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Question Number 134623 by bobhans last updated on 05/Mar/21

$$\mathcal{PR}\mathrm{obability}$$An urn contains 3 red balls, 2 green balls and 1 yellow ball. Three balls are selected at random and without replacement from the urn. What is the probability that at least 1 color is not drawn?

by : Bobhans

Answered by EDWIN88 last updated on 05/Mar/21

$$\mathrm{Total}\mathrm{of}\mathrm{balls}=3+2+1=6$$$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{selecting}3\mathrm{balls}\mathrm{from}$$$$\mathrm{these}6\mathrm{balls}=\left(\begin{array}{c}6\\ 3\end{array}\right)=20$$$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{of}\mathrm{selecting}1\mathrm{ball}\mathrm{of}$$$$\mathrm{each}\mathrm{color}=\left(\begin{array}{c}3\\ 1\end{array}\right)\times \left(\begin{array}{c}2\\ 1\end{array}\right)\times \left(\begin{array}{c}1\\ 1\end{array}\right)=6$$$$\mathrm{Therefore}\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{selecting}3\mathrm{balls}$$$$\mathrm{of}\mathrm{different}\mathrm{color}=\mathrm{p}\left(\mathrm{A}\right)=\frac{6}{20}=\frac{3}{10}$$$$\mathrm{so}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{least}1\mathrm{color}$$$$\mathrm{is}\mathrm{not}\mathrm{drawn}\mathrm{is}\mathrm{p}\left({\mathrm{A}}^{\mathrm{c}}\right)=1-\mathrm{p}\left(\mathrm{A}\right)=\frac{7}{10}$$

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