MHB -aux07.Venn diagram -about 120 students in a school

[karush]

(a) since $n(U)=120$ and the intersections showing $8,7$ and $4$ added is $19$ then $\displaystyle\frac{19}{120}$ is the probability of studying exactly $2$ languages

(b) since $n(J)=35$ then subtracting $8,5,7$ would be $15$ so $\displaystyle\frac{15}{120}=\frac{1}{8}$ would be probability of studying only Japanese

(c) $\displaystyle\frac{n(C)\ \cup\ n(J)\cup \ n(S)}{n(U)}=
\frac{19}{30}$

 

[Evgeny.Makarov]

(a) and (b) are correct. As for (c), you need to subtract your answer from 1 because the question asks for the probability that the student does not study any of these languages. The expression $n(C)\cup n(J)$ is a type error because $n(C)$ and $n(J)$ are numbers, while $\cup$ acts on sets. I agree that $n(C\cup J\cup S)=19\cdot4=76$, but it is strange that you did not explain this more difficult part while you explained the easier ones.

 

[karush]

(a) and (b) are correct. As for (c), you need to subtract your answer from 1 because the question asks for the probability that the student does not study any of these languages. The expression $n(C)\cup n(J)$ is a type error because $n(C)$ and $n(J)$ are numbers, while $\cup$ acts on sets. I agree that $n(C\cup J\cup S)=19\cdot4=76$, but it is strange that you did not explain this more difficult part while you explained the easier ones.

I probably misunderstood the notation. so $\frac{19}{30}$ is the probability that a student will study a language. and $\frac{11}{30}$ is the probability that a student does not study a language.

so saying $14\cup 16$ is improper it has to a $A\cup B$ etc

 

[Evgeny.Makarov]

I probably misunderstood the notation. so $\frac{19}{30}$ is the probability that a student will study a language. and $\frac{11}{30}$ is the probability that a student does not study a language.

Yes.

so saying $14\cup 16$ is improper it has to a $A\cup B$ etc

Yes.