Imagine that, somewhere inside an origin-centered, unit-radius sphere
$S$ in $\mathbb{R}^3$,
sits a convex body $K$ of volume vol$(K)=\alpha (\frac{4}{3} \pi)$,
with $\alpha < 1$ the fraction of the volume of $S$.
$K$ is inside $S$ at an unknown but fixed location and orientation.
My question is: How many line-probes are needed to detect
its presence?
A *line-probe* is a line $L$ whose intersection
with $K$ includes a point strictly interior to $K$.
One might need many probes to certainly detect the presence of a small-volume $K$.

Let $f(k)$ be the volume fraction $\alpha$ such that (a) there is some body $K$ that is not detected by any fixed set of $k$ probes, and (b) every body with vol$(K) > \alpha$ is detectable by $k$ probes.

I believe $f(1)=\frac{1}{2}$: If $K$ fills a hemisphere, it could
"hide" in $S$ from any single probe. But any $K$ with more than half
the volume of $S$ necessarily includes the origin, and so a line
through
the origin would detect it.

It may be that $f(2)=\frac{1}{3}$ by two orthogonal probes that partition $S$
into two spherical caps and the sandwich between, each of
$\frac{1}{3}$
the volume of $S$.
And perhaps $f(3)=\frac{1}{4}$ via three probes through the origin.
But I am uncertain of these values of $f()$. If anyone can hide bodies of larger volumes
from these probes, please let me know!

This feels like a question that was likely considered before;
if so, a pointer would be appreciated.
Of course, the question generalizes to $\mathbb{R}^d$, with various dimensional probes.
In $\mathbb{R}^1$ with point-probes, $f(k)=\frac{1}{k+1}$.
*Edit*: Michael Biro suggests in the comments that the $f(2)$ example above could be
generalized to establish that also $f(k)=\frac{1}{k+1}$ in $\mathbb{R}^3$.

**Update**.
Here is an illustration of Ilya Bogdanov's argument that my 2nd example does *not*
establish that $f(2)=\frac{1}{3}$:

`$\{-1,1\}$`

, not`$\{0,1\}$`

. $\:$ $\endgroup$ – user5810 Sep 10 '12 at 0:49