Let $\pi:X = \mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^n$ be a projective bundle, where $\mathcal{E}$ is a rank two vector bundle over $\mathbb{P}^n$.
If $n = 0$ then $X = \mathbb{P}^1$, and for $n = 1$ we have that $X$ is a Hirzebruch surface. In both cases $-K_X$ is effective.
Is there an example, with $n\geq 2$, of a projective bundle $X = \mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^n$ such that $-K_X$ is not effective?
There is a formula, $K_X\simeq \mathcal{O}_X(-2)\otimes \pi^*\det \mathcal{E}^{\vee}\otimes \pi^*K_{\mathbb{P}^n}$.
(More generally, for rank $r+1$ bundle, replace -2 by $-r-1$ in the above formula.)
Now $\pi^*K_{\mathbb{P}^n}\simeq \pi^*\mathcal{O}_{\mathbb{P}^n}(-n-1).$ So $$K_X\simeq \mathcal{O}_X(-2)\otimes \pi^*\det \mathcal{E}^{\vee}\otimes\pi^*\mathcal{O}_{\mathbb{P}^n}(-n-1)=\mathcal{O}_X(-2)\otimes \pi^*(\det \mathcal{E}^{\vee}\otimes\mathcal{O}_{\mathbb{P}^n}(-n-1))$$
Hence $$-K_X\simeq \mathcal{O}_X(2)\otimes \pi^*(\det \mathcal{E}\otimes\mathcal{O}_{\mathbb{P}^n}(n+1))$$
So $-K_X$ is always effective as long as the line bundle on the RHS above has a non-zero section.
Added (I will leave this here for future readers): When I wrote this, I wasn't aware of Hartshorne's conjecture for rank 2 bundles. Thanks to @YosemiteStan for the enlightenment! See YosemiteStan's and abx's comment below for more details.
