Let $E$ be the total space of a fiber bundle with base $B$ and fiber $F$, where $B$ and $F$ are smooth manifolds.
Under what condition is $E$ unoriented cobordant to $B\times F$?

And what happens if we take $B$, $F$, and $E$ oriented. Is $E$ oriented cobordant to $B\times F$?

Let $E$ be the total space of a fiber bundle with base $B$ and fiber $F$, where $B$ and $F$ are smooth manifolds.
Under what condition is $E$ unoriented cobordant to $B\times F$?

And what happens if we take $B$, $F$, and $E$ oriented. Is $E$ oriented cobordant to $B\times F$?

Edit : As mentioned in the comment, there are classical counterexamples, due to Atiyah, in the oriented case when the base is not simply connected.

Furthermore Dold showed that the unoriented cobrodism ring is a polynomial ring over $\mathbb{Z}/2$ with generator the $\mathbb{R}\mathbb{P}^{i}$ for $i$ even, and some fiber bundle with base $\mathbb{R}\mathbb{P}^{2^r-1}$ and fiber $\mathbb{C}\mathbb{P}^{s2^r}$ with $s,r\geq 1$. see http://www.map.mpim-bonn.mpg.de/Unoriented_bordism or http://link.springer.com/article/10.1007%2FBF01473868. Those bundles have $\mathbb{R}\mathbb{P}^{2^r-1}$ as base spaces which are null-bordant, so they provide counterexample.

However, in both those cases, counterexamples (seem to) come from the fact that the base space is not simply connected. In fact, signature is known to be multiplicative in the simply-connected case, see http://www.maths.ed.ac.uk/~aar/papers/chs.pdf.

So I will try to reformulate my question with those facts in mind :

Given that $B$ is simply connected, is there some reasonable hypothesis on $F$ assuring that $E$ is cobordant to $B\times F$?