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Problem 103: Among the 13 countries that an international trade federation is considering for their next 4 annual conferences, 6 are in Asia. To avoid arguments, the selection is left to chance. If none of the countries can be selected more than once, what are the probabilities that
(a) all the conferences will be held in Asia?
(b) none of the conferences will be held in Asia?

Problem 103: Among the 13 countries that an international trade federation is considering for their next 4 annual conferences, 6 are in Asia. To avoid arguments, the selection is left to chance. If none of the countries can be selected more than once, what are the probabilities that
(a) all the conferences will be held in Asia?
(b) none of the conferences will be held in Asia?


Solution:


a. all the conferences will be held in Asia
Given that $x=4,n=4,a=6, N=13$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=4) = h(4;4,6,13)=\frac{\left(\begin{array}{c}6\\ 4\end{array}\right)\left(\begin{array}{c}13-6\\ 4-4\end{array}\right)}{\left(\begin{array}{c}13\\ 4\end{array}\right)} = 0.020979.$

b. none of the conferences will be held in Asia
Given that $x=0,n=4,a=6, N=13$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=0) = h(0;4,6,13)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}13-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}13\\ 4\end{array}\right)} = 0.0489.$

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