If the graph is known to be connected: The graph has no cycles iff it is a tree.
By Cayley's formula, the number of trees that can be formed with $v$ vertices is $v^{v-2}$
A tree with $v$ vertices has $v-1$ edges. The probability that a specific tree can be formed is
$P^{v-1}\cdot(1-P)^{k-v+1}$
where
$k=v\cdot(v-1)/2$,
adding up to
$v^{v-2}\cdot P^{v-1}\cdot(1-P)^{k-v+1}$.
The case that the graph can be disconnected is complicated. It takes the distinct combinations of disconnected graphs. I havn't seen a study on this and don't have one mine here.
EDIT: joriki's solution seems to be covering it for the general case.