We have the following cases to consider (1,1), (1,3), (1,8), (3,1) ,(3.3) (3,8), (8,1) (8,3), (8.8). Well we know three of our samples will have a variance of 0. The variance for the case when we have 1 and 3 is 3.3; likewise, the variance for when we have 8 and 3 is 6.3 and the variance for when we have 1 and 8 is 49. The possible variances are 3.3, 1, 6.3, and 49. The probability of having a variance of 1 is 3/9. For a variance of 3.3 its 2/9, for 6.3 its 2/9 and lastly, for a variance of 49 its 2/9. As you can see, 3/9 + 2/9 + 2/9 + 2/9 = 1.
I had this same question and this was the answer....see if you plug this in if its works.