Question: 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**
> 
> 
> 1. Two balls are drawn without replacement. What is the probability that the sum is 6?
> 2. 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

Answer: 
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?
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