# probability - a bag contains 10 blue marbles

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.

$$ \ \\ \ \\ \ \\ \ \\ \ \\ \ \\$$

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?

• In e.g. scenario RGR two marbles of the same color are drawn. – drhab Dec 30 '18 at 15:17
• math.stackexchange.com/questions/1848392/… – kludg Dec 30 '18 at 15:25
• 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. – David K Dec 30 '18 at 15:39

• 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