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Question Number 158995 by Raxreedoroid last updated on 11/Nov/21

A population of 15 items. 4 red, 6 black and 5 green. a sample of 4 items are selected. what is the probability that the sample contains 3 black and 1 green?\\n

Commented bytounghoungko last updated on 14/Nov/21

p=((C_3 ^6 ×C_1 ^5 )/C_4 ^(15) ) = ((20×5)/((15 ^3 ×14 ^7 ×13×12)/(4×3×2×1)))   p= ((20)/(21×13))

$${p}=\frac{{C}_{\mathrm{3}} ^{\mathrm{6}} ×{C}_{\mathrm{1}} ^{\mathrm{5}} }{{C}_{\mathrm{4}} ^{\mathrm{15}} }\:=\:\frac{\mathrm{20}×\cancel{\mathrm{5}}}{\frac{\cancel{\mathrm{15}}\color{mathred}{\:}^{\mathrm{\color{mathred}{3}}} ×\cancel{\mathrm{14}}\:^{\mathrm{\color{mathred}{7}}} ×\mathrm{13}×\cancel{\mathrm{12}}}{\cancel{\mathrm{4}×\mathrm{3}}×\cancel{\mathrm{2}×\mathrm{1}}}}\: \\ $$ $${\color{mathred}{p}}\color{mathred}{=}\color{mathred}{\:}\frac{\mathrm{\color{mathred}{2}\color{mathred}{0}}}{\mathrm{\color{mathred}{2}\color{mathred}{1}}\color{mathred}{×}\mathrm{\color{mathred}{1}\color{mathred}{3}}} \\ $$

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