In a game there is a paper bag containing money: 2 ten dollar bills, 3 five dollar bills and 4 one dollar bills. You reach into the bag and pull out two bills.
List the amount of money you could have in your hand from low to high. 2,6,10,11,15,20
List the probabilities of these events in the same order (implified fraction form)
+591 2:4x3 possible ways so 12/72=1/6
6:4x3 possible ways so 12/72=1/6
10:3x2 possible ways so 6/72=1/12
11:4x2 possible ways so 8/72=1/9
15:3x2 possible ways, so 6/72=1/12
20:2x1 possible ways, so 2/72=1/36
Answer:$\boxed{\dfrac16,\dfrac16,\dfrac1{12},\dfrac19,\dfrac1{12},\dfrac1{36}}$
I don't think thats correct. Since those are all the possible probabilites in the solution space, their sum should equal one, however your answer does not sum up to one.
+118608 Sparklingwater2 is almost correct. 
To get 2 you have to have 1,1 what you have done is perfect.
But to get 6 you have to have a 1 and a 6. OR a 6 and a 1. Your answer for that bit will need to be doubled.
etc
I have done it and I also checked that my probabilities add to 1. (And they do)
