Question

In a class of 78 students 41 are offering French, 22 are offering German. Of the students offering French or German, 9 are offering both courses. What is the probability that none of the students are enrolled in either course?

Answer 1

$\text{Let A and B be the set of students offering French}$

$\text{and German respectively}$

$\textbf{Given:}$

$n(A)=41$

$n(B)=22$

$n(A{\cap}B)=9$

$\textbf{To find:}$

$\text{Probability of none of the students enrolled in either course}$

$\textbf{Solution:}$

$\text{Using the formula}$

$\boxed{\bf\,n(A{\cup}B)=n(A)+n(B)-n(A{\cap}B)}$

$n(A{\cup}B)=41+22-9$

$n(A{\cup}B)=73-9$

$n(A{\cup}B)=64$

$\text{Now,}$

$P(\text{None of students enrolled in either course})$

$=P((A{\cup}B)')$

$=1-P(A{\cup}B)$

$=1-\dfrac{n(A{\cup}B)}{n(S)}$

$=1-\dfrac{64}{78}$

$=1-\dfrac{32}{39}$

$=\dfrac{39-32}{39}$

$=\dfrac{7}{39}$

$\therefore\textbf{Probability that none of the students are}$

$\textbf{enrolled in either course is \bf\dfrac{7}{39} }$

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