Question

There are 6 brown, 5 blue, and 2 orange marbles in a hat. What is the probability of picking two orange marbles in a row without returning the marbles back to the hat? (Write your answer as a fraction.)

Answer 1

You start out with (6 + 5 + 2) = 13 marbles in the hat.
You accidentally knock the hat off the table, they all roll out
onto the floor, and you have lost all of your marbles. 
You pick them up, put them back in the hat, put the hat back
on the table, and you still have 13 marbles in the hat.

The probability of pulling an orange marble without looking is

     (number of orange ones) / (total number of marbles) =  2/13 .

Now there are only 12 marbles left in the hat.  If the first draw
was successful, then there's only 1 orange marble left in there.
The probability of pulling that one without looking is

   (number of orange ones) / (total number of marbles) =  1/12 .

So the probability of BOTH events happening is

                ( 2/13 ) x ( 1/12 )  =  2/156  =  1/78 .    

That's the answer in fraction form, like the question wants it.

That fraction is roughly equal to  1.28 %  .  

Answer 2

Lets look at the information we have-

6 brown marbles
5 blue marbles 
2 orange marbles. 

Question:- 
What is the probability of picking two orange marbles in a row without returning the marbles back to the hat?
Solve:-

First add all the marbles:-

6 + 5 + 2 = 13

13 marbles in total. 

Now lets write the 1st probability:- 

2 orange
13 together
$\frac{2}{13}$

2nd probability:- 

13 - 1 = 12

1 orange (left)
12 total (without putting back)
$\frac{1}{12}$

Hope i helped ya!! xD