Michael F.'s answer is correct. He used "counting" formulas to get the answer. It is always nice to be able to check your answer by a different technique, though.
Another approach to this problem is to do it directly, using conditional probabilities. The probability of doing A first, and then B afterward, is P(A)×P(B|A), where the 2nd is the conditional probability of doing B after A has been done. [NOTE: a lot of incorrect answers are obtained when forgetting to use the conditional probability for the second factor!]
So the probability of selecting G, a girl first, is P(G) = 4/9. That would then leave 3 girls and 5 boys, so the probability of selecting a boy second after a girl, P(B|G), is 5/8. Then the probability of doing both (in that order), is found by multiplying 4/9 ×5/8 = 5/18.
In a similar manner, the probability of selecting a boy first and then a girl second after the boy would be
5/9×4/8 = 5/18.
Since the event of selecting one boy and one girl could happen either way, we would have to add the two probabilities to get 5/18 + 5/18 = 5/9 for the probability of getting one student of each in drawing two.
The answer is 5/9, as seen in the answer from the other approach with the Combination formulas.
The counting formulas for combinations and permutations come in handiest when dealing with larger sets. This problem is done just as easily by either approach (as long as done correctly).