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Question Number 113605 by bobhans last updated on 14/Sep/20

A bag contains 6 red, 5 white and 4 black balls  If twl balls are drawn , what is the probability  that none of them are red ?

$$\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{6}\:\mathrm{red},\:\mathrm{5}\:\mathrm{white}\:\mathrm{and}\:\mathrm{4}\:\mathrm{black}\:\mathrm{balls} \\ $$$$\mathrm{If}\:\mathrm{twl}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{drawn}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{none}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{red}\:? \\ $$$$ \\ $$

Commented by bemath last updated on 14/Sep/20

method_2  p(A)=(9/(15))×(8/(14)) =(3/5)×(4/7)=((12)/(35))

$${method\_}\mathrm{2} \\ $$$${p}\left({A}\right)=\frac{\mathrm{9}}{\mathrm{15}}×\frac{\mathrm{8}}{\mathrm{14}}\:=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{4}}{\mathrm{7}}=\frac{\mathrm{12}}{\mathrm{35}} \\ $$

Answered by bemath last updated on 14/Sep/20

method_1  n(S)= C_2 ^(15)  = ((15×14)/(2×1)) = 105  none of them are red→(2W),(2B),(1W,1B)  n(A)=C_2 ^5 +C_2 ^4 +C_1 ^5 ×C_1 ^4   n(A)= 10+6+20=36  so p(A) = ((36)/(105)) = ((12)/(35))

$${method\_}\mathrm{1} \\ $$$${n}\left({S}\right)=\:{C}_{\mathrm{2}} ^{\mathrm{15}} \:=\:\frac{\mathrm{15}×\mathrm{14}}{\mathrm{2}×\mathrm{1}}\:=\:\mathrm{105} \\ $$$${none}\:{of}\:{them}\:{are}\:{red}\rightarrow\left(\mathrm{2}{W}\right),\left(\mathrm{2}{B}\right),\left(\mathrm{1}{W},\mathrm{1}{B}\right) \\ $$$${n}\left({A}\right)={C}_{\mathrm{2}} ^{\mathrm{5}} +{C}_{\mathrm{2}} ^{\mathrm{4}} +{C}_{\mathrm{1}} ^{\mathrm{5}} ×{C}_{\mathrm{1}} ^{\mathrm{4}} \\ $$$${n}\left({A}\right)=\:\mathrm{10}+\mathrm{6}+\mathrm{20}=\mathrm{36} \\ $$$${so}\:{p}\left({A}\right)\:=\:\frac{\mathrm{36}}{\mathrm{105}}\:=\:\frac{\mathrm{12}}{\mathrm{35}} \\ $$

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