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A bag contain 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated 3 times. the probability that no two of the marbles drawn have the same color is____.

I considered different combinations of above scenario.

$ <R,B,G> \ <R,G,R>\\ <R,G,B> \ <R,B,R>\\ <B,R,G> \ <B,G,B>\\ <B,G,R> \ <B,R,B>\\ <G,R,B> \ <G,R,G>\\ <G,B,R> \ <G,B,G>\\ $

and my sample space will be = 3*3*3 -> each ball selection could be off 3 colors. so probability becomes = $\frac{12}{27}$ = $\frac{4}{9}$ but it is not the correct answer.

I know I haven't counted no. of the given balls So I though of this approach: = $\frac{12}{60_{C_3}}$

but no answer was still wrong. What should have been the correct way of solving it, and where I am making mistake?

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    $\begingroup$ In e.g. scenario RGR two marbles of the same color are drawn. $\endgroup$ – drhab Dec 30 '18 at 15:17
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    $\begingroup$ math.stackexchange.com/questions/1848392/… $\endgroup$ – kludg Dec 30 '18 at 15:25
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    $\begingroup$ Taking the total number of marbles into account was a good idea, but ${}^{60}C_3$ is the number of ways to draw three marbles without putting each marble back in the bag before drawing the next one. $\endgroup$ – David K Dec 30 '18 at 15:39
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Two issues:

  • Your left hand column has the six cases with "no two of the marbles drawn have the same color", but your right hand column does not: $R,B,G$ are all different but $R,G,R$ has two $R$s

  • The probability of drawing $R,B,G$ in that order is $\frac{30}{60} \times \frac{10}{60} \times \frac{20}{60} = \frac1{36}$. Each of the others in the left hand column have the same probability and adding these up gives $6 \times \frac1{36}= \frac16$, which I would expect to be the answer

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  • $\begingroup$ Thank you, I got it. $\endgroup$ – swapnil Dec 30 '18 at 15:31

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