Random selection probability Hello and thanks for looking at my question.
I'm having trouble figuring out how to do this problem:

An urn contains ten numbered balls- four 1's three 2's two 3's and one 4

*

*Two balls are drawn without replacement. What is the probability that the sum is 6?


*Two balls are drawn with replacement. What is the probability that the sum is 6?

I know the answers are 8/90th and 1/10th but I don't know how they got them. Any help would be appreciated
 A: Let's try the first question.  The experiment "two balls are drawn without replacement" has $90$ equally likely outcomes:  $10$ possibilities for the first ball drawn, and $9$ possibilities for the second ball drawn.  The sample space would be $\{1_11_2,1_11_3,1_11_4,1_12_1,1_12_2,1_12_3,\ldots,4_12_3,4_13_1,4_13_2\}$.  The event you're interested in is composed of all the outcomes that have a sum of $6$: $\{2_14_1,2_24_1,2_34_1,3_13_2,3_23_1,4_12_1,4_12_2,4_12_3\}$.  This set has $8$ outcomes.  Thus, the probability for the first question is $8/90$.  Can you now answer the second question?
A: Answer:
Without replacement:  Number of ways you can draw two balls = 10*9 meaning, 10 ways to choose the first ball and 9 ways to choose the second ball.
For the sum to be 6, the possible ways are {3,3},{4,2},{2,4}.
For {3,3}, Number of ways you can draw a 3 for the first ball is 2 and the number of ways you can draw a 3 for the second ball is 1.  So the total will be 2*1 = 2
For {4,2}, Number of ways you can draw a 4 for the first ball is 1 and the number of ways you can draw a 2 for the second ball is 3.  So the total will be 1*3 = 3
For {2,4}, Number of ways you can draw a 2 for the first ball is 3 and the number of ways you can draw a 4 for the second ball is 1.  So the total will be 3*1 = 3
Total outcomes = 2+3+3 = 8 and thus the required probability = 8/90
With replacement Number of ways you can draw two balls = 10*10 meaning, 10 ways to choose the first ball and 10 ways to choose the second ball.
For the sum to be 6, the possible ways are {3,3},{4,2},{2,4}.
For {3,3}, Number of ways you can draw a 3 for the first ball is 2 and the number of ways you can draw a 3 for the second ball is 2.  So the total will be 2*2 = 4
For {4,2}, Number of ways you can draw a 4 for the first ball is 1 and the number of ways you can draw a 2 for the second ball is 3.  So the total will be 1*3 = 3
For {2,4}, Number of ways you can draw a 2 for the first ball is 3 and the number of ways you can draw a 4 for the second ball is 1.  So the total will be 3*1 = 3
Total outcomes = 4+3+3 = 10 and thus the required probability = 10/100 = 1/10
