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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 87 students in the school. There are 34 in the Spanish class, 31 in the French class, and 19 in the German class. There are 14 students that in both Spanish and French, 5 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 two students are chosen randomly, what is the probability that neither of them are taking Spanish?

My thought process was to first find the probability that the first and second student was taking Spanish, which would be 34/87 * 33/86. We have to find the complement of this, which would be 1-(187/1247) = 1060/1247. Is this correct? If not, can someone tell me how to go about the problem?

papayaaa
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  • Neither Spanish: $\frac{53}{87}\cdot \frac{52}{86}\approx \large{36.8%}$ Comprehensible? – callculus42 Aug 31 '23 at 03:25
  • In my opinion, you skipped a step. My standard Step 1, in a problem like this, would be to create a table with 8 rows, which represent the 8 possibilities for each of the students. That is, for each student, and each language, they either are or are not taking the corresponding class. See this response for an illustration. – user2661923 Aug 31 '23 at 05:45

2 Answers2

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Taking the complement isn't the right move, since there are scenarios in the complement in which one of the two students is taking Spanish while the other isn't (since we are taking the complement of both students taking Spanish).

Instead, I would take the complement of the first student not taking Spanish (1-34/87=53/87) and the complement of the second student not taking Spanish (1-33/86=53/86). Multiplying those two gives you the specific case for which both students don't take Spanish : 53/87*52/86=1378/3741≈36,84%. As you can see, it's a much smaller probability.

I'm not sure how all the rest of the superfluous information in the problem comes into play. Why are we told the number of students who take two classes? Why is it that, when adding the number of students in each class, we get 84 instead of 87? Maybe there is something that I'm missing, but this is what I got.

Etienne
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  • I thought the same thing when it came to the 84 vs 87 students then I realized there are students not taking any of these classes. The problem that papayaaa is asking might be one of many problems that use the same information. There could be other questions which requires the inclusion-exclusion principal. Then you need all of the information above. – quantus14 Aug 31 '23 at 03:29
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Truth Table Approach

$\underline{\text{Variables}}$

T : Taking Course = True, F : Taking Course = False.

\begin{array}{| l | l | l | l| } \hline \text{Spanish} & \text{French} & \text{German} & \text{Variable} \\ \hline \text{T} & \text{T} & \text{T} & x_1 \\ \hline \text{T} & \text{T} & \text{F} & x_2 \\ \hline \text{T} & \text{F} & \text{T} & x_3 \\ \hline \text{T} & \text{F} & \text{F} & x_4 \\ \hline \text{F} & \text{T} & \text{T} & x_5 \\ \hline \text{F} & \text{T} & \text{F} & x_6 \\ \hline \text{F} & \text{F} & \text{T} & x_7 \\ \hline \text{F} & \text{F} & \text{F} & x_8 \\ \hline \end{array}

$\underline{\text{Equations}}$

            Variables 
Equations : x1 x2 x3 x4 x5 x6 x7 x8 
     EQ-1 :  1  1  1  1  1  1  1  1 = 87
     EQ-2 :  1  1  1  1             = 34
     EQ-3 :  1  1        1  1       = 31  
     EQ-4 :  1     1     1     1    = 19
     EQ-5 :  1  1                   = 14
     EQ-6 :  1     1                =  5
     EQ-7 :  1           1          =  6
     EQ-8 :  1                      =  2

$\underline{\text{Analysis}}$

The desired computation is

$$\frac{x_5 + x_6 + x_7 + x_8}{x_1 + \cdots + x_8} \times \frac{x_5 + x_6 + x_7 + x_8 - 1}{x_1 + \cdots + x_8 - 1}. \tag1 $$

In (1) above, since the two denominators are known to be $~87~$ and $~86~$ respectively, the problem reduces to determining $~x_5, ~x_6, ~x_7, ~$ and $~x_8.$

Using EQ-5 through EQ-8 :

Known Values :  x1  x2  x3   x4  x5  x6  x7  x8   
                 2  12   3        4

Then, using EQ-2 through EQ-4 :

Known Values :  x1  x2  x3   x4  x5  x6  x7  x8   
                 2  12   3   17   4  13  10

Then, using EQ-1 :

Known Values :  x1  x2  x3   x4  x5  x6  x7  x8   
                 2  12   3   17   4  13  10  26

Consequently, the computation in (1) above is represented by

$$\frac{4 + 13 + 10 + 26}{87} \times \frac{4 + 13 + 10 + 26 - 1}{87 - 1}.$$

$\underline{\text{Addendum}}$

As indicated in the answer of Etienne, much of the work shown in my response isn't necessary to solve the specific problem. That is, since

  • $x_1 + \cdots + x_8 = 87$

  • $x_1 + \cdots + x_4 = 34,$

you can infer that $~x_5 + x_6 + x_7 + x_8 = (87 - 34),~$ without bothering to compute the values of the individual variables $~x_5, ~x_6, ~x_7, ~$ and $~x_8.$

However, this answer is (also) intended as a guide to solving other, similar, more complicated problems.

user2661923
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