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](https://tex.z-dn.net/?f=N%20%3D%2097%5C%5Cn%28S%29%3D33%5C%5Cn%28F%29%3D36%5C%5Cn%28G%29%3D17%5C%5Cn%28S%5Ccap%20F%29%3D13%5C%5Cn%28S%5Ccap%20G%29%3D4%5C%5Cn%28F%5Ccap%20G%29%3D6%5C%5Cn%28S%5Ccap%20F%5Ccap%20G%29%3D2)
(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](https://tex.z-dn.net/?f=P%28At%5C%20least%5C%201%5C%20class%29%3D1-P%28%28S%5Ccup%20F%5Ccup%20G%29%5E%7Bc%7D%29%5C%5C%3D1-%5B1-P%28S%5Ccup%20F%5Ccup%20G%29%5D%5C%5C%3DP%28S%5Ccup%20F%5Ccup%20G%29%5C%5C%3DP%28S%29%2BP%28F%29%2BP%28G%29-P%28S%5Ccap%20F%29-P%28S%5Ccap%20G%29-P%28F%5Ccap%20G%29%20%2B%20P%28S%5Ccap%20F%5Ccap%20G%29%5C%5C%3D%5Cfrac%7B33%7D%7B97%7D%20%2B%5Cfrac%7B36%7D%7B97%7D%20%2B%5Cfrac%7B17%7D%7B97%7D%20-%5Cfrac%7B13%7D%7B97%7D%20-%5Cfrac%7B4%7D%7B97%7D%20-%5Cfrac%7B6%7D%7B97%7D%20%2B%5Cfrac%7B2%7D%7B97%7D%20%5C%5C%3D%5Cfrac%7B33%2B36%2B17-13-4-6%2B2%7D%7B97%7D%20%5C%5C%3D%5Cfrac%7B65%7D%7B97%7D%20%5C%5C%3D0.6701)
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](https://tex.z-dn.net/?f=%7B97%5Cchoose%202%7D%3D%5Cfrac%7B97%21%7D%7B2%21%2897-2%29%21%7D%20%3D%5Cfrac%7B97%21%7D%7B2%21%5Ctimes95%21%7D%20%3D%204656)
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](https://tex.z-dn.net/?f=%3D0.6701%5Ctimes97%5C%5C%3D64.9997%5C%5C%5Capprox65)
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](https://tex.z-dn.net/?f=%7B65%5Cchoose%202%7D%3D%5Cfrac%7B65%21%7D%7B2%21%2865-2%29%21%7D%20%3D%5Cfrac%7B65%21%7D%7B2%21%5C63%21%7D%20%3D%202080)
Then the number of students who does not takes any of the 3 classes is
![=97-65\\=32](https://tex.z-dn.net/?f=%3D97-65%5C%5C%3D32)
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](https://tex.z-dn.net/?f=%3D65%5Ctimes32%5C%5C%3D2080)
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](https://tex.z-dn.net/?f=P%20%28At%5C%20least%5C%201%5C%20of%5C%202%5C%20takes%5C%20the%5C%20classes%29%3D%5Cfrac%7B2080%2B2080%7D%7B4656%7D%20%3D0.8935)
Thus, the probability that at least one of the two students is taking a language class is 0.8935.