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 103 students in the school. There are 37 in the Spanish class, 38 in the French class, and 21 in the German class. There are 17 students that in both Spanish and French, 7 are in both Spanish and German, and 10 are in both French and German. In addition, there are 4 students taking all 3 classes.
Required:
a. If a student is chosen randomly, what is the probability that he or she is not in any of the language classes?
b. If a student is chosen randomly, what is the probability that he or she is taking exactly one language class?
c. If 2 students are chosen randomly, what is the probability that at least 1 is taking a language class?

Answer 1

Answer:

0.3689 ; 0.3592 ; 0.6301

Step-by-step explanation:

Spanish = S

French = F

German = G

n(S) = 37 ; n(F) = 38 ; n(G) = 21

n(SnF) = 17 ; n(SnG) = 7 ; n(FnG) = 10

n(SnFnG) = 3

n(SnF) only = 17 - 3 = 14 ;

n(SnG) only = 7 - 3 = 4 ;

n(FnG) only = 10 - 3 = 7

n(S) only = 37 - (14 + 3 + 4) = 16 ;

n(F) only = 38 - (14 + 3 + 7) = 14;

n(G) only = 21 - (7 + 3 + 4) = 7

Probability of not being in any language class :

Number of students not in any class (x)

(16+7+14+14+4+7+3+x) = 103

65 + x = 103

x = 103 - 65

x = 38

Probability, P = required outcome / Total possible outcomes

P(not in any class) = 38 / 103

= 0.3689

2.) probability of taking exactly one class :

Either Spanish or German or French

16/103 + 7/103 + 14/103 = 37/103 = 0.3592

3.) probability that atleast one of two students is taking a language class :

Probability that a student chosen is taking a language class :

Number of students taking a language class / total number of students

= 65 / 103

= 0.631

[P(takes class) * p(does not take class)] + [p(takes class) * p(takes class)]

(0.631 * 0.3689) + (0.631 * 0.631) = 0.6301