Given a point $Q = (X, Y)$ which is uniformly randomly chosen within the range of $0 \leq X, Y \leq 1$, it forms a rectangle with both the $x, y$ axis whose area is $A = XY$. For some $r$ such that $0 \leq r \leq 1$, find the probability that $A \leq r$.

My thoughts: Since the point is randomly chosen, $\mathrm{P}(A \leq r)$ is simply the area under $xy = r$, or $y = \frac{r}{x}$ within the range, divided by the area of the whole range, which is $1$.

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To calculate the area, I tried to divide the area into two pieces: the rectangle whose upper-right corner is point $M$, and the area at the right.

The left-hand side is easy: since point $M$ should be at where equations $\begin{cases}y = 1 \\ xy = r\end{cases}$ both is satisfied, the area should be $r$.

The right-hand side, i think, should be the result of $\displaystyle\int^{1}_{r}\frac{r}{x}dx = -r \ln{(r)}$.

So $\mathrm{P}(A \leq r) = \frac{r -r \ln{(r)}}{1} = r -r \ln{(r)}$.

I think I got it right, but without an answer, I couldn't confirm. Also, I wonder is there any better approach. Anything helps, and thanks in advance.

  • $\begingroup$ Your answer is correct. $\endgroup$ Jul 20, 2022 at 7:30
  • $\begingroup$ @geetha290krm Thank god I finally got something correct. $\endgroup$
    – Uduru
    Jul 20, 2022 at 7:32


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