A number of five rabbits are randomly selected for a behavioral test out of a group of 10 rabbits. If 60% of the rabbits are male, what is the probability that it will take at least 4 selection to find a male? a. 2.25% b. 3.33% c. 5.66% d. 8.50%
+118609 Ok so there are 6 male and 4 female to choose from.
P(at least 4 males) = 5*(6/10)*(5/9)*(4/8)*(3/7)*(4/6) + (6/10)*(5/9)*(4/8)*(3/7)*(2/6)
$${\mathtt{5}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{10}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{5}}}{{\mathtt{9}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{6}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{10}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{5}}}{{\mathtt{9}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{6}}}}\right) = {\frac{{\mathtt{11}}}{{\mathtt{42}}}} = {\mathtt{0.261\: \!904\: \!761\: \!904\: \!761\: \!9}}$$
+118609 Ok so there are 6 male and 4 female to choose from.
P(at least 4 males) = 5*(6/10)*(5/9)*(4/8)*(3/7)*(4/6) + (6/10)*(5/9)*(4/8)*(3/7)*(2/6)
$${\mathtt{5}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{10}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{5}}}{{\mathtt{9}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{6}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{10}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{5}}}{{\mathtt{9}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{6}}}}\right) = {\frac{{\mathtt{11}}}{{\mathtt{42}}}} = {\mathtt{0.261\: \!904\: \!761\: \!904\: \!761\: \!9}}$$
