Question

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 97 students in the school. There are 33 in the Spanish class, 36 in the French class, and 17 in the German class. There are 13 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 at least one of them is taking a language class?

Answer 1

Answer:

(a) P (A student takes at least 1 class) = 0.6701

(b) P (At least one of the two is taking a class) = 0.8935

Step-by-step explanation:

Let S = a student takes a Spanish class, F = a student takes a French class and G = a student takes a German class.

Given:

$N = 97\\n(S)=33\\n(F)=36\\n(G)=17\\n(S\cap F)=13\\n(S\cap G)=4\\n(F\cap G)=6\\n(S\cap F\cap G)=2$

(a)

Compute the probability that a randomly selected student takes at least one language class as follows:

P (Student takes at least 1 class) = 1 - P (Student does not takes any class)

$P(At\ least\ 1\ class)=1-P((S\cup F\cup G)^{c})\\=1-[1-P(S\cup F\cup G)]\\=P(S\cup F\cup G)\\=P(S)+P(F)+P(G)-P(S\cap F)-P(S\cap G)-P(F\cap G) + P(S\cap F\cap G)\\=\frac{33}{97} +\frac{36}{97} +\frac{17}{97} -\frac{13}{97} -\frac{4}{97} -\frac{6}{97} +\frac{2}{97} \\=\frac{33+36+17-13-4-6+2}{97} \\=\frac{65}{97} \\=0.6701$

Thus, the probability that a randomly selected student takes at least one language class is 0.6701.

(b)

First determine the number of combinations of selecting 2 students from 97:

Number of ways of selecting 2 students from 97 = ${97\choose 2}=\frac{97!}{2!(97-2)!} =\frac{97!}{2!\times95!} = 4656$

Compute the total number of students taking any of the classes.

Number of students classes = P (Student takes at least 1 class) × 97

                                               $=0.6701\times97\\=64.9997\\\approx65$

The number of ways to select two students from those who takes the classes is:

Both students takes classes = ${65\choose 2}=\frac{65!}{2!(65-2)!} =\frac{65!}{2!\63!} = 2080$

Then the number of students who does not takes any of the 3 classes is

 $=97-65\\=32$

The number of ways to select one student who takes a class and one who does not is:

Only one student takes the class=

 $=65\times32\\=2080$

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

$P (At\ least\ 1\ of\ 2\ takes\ the\ classes)=\frac{2080+2080}{4656} =0.8935$

Thus, the probability that at least one of the two students is taking a language class is 0.8935.