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There-are-20-white-and-10-black-balls-in-the-box-In-turn-draw-4-balls-and-each-draw-ball-is-returned-to-the-box-before-the-next-ball-is-drawn-and-the-balls-in-the-box-are-mixed-What-is-the-probabilit




Question Number 108377 by 1549442205PVT last updated on 16/Aug/20
There are 20 white and 10 black balls  in the box.In turn draw 4 balls and  each draw ball is returned to the box  before the next ball is drawn and the   balls in the box are mixed.What is the  probability that out of four balls  taken out there will be two white balls?
$$\mathrm{There}\:\mathrm{are}\:\mathrm{20}\:\mathrm{white}\:\mathrm{and}\:\mathrm{10}\:\mathrm{black}\:\mathrm{balls} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{box}.\mathrm{In}\:\mathrm{turn}\:\mathrm{draw}\:\mathrm{4}\:\mathrm{balls}\:\mathrm{and} \\ $$$$\mathrm{each}\:\mathrm{draw}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{returned}\:\mathrm{to}\:\mathrm{the}\:\mathrm{box} \\ $$$$\mathrm{before}\:\mathrm{the}\:\mathrm{next}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{balls}\:\mathrm{in}\:\mathrm{the}\:\mathrm{box}\:\mathrm{are}\:\mathrm{mixed}.\mathrm{What}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{out}\:\mathrm{of}\:\mathrm{four}\:\mathrm{balls} \\ $$$$\mathrm{taken}\:\mathrm{out}\:\mathrm{there}\:\mathrm{will}\:\mathrm{be}\:\mathrm{two}\:\mathrm{white}\:\mathrm{balls}? \\ $$
Answered by bobhans last updated on 16/Aug/20
       ((bobhans)/(⋰⋱))  by binomial distribution probability  p = ((20)/(30)) = (2/3) , q = 1−(2/3)=(1/3)  let X is randomly variable . we want compute  P(X=2) = C_( 2) ^( 4)  p^2  q^2  = ((4×3)/(2×1)). ((2/3))^2 .((1/3))^2                      = 6×(4/(81)) = (8/(27))
$$\:\:\:\:\:\:\:\frac{\boldsymbol{{bobhans}}}{\iddots\ddots} \\ $$$${by}\:{binomial}\:{distribution}\:{probability} \\ $$$${p}\:=\:\frac{\mathrm{20}}{\mathrm{30}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:,\:{q}\:=\:\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${let}\:{X}\:{is}\:{randomly}\:{variable}\:.\:{we}\:{want}\:{compute} \\ $$$${P}\left({X}=\mathrm{2}\right)\:=\:{C}_{\:\mathrm{2}} ^{\:\mathrm{4}} \:{p}^{\mathrm{2}} \:{q}^{\mathrm{2}} \:=\:\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}×\mathrm{1}}.\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} .\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}×\frac{\mathrm{4}}{\mathrm{81}}\:=\:\frac{\mathrm{8}}{\mathrm{27}} \\ $$
Commented by 1549442205PVT last updated on 17/Aug/20
Thank Sir.The result is very correct
$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{The}\:\mathrm{result}\:\mathrm{is}\:\mathrm{very}\:\mathrm{correct} \\ $$

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