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Question Number 108377 by 1549442205PVT last updated on 16/Aug/20

$$\mathrm{There}\mathrm{are}20\mathrm{white}\mathrm{and}10\mathrm{black}\mathrm{balls}$$$$\mathrm{in}\mathrm{the}\mathrm{box}.\mathrm{In}\mathrm{turn}\mathrm{draw}4\mathrm{balls}\mathrm{and}$$$$\mathrm{each}\mathrm{draw}\mathrm{ball}\mathrm{is}\mathrm{returned}\mathrm{to}\mathrm{the}\mathrm{box}$$$$\mathrm{before}\mathrm{the}\mathrm{next}\mathrm{ball}\mathrm{is}\mathrm{drawn}\mathrm{and}\mathrm{the}$$$$\mathrm{balls}\mathrm{in}\mathrm{the}\mathrm{box}\mathrm{are}\mathrm{mixed}.\mathrm{What}\mathrm{is}\mathrm{the}$$$$\mathrm{probability}\mathrm{that}\mathrm{out}\mathrm{of}\mathrm{four}\mathrm{balls}$$$$\mathrm{taken}\mathrm{out}\mathrm{there}\mathrm{will}\mathrm{be}\mathrm{two}\mathrm{white}\mathrm{balls}?$$

Answered by bobhans last updated on 16/Aug/20

$$\frac{\mathit{b}\mathit{o}\mathit{b}\mathit{h}\mathit{a}\mathit{n}\mathit{s}}{\text{iddots}\ddots }$$$$bybinomialdistributionprobability$$$$p=\frac{20}{30}=\frac{2}{3},q=1-\frac{2}{3}=\frac{1}{3}$$$$letXisrandomlyvariable.wewantcompute$$$$P(X=2)=C_2^4{p}^2{q}^2=\frac{4\times 3}{2\times 1}.{\left(\frac{2}{3}\right)}^2.{\left(\frac{1}{3}\right)}^2$$$$=6\times \frac{4}{81}=\frac{8}{27}$$

Commented by 1549442205PVT last updated on 17/Aug/20

$$\mathrm{Thank}\mathrm{Sir}.\mathrm{The}\mathrm{result}\mathrm{is}\mathrm{very}\mathrm{correct}$$

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