按公司分类的量化面试真题

The table stands exactly when the centre of the tabletop lies inside the triangle formed by the three legs.

Fix the first leg. Measure the second leg by the angle $\theta$ between them along the shorter arc. Where can the third leg go?

When does it stand? The table's weight acts at the centre $O$. It stands exactly when $O$ is inside the triangle formed by the three legs. (Cases where $O$ lands exactly on an edge have probability $0$.)

Fix the first two legs. Put leg $A$ anywhere. Measure leg $B$ by the angle $\theta$ from $A$ along the shorter arc, so $0 \le \theta \le \pi$. Whether $B$ is clockwise or anticlockwise of $A$ doesn't matter by symmetry, so $\theta$ is uniform on $[0, \pi]$.

Where must the third leg go? Let $A'$ and $B'$ be the points directly opposite $A$ and $B$. The triangle contains $O$ exactly when leg $C$ lands on the arc between $A'$ and $B'$, the one facing the short arc $AB$. If $C$ lands anywhere else, all three legs fit in one half of the table and $O$ is outside the triangle. That arc has angle $\theta$, so

Average over $\theta$. Since $\theta$ is uniform on $[0, \pi]$,

so

Answer: $0.25000$.

Quick check: the table falls exactly when all three legs fit in some half of the table. For $n$ random points on a circle that happens with probability $n/2^{n-1}$, which is $3/4$ for $n = 3$. That leaves $1/4$.