There are two ways to get a full house in this situation: (1) draw a three of a kind or (2) draw one more to the pair and another pair. I'm going to assume you discard three singletons.
First, lets work out the number of combinations under (1). There are 3 ranks with only 3 suits left (remember you discarded 3 singletons) and 9 ranks with 4 suits left. The number of combinations is thus 3*combin(3,3)+9*combin(4,3) = 3*1 + 9*4 = 39.
Next, let's work out the number of combinations under (2). There are 2 suits left to add to the existing pair. There are combin(3,2) ways to form a pair from the 3 ranks with 3 cards left and combin(4,2) ways to form a pair to from the ranks with 4 cards left. So the total combinations under 2 is 2*(3*combin(3,2)+9*combin(4,2)) = 2*(3*3 + 9*6) = 126. The total number of ways to arrange a full house is the sum under (1) and (2), or 39+126=165. There are combin(47,3)=16,215 ways to arrange the 3 cards on the second draw. The probability of drawing a full house is the number of ways to draw a full house divided by the total combinations, or 165/16,215 = 0.0101758, or about 1 in 98.
For more information on the combin() function, please see my section on probabilities in poker page.