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Six balls are to be randomly chosen from an urn containing 8 red, 10 green, and 12 blue balls.

What is the probability at least one red ball, one blue and one green ball is chosen?

3 Answers

Best answer
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Method 1

The probability of R, G + Probability of R, B + Probability of B, G

Above equation covers All red, All blue and All Green twice so we subtract it once, Thus we get:

The probability of R, G + Probability of R, B + Probability of B, G  - ALL R - ALL G - ALL B

1-$\frac{C(18,6)+C(20,6)+C(22,6)-C(8,6)-C(10,6)-C(12,6)}{C(30,6)}$=0.7797

 

1- ( all red + all green + all blue + red,green + red,blue+ green,blue) this is incorrect. It has several repetition

 

Method 2

let <R, G, B> be a triplet

now we first take One R, One G, One B now we have 3 balls left to pick

<3,0,0>, <2,0,1>, <2,1,0>,<1,2,0>,<1,0,2>,<1,1,1>,<0,2,1>,<0,1,2>,<0,3,0>,<0,0,3>

So we have 10 triplets which are same if we solve generating function

Xr+Xg+Xb=6

($x+x^{2}+...+x^{8}$)*($x+x^{2}+...+x^{10}$)*($x+x^{2}+...+x^{12}$)=6

($1+x^{1}+...+x^{7}$)*($1+x^{1}+...+x^{9}$)($1+x^{1}+...+x^{11}$)=3

$\left ( \frac{1-x^{8}}{1-x} \right )$*$\left ( \frac{1-x^{10}}{1-x} \right )$*$\left ( \frac{1-x^{12}}{1-x} \right )$=3

$\left ( 1-x^{8} \right )$*$\left ( 1-x^{10} \right )$*$\left ( 1-x^{12} \right )$*$\left ( 1-x \right )^{-3}$=3

$\rightarrow$C(5,3)=10

Initially, we took 1 R, 1 G, 1 B adding it to triplet we get

<4,1,1>,<3,1,2>,<3,2,1>,<2,3,1>,<2,1,3>,<2,2,2>,<1,3,2>,<1,2,3>,<1,4,1>,<1,1,4>

so total favorable case is 4,1,1=C(8,4)*C(10,1)*C(12,1)=8400
3,1,2=36960
3,2,1=30240
2,3,1=40320
2,1,3=61600
2,2,2=83160
1,3,2=63360
1,2,3=79200
1,4,1=20160
1,1,4=39600

=463000 combination in which we have at least 1 R, 1 G, 1 B

Probability=$\frac{463000}{C(30,6)}$=0.7797
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atleast 1‘ red ball, ‘atleast 1‘ green ball, ‘atleast 1‘ blue ball
atleast 1‘ here strongly suggests to use $\text{‘total – no red ball’, ‘total – no green ball’, ‘total – no blue ball’}.$

But then $\text{‘no red ball’}$, $\text{‘no green ball’}$, $\text{‘no blue ball’}$ will have common elements among them. 

∴ We'll be using the concepts of Set Theory.

Let $\text{A = selections such that no red ball is selected,} $

$\text{B = selections such that no green ball is selected, }$

$\text{C = selections such that no blue ball is selected}$

$\color{blue}{p(A \cup B \cup C) = p(A) + p(B) + p(C) – p(A \cap B) – p(B \cap C) – p(A \cap C) + p(A \cap B \cap C)} $

$\qquad\qquad \qquad =\dfrac{^{22}C_{6}\; +\;^{20}C_{6}\; +\;^{18}C_{6}\; – \;^{12}C_{6}\; – \;^{8}C_{6}\; – \;^{10}C_{6}\; + \;0}{^{30}C_{6}}$

$\qquad\qquad \qquad =0.220243358$

$∴\color{Green}{\text{Required Probability} = 1- 0.220243358}$

$\qquad\qquad \qquad\qquad\qquad = \color{orange}{0.779756642}$ 

edited by
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We need to find

" the probability at least one red ball, one blue and one green ball is chosen "

So, among 6 balls we need to select 1 red,1 blue and 1 green ball

Now how many ways we can select? 6 ways

i.e.(R,G,B),(R,B,G) , (B,G,R),(B,R,G) , (G,B,R),(G,R,B)

Now among remaining 3 balls we need select 3 balls

How many ways we can select?27 ways

like this

(R,R,R)-1 ways

(R,G,G) =3!/2!=3 ways

(R,R,G)=3 ways

(R,R,B)=3 ways

(R,B,B)=3 ways

(B,B,B)=1 ways

(B,B,G)=3 ways

(B,G,G)=3 ways

(G,G,G)=1 ways

and (R,G,B),(R,B,G) , (B,G,R),(B,R,G) , (G,B,R),(G,R,B)

Now, total probability will be =$\frac{6\times 27}{\binom{30}{6}}$=$\frac{18}{5075}$

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