I believe the following is true:
Let $F\rightarrow E \rightarrow B$ be a fiber bundle. If $F$ is contractible, then $H^*(E;R) \cong H^*(B;R)$ for a commutative ring $R$ with identity.
I will apply Theorem 4D.1, Hatcher's. There are two conditions to check.
(a) $H^n(F;R)$ is f.g. free $R$-module for all $n$ - which is trivially true.
(b) that there exists classes $c \in H^k(E;R)$ such that $i^*(c)$ generates $H^*(F;R)$ for each fiber.
But this follows as $$i^*: H^*(E;R) \rightarrow H^*(F;R) \cong R $$ is a ring homomorphism, mapping $1$ to $1$.
The result of theorem states that $H^*(B;R) \otimes_R R \rightarrow H^*(E;R)$ is an isomorphism of groups.
Is the statement true, and is my proof correct? Are there any details that I may have missed?