1
$\begingroup$

An elementary school is offering 3 language classes: one in Spanish, one in French, and one in German. These classes are open to any of the 76 students in the school. There are 28 in the Spanish class, 29 in the French class, and 15 in the German class. There are 11 students that in both Spanish and French, 4 are in both Spanish and German, and 6 are in both French and German. In addition, there are 2 students taking all 3 classes.

If one student is chosen randomly, what is the probability that he or she is taking at least one language class?

If two students are chosen randomly, what is the probability that both of them are taking French?

$\endgroup$
2
$\begingroup$

The best way to tackle the problem is to draw a Venn diagram. So draw three mutually intersecting ovals, and label the ovals French, German, and Spanish, or more neatly $F$, $G$, and $S$.

Look at the little region which is in common between $F$, $G$, and $S$. There are $2$ people in that region. Write a $2$ there.

There are $11$ people who are in both $F$ and $S$. Of these, $2$ are taking all three languages. So there are $9$ taking $F$ and $S$ but not $F$. Write a $9$ in the little region that represents $F$ and $S$ but not $G$. Do something similar for the $4-2=2$ people taking $S$ and $G$ but not all three. Also do the same for the $4$ who are in $F$ and $G$ but not all three.

Keep on working outwards. Now you should be able to fill in how many people are taking $F$ and nothing else. And so on.

Another way: Or else we can do the counting by formula. For any set $S$, let $|S|$ be the number of elements in $S$. We have the Inclusion/Exclusion formula $$|F\cup G\cup S|=|F|+|G|+|S|-|F\cap G|-|G\cap S|-|S\cap F|+|F\cap G\cap S|.$$

You have been told all the numbers on the right-hand side. So you now know the left-hand side, which is the number of people taking at least one language. If I am doing my additions/subtractions right, that number is $53$.

You can keep on using formulas, for example the simpler $|X\cup Y|=|X|+|Y|-|X\cap Y|$ to figure out whatever else you need. Use a picture (Venn diagram) to stay on the right track.

The probabilities: The probability that a student is taking at least one language is $\dfrac{53}{76}$. This is because we found that $53$ students are taking at least one language.

For the probability both students are taking French, we don't need any of the above calculations. Out of $76$ students, $29$ are taking French. Pick the students out one at a time, ask them whether they are taking French. The probability the first one says yes is $\dfrac{29}{76}$. Given the first said yes, the probability the second says yes is $\dfrac{28}{75}$. So the probability they both say yes is the product $\dfrac{29}{76}\cdot\dfrac{28}{75}$. One can simplify a bit.

$\endgroup$
0
$\begingroup$

Hint #1: There are in total $28+29+15=72$ students, who attend a language class.
Hint #2: There are $C_{76}^2$ ways of chosing 2 students from 76, and $C_{29}^2$ ways of chosing 2 students from 29.

Here $C_n^m=\frac{n!}{m!(n-m)!}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.