Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$
Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.
So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.
Therefore the probability of the triangle area being greater than 2 is $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$
However the solution says the answer is $\frac{8}{9}$, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!
Best Answer
The area of the triangle is given by the function $\displaystyle A(X) = \frac{3}{2}(X+1)$, where X is uniformly distributed in $[0,3]$, so \begin{equation} \mathbb{P}(A(X) \geq 2) = \mathbb{P}\left(X\geq\frac{1}{3} \right) = \frac{1}{3} \left(3 - \frac{1}{3} \right) = \frac{8}{9} \end{equation}