Year: 2012
Paper: 3
Question Number: 12
Course: LFM Stats And Pure
Section: Geometric Probability
No solution available for this problem.
The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1469.4
Banger Comparisons: 2
\begin{questionparts}
\item A point $P$ lies in an equilateral
triangle $ABC$ of height 1. The perpendicular
distances from $P$ to the sides $AB$, $BC$ and $CA$ are
$x_1$, $x_2$ and $x_3$, respectively. By considering the
areas of triangles with one vertex at $P$, show
that $x_1+x_2+x_3=1$.
Suppose now that $P$ is placed at random in the equilateral triangle
(so that the probability of it lying in any given region of the triangle is
proportional to the area of that region). The perpendicular
distances from $P$ to the sides $AB$, $BC$ and $CA$ are
random variables $X_1$, $X_2$ and $X_3$, respectively.
In the case $X_1= \min(X_1,X_2,X_3)$, give a sketch showing
the region of the triangle in which $P$ lies.
Let $X= \min(X_1,X_2,X_3)$. Show that
the probability density function for $X$ is
given by
\[
\f(x) =
\begin{cases}
6(1-3x) & 0 \le x \le \frac13\,, \\
0 & \text{otherwise}\,.
\end{cases}
\]
Find the expected value of $X$.
\item
A point is chosen at random in a regular tetrahedron of height 1.
Find the expected value of the distance from the point to the
closest face.
\newline
[The volume of a tetrahedron is
$\frac13 \times \text{area of base}\times\text{height}$ and its centroid
is a distance $\frac14\times \text{height}$ from the base.]
\end{questionparts}
Under 9% of candidates attempted this, though the level of success was comparable with that achieved in questions 3 to 6, 10 and 11. The derivation of the pdf was, in many cases, the stumbling point whether being found directly, or via the cdf, lacking clear explanation. The expectation caused few problems. The second part reflected the first in each respect.