4 French and German, but not Spanish.
4 that are in both Spanish and German
This means that:
[tex](A \cap C) + (A \cap B \cap C) = 4[/tex]
So
[tex](A \cap C) = 2[/tex]
2 Spanish and German, but not French
12 students that are in both Spanish and French
This means that:
[tex](A \cap B) + (A \cap B \cap C) = 12[/tex]
So
[tex](A \cap B) = 10[/tex]
10 Spanish and French, but not German
16 in the German class.
This means that:
[tex](C - B - A) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 16[/tex]
[tex](C - B - A) + 2 + 4 + 2 = 16[/tex]
[tex](C - B - A) = 8[/tex]
8 in only German.
26 in the French class
[tex](B - C - A) + (A \cap B) + (B \cap C) + (A \cap B \cap C) = 26[/tex]
[tex](B - C - A) + 10 + 4 + 2 = 26[/tex]
[tex](B - C - A) = 10[/tex]
10 only French
28 students in the Spanish class
[tex](A - B - C) + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 16[/tex]
[tex](A - B - C) + 10 + 2 + 2 = 28[/tex]
[tex](A - B - C) = 14[/tex]
14 only Spanish
At least one of them:
The sum of all the above values. So
[tex](A \cup B \cup B) = 14 + 10 + 8 + 10 + 2 + 4 + 2 = 50[/tex]
None of them:
100 total students, so:
[tex]100 - (A \cup B \cup B) = 100 - 50 = 50[/tex]
(a) If a student is chosen randomly, what is the probability that he or she is not in any of the language classes?
50 out of 100. So
50/100 = 0.5 = 50% probability that he or she is not in any of the language classes.
