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?
Best Answer
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