What is the probability that sum of two randomly generated numbers is 12 given that they are distinct? Numbers are generated from [1,10] inclusive. I am required to find probability that sum of two numbers is 12 given that they are distinct.
My problem is not the question but the correct method of answering it. Since there are two numbers being generated, I have nine outcomes that add to 12: 
A ={(2,10),(3,9),(4,8),(5,7),(6,6),(10,2),(9,3),(8,4),(7,5)}
Please either confirm which of the following answer is correct or provide the right one.
Method 1)
Simply take distinct numbers adding to sum = 8.
Divide by total combinations = 100
*Probability is 0.08.*

Method 2)
Apply Bayes' Theorem with 
> A = Sum is 12
>  A' = Sum isn't 12
>  B = They are distinct

Find there probabilities and put them in formula to find 

P(A|B) = 8/90 = 0.08889

Which is correct?
 A: I think this is the same as randomly selecting two chips without replacement
from an urn with 10 chips numbered from 1 through 10. You want the
probability of getting a total of 12. The total doesn't
arise from outcome $(6,6)$ because sampling is without replacement. 
The number of possible choices is $10 \times 9 = 90.$ (Ten choices on the first draw and 9 on the second.)
There are eight satisfactory choices of two chips: $\{(2,10), (3,9), (4,8), \dots (10,2)\}.$ 
So the desired probability is $8/90.$ I will leave it to you to
supply 'formulas' based on your text and class notes.

Note: Because you pose this as a simulation problem, I will show results from a
million 2-chip draws, using R statistical software. The random function sample (without extra
parameters) does sampling without replacement. The vector tot == 12
is a logical m-vector taking values TRUE and FALSE; its mean
is its proportion of TRUEs. (With a million iterations, one expects 95% of such simulated
proportions will be within $\pm 0.0003$ of $8/90;$ that is, the third decimal place could be off by one digit.) This particular simulation happens to give three place accuracy.
m = 10^6;  chips = 1:10
tot = replicate( m, sum(sample(chips, 2)) )
mean(tot==12);  8/90
## 0.089215               # aprx P(T=12)
## 0.08888889             # exact P(T=12)
sd(tot==12)/sqrt(m)
## 0.000285054            # 95% margin of simulation error

A: There are $10^2=100$ equiprobable pairs, $90$ of them distinct. According to your count  there are $8$ distinct pairs summing to $12$. The probability you are looking for then is
$${8\over90}={4\over45}=0.08889\ .$$
