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Question Number 135181 by liberty last updated on 11/Mar/21
$$\mathrm{A}\:\mathrm{bag}\:\mathrm{has}\:\mathrm{4}\:\mathrm{red}\:\mathrm{marbles},\:\mathrm{5}\:\mathrm{white}\: \\ $$$$\mathrm{marbles}\:,\:\mathrm{and}\:\mathrm{6}\:\mathrm{blue}\:\mathrm{marbles}.\:\mathrm{Three} \\ $$$$\mathrm{marbles}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bag},\:\left(\mathrm{without}\right. \\ $$$$\left.\mathrm{replacement}\right)\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{they}\:\mathrm{are}\:\mathrm{all}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color}\:?\: \\ $$
Answered by EDWIN88 last updated on 11/Mar/21
![p(A) = ((C_3 ^4 +C_3 ^5 +C_3 ^6 )/C_3 ^(15) ) = ((4+10+20)/( [((15.14.13)/(3.2.1))])) = ((34)/(5.7.13)) = ((34)/(35.13))=((34)/(455))](Q135189.png)
$$\mathrm{p}\left(\mathrm{A}\right)\:=\:\frac{\mathrm{C}_{\mathrm{3}} ^{\mathrm{4}} +\mathrm{C}_{\mathrm{3}} ^{\mathrm{5}} +\mathrm{C}_{\mathrm{3}} ^{\mathrm{6}} }{\mathrm{C}_{\mathrm{3}} ^{\mathrm{15}} }\:=\:\frac{\mathrm{4}+\mathrm{10}+\mathrm{20}}{\:\left[\frac{\mathrm{15}.\mathrm{14}.\mathrm{13}}{\mathrm{3}.\mathrm{2}.\mathrm{1}}\right]} \\ $$$$=\:\frac{\mathrm{34}}{\mathrm{5}.\mathrm{7}.\mathrm{13}}\:=\:\frac{\mathrm{34}}{\mathrm{35}.\mathrm{13}}=\frac{\mathrm{34}}{\mathrm{455}} \\ $$$$ \\ $$