# Thread: Dividing students up.

1. ## Dividing students up.

Could someone double check my answers for me? They differ substantially from the text and I'm not sure why.

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 that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that 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?

the probability that someone is in a language class should just be the total number of students in each class over the total number of students,

$\frac{28+26+16}{100}=\frac{70}{100}$

The question wants the compliment of this, so there's a 30% chance of selecting a non-language student from the school.

The book says .5

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

This answer should just be the quotient, of the sum of each class, minus duplicates, over the total. So take for example the Spanish class, of the 28 total students, 12 are taking French too, 4 are taking German as well, and 2 are in all three classes; so there are only 10 students taking only Spanish.

$\frac{(28-12-4-2)+(26-12-6-2)+(16-6-4-2)}{100}=\frac{10+6+4}{100}=\frac{20}{100}$

The given answer is .32

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

So if you randomly select 2 students there are 4 outcomes only one of which involves both students not in a language class. I should just be able to subtract the probability of both students not being in a language class, in decimal forum, from 1 and have my answer?

$1-0.2^2=0.96$

Here the text gives a solution of 149/198

2. ## Re: Dividing students up.

Originally Posted by bkbowser
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 that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that 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?

the probability that someone is in a language class should just be the total number of students in each class over the total number of students The book says .5
$28+26+16-12-4-6+2=~?$

3. ## Re: Dividing students up.

I'm afraid that they have thrown in some confusion with their terminology. When they say there are 12 students taking both Spanish and French that number includes the 2 students taking all three languages. Thus the number of students taking Spanish and French but not German is 12-2 = 10.

There are 8 sets of distinct students:

All three languages: 2
Spanish and French only: 12-2 = 10
Spanish and German only: 4-2 = 2
French and German only: 6-2 = 4
Spanish only: 28-10-2-2 = 14
French only: 26-10-4-2 = 10
German only: 16-2-4-2 = 8
No language at all: 100 - 2-10-2-4-14-10-8 = 50

So part a is 50/100 = 0.5
Part b is (14+10+8)/100 = 0.32
Part c: remember that you choose two students without replacement, and so the probability of both being non-language students is 50/100 x 49/99 = 49/198, and the probability that at least one of them is a language student is 1 - 49/198 = 149/198.

4. ## Re: Dividing students up.

Originally Posted by ebaines
I'm afraid that they have thrown in some confusion with their terminology. When they say there are 12 students taking both Spanish and French that number includes the 2 students taking all three languages. Thus the number of students taking Spanish and French but not German is 12-2 = 10.

There are 8 sets of distinct students:

All three languages: 2
Spanish and French only: 12-2 = 10
Spanish and German only: 4-2 = 2
French and German only: 6-2 = 4
Spanish only: 28-10-2-2 = 14
French only: 26-10-4-2 = 10
German only: 16-2-4-2 = 8
No language at all: 100 - 2-10-2-4-14-10-8 = 50

So part a is 50/100 = 0.5
Part b is (14+10+8)/100 = 0.32
Part c: remember that you choose two students without replacement, and so the probability of both being non-language students is 50/100 x 49/99 = 49/198, and the probability that at least one of them is a language student is 1 - 49/198 = 149/198.
Ah, OK good I was afraid that I was doing something very wrong when I'm just double counting a bunch of people.

Thanks.