In this problem all that is essential is that there are 3 o's and 7 "not o's".
Let D = 10C3, then number of ways of picking 3 things from 10. This is the denominator in the hypergeometric distribution.
There are 7C3-k * 3Ck ways of picking k o's and 3-k "not o's" from a group of 10...and in in this case we are not interested in k=0. Call this number Nk.
The probability of an o from a group 3 containing 1 o is 1/3, 2 o's 2/3 and 3 o's 1.
Therefore the probability you seek is the sum of (k/3)*Nk/D for k=1 to 3.