Question Number 134623 by bobhans last updated on 05/Mar/21
![PRobability 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](https://www.tinkutara.com/question/Q134623.png)
$$\mathscr{PR}\mathrm{obability} \\ $$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
![Total of balls = 3+2+1 = 6 The number of ways of selecting 3 balls from these 6 balls = ((6),(3) ) = 20 The number of ways of selecting 1 ball of each color = ((3),(1) )× ((2),(1) )× ((1),(1) ) = 6 Therefore the probability of selecting 3 balls of different color =p(A)=(6/(20)) = (3/(10)) so the probability that least 1 color is not drawn is p(A^c ) =1−p(A)=(7/(10))](https://www.tinkutara.com/question/Q134624.png)
$$\mathrm{Total}\:\mathrm{of}\:\mathrm{balls}\:=\:\mathrm{3}+\mathrm{2}+\mathrm{1}\:=\:\mathrm{6}\: \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{selecting}\:\mathrm{3}\:\mathrm{balls}\:\mathrm{from} \\ $$$$\mathrm{these}\:\mathrm{6}\:\mathrm{balls}\:=\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:=\:\mathrm{20} \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{selecting}\:\mathrm{1}\:\mathrm{ball}\:\mathrm{of} \\ $$$$\mathrm{each}\:\mathrm{color}\:=\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\:=\:\mathrm{6} \\ $$$$\mathrm{Therefore}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{selecting}\:\mathrm{3}\:\mathrm{balls} \\ $$$$\mathrm{of}\:\mathrm{different}\:\mathrm{color}\:=\mathrm{p}\left(\mathrm{A}\right)=\frac{\mathrm{6}}{\mathrm{20}}\:=\:\frac{\mathrm{3}}{\mathrm{10}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{least}\:\mathrm{1}\:\mathrm{color}\: \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{drawn}\:\mathrm{is}\:\mathrm{p}\left(\mathrm{A}^{\mathrm{c}} \right)\:=\mathrm{1}−\mathrm{p}\left(\mathrm{A}\right)=\frac{\mathrm{7}}{\mathrm{10}} \\ $$