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#### karush

##### Well-known member

- Jan 31, 2012

- 3,066

(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}$

(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}$

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