The answer depends on whether you want to count only the throws that satisfy only the requirements for a full-house/four-of-a-kind entry, or also the throws with all five dice the same, which you could use either in the full-house/four-of-a-kind row or in the Yahtzee (five-of-a-kind) row. The difference is the probability of a Yahtzee, which is $1$ in $6^4=1296$; I'll calculate the probabilities excluding five of a kind.
In both cases, full house and four of a kind, there are $6\cdot5$ different choices for the numbers. For full house, there are $\binom52$ choices for the positions and for four of a kind there are $\binom51$. Thus the probability for a full house (excluding five of a kind) is
and for four of a kind
Note that these are just the probabilities for the first throw, which I think is what you asked for; the more interesting part of the game is deciding which dice to keep to optimize the chances on subsequent throws.