Probability of no two of the marbles drawn have same color A bag contains 10 blue marbles ,20 black marbles , 30 red marbles.A marble is drawn from the bag, its colour 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?
If 3 repeated events are independent then $\frac {10}{60}\times\frac {20}{60}\times\frac {30}{60}=\frac 1{36}$
But the answer is $\frac 1 6$. How do you get this?
 A: You have 3 colors, so you have $3!=6$ possible combinations of total outcomes.
Your calculation was almost correct, but you forgot to multiply by this number.
Edit - to clear up confusion
The question is asking you to pull a marble 3 times, and each time it needs to be a different color. You can do that in these ($3!=6$) ways:


*

*Blue, Black, Red

*Blue, Red, Black

*Black, Blue, Red

*Black, Red, Blue

*Red, Blue, Black

*Red, Black, Blue


As you stated, the probabilities of drawing marbles are:


*

*Blue = $\frac {10}{60}$

*Black = $\frac {20}{60}$

*Red= $\frac {30}{60}$


So let's look at order #1, Blue, Black, Red. You have a $\frac {10}{60}\times\frac {20}{60}\times\frac {30}{60}=\frac 1{36}$ chance of this happening. But since each of the 6 options above include the same variables, each of these options has the same probability of occuring. So you have $6$ events with $\frac 1 {36}$ probability of occurring. That means that the probability of any of these outcomes is $6\times\frac 1 {36}=\frac 1 6$
A: First, you may pick blue-black-red. Arrangement of this colours is 3!. Total probability is Your answer * 3!. i.e 1/6. 
