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Problem 99: Among the 16 cities that a professional society is considering for its next 3 annual conventions, 7 are in the western part of the United States. To avoid arguments, the selection is left to chance. If none of the cities can be chosen more than once, what are the probabilities that
a. none of the conventions will be held in the western part of the United States?
b. all of the conventions will be held in the western part of the United States?

Problem 99: Among the 16 cities that a professional society is considering for its next 3 annual conventions, 7 are in the western part of the United States. To avoid arguments, the selection is left to chance. If none of the cities can be chosen more than once, what are the probabilities that
a. none of the conventions will be held in the western part of the United States?
b. all of the conventions will be held in the western part of the United States?


Solution:


a. none of the conventions will be held in the western part of the United States?
Given that $x=0,n=3,a=7, N=16$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=0) = h(0;3,7,16)=\frac{\left(\begin{array}{c}7\\ 0\end{array}\right)\left(\begin{array}{c}16-7\\ 3-0\end{array}\right)}{\left(\begin{array}{c}16\\ 3\end{array}\right)} = 0.15.$

b. all of the conventions will be held in the western part of the United States?
Given that $x=3,n=3,a=7, N=16$
Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$
$P(x=3) = h(3;3,7,16)=\frac{\left(\begin{array}{c}7\\ 3\end{array}\right)\left(\begin{array}{c}16-7\\ 3-3\end{array}\right)}{\left(\begin{array}{c}16\\ 3\end{array}\right)} = 0.0625.$

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