We have a fibration $F \rightarrow X \rightarrow B$. If $X = B \times F$, then combining Künneth formulas and universal coefficient theorem gives an isomorphism $H_n(X;G) \simeq \oplus H_p (B;H_{n-p}(F;G))$.
Why is this result true? This sentence in the Spectral sequence of Allen Hatcher, page 8.
The Künneth formula is $$H_n(B\times F;G) \simeq \bigoplus_{p=0}^n H_p (B;G)\otimes H_{n-p}(F;G)$$ The Universal Coefficient Theorem gives for each $p$ $$H_p (B;\Bbb Z)\otimes H_{n-p}(F;G) \simeq H_p (B;H_{n-p}(F;G))$$ So we are almost there, just short a coefficient group $G$ instead of $\Bbb Z$ on the left of the second formula.
It turns out that since $H_{n-p}(F;G)$ is a $G$-module, it is isomorphic to $H_{n-p}(F;G)\otimes G$.
Therefore $$H_p (B;\Bbb Z)\otimes H_{n-p}(F;G) \simeq H_p (B;\Bbb Z)\otimes G\otimes H_{n-p}(F;G) $$ and now all you need is $$H_p (B;\Bbb Z)\otimes G\simeq H_p (B; G)$$ which, again by the universal coefficient theorem, is true whenever Tor$(H_{p-1}(B),G)=0$. This happens for instance when $H_{p-1}(B,\Bbb Z)$ is free or when $G$ is free.