5
$\begingroup$

Two points are selected randomly on a line of length $L$, so as to be on opposite sides of the midpoint of the line[In other words, the two points $X$ and $Y$ are independent random variables with a uniform distribution over $(0, L/2)$ and $(L/2, L)$ respectively. Find the probability that the distance between the two points is greater than $L/3$.

Attempt: We have $f_X(x) = f_Y(y) = 2/L => f(x,y) = 4/L^2 $ The probability I want is $P(Y-X > L/3) = P(Y> L/3 + X)$. So I need to evaluate: $ \iint_{(x,y): y > L/3 + x} f(x,y)\,dy\,dx.$ I said the limits on $y$ were from $L/3 + x$ to $L$ and $x$ from $0$ to $L/2$, so I evaluate $\int_0^{L/2} \int_{L/3 +x}^{L} \frac{4}{L^2}\,dy\,dx$ which gives $5/6$, but that is incorrect. Where did I go wrong?

  • 0
    Wlog. $L=2$ and you want the area of the unit square $[0,1]^2$ above the line $y=x-\frac23$, that is a triangle of size $\frac1{18}$ is cut off.2012-12-10

2 Answers 2

2

The problem is that for $x\lt L/6$ you're allowing $y\lt L/2$.

  • 0
    So you essentially said there are 3! Arrangements of the $X_i$ and 2 of which result in $X_2$ between X_1, X_3, => 2/6 = 1/3. In terms of my integrand, is it okay to have the limits on the second integral from $x_1 $ to $ x_3$? It makes sense but then I don't get an actual probability (number) in the end.2012-12-11
3

Since you are integrating a constant function, it's going to be so much easier if you draw yourself a picture of the area and use simple geometry, rather than integrating an analytic expression:

enter image description here

And dividing the hatched area by the total square area, the resulting probability is $\frac{7}{36}/\frac{1}{4} = \frac{7}{9}$

  • 0
    OP CAF has been advised previously (see e.g. the comments on [this problem](http://math.stackexchange.com/q/255429/15941)) to use geometry, but it is unlikely that he will heed that advice (or yours).2012-12-10