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Choose independently two numbers B and C from the interval [0,1] with unifrom density. Note that the point $(B,C)$ is then chosen at random in the unit square. Find the following probabilities:

a)$B+C < \frac{1}{2}$

b)$BC < \frac{1}{2}$

c)$ \lvert{B-C} \rvert < \frac{1}{2}$

I have solved the first one. It is simply the area of the triangle formed in the unit square and the half plane $B+C < \frac{1}{2}$. I know I should be able to solve the other ones using geometry as well but I am assuming there are other ways as well and I am having a hard time starting the next two. I know the answers are $\frac{1}{8}$ ; $\frac{1}{2}ln(2)$ ; $\frac{3}{4}$

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  • $\begingroup$ Perhaps I am misreading, but for $b$ I'd have thought the value was higher. After all $B<\frac 12$ implies any $C$ will work so the probability is at least $\frac 12$. I'd solve it by looking at $C<\frac 1{2B}$. $\endgroup$
    – lulu
    Sep 12 '17 at 1:08
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    $\begingroup$ Why not "by geometry". Writing the inequalities as equalities gives you the boundary of the area you want. Then you may have to integrate. $\endgroup$ Sep 12 '17 at 1:09
  • $\begingroup$ Yes, I get $\frac 12\times (1+\ln 2)$ for $b$. There's a simple integral involved in that one. For $c$ there is no need for an integral, you can do the geometry directly. $\endgroup$
    – lulu
    Sep 12 '17 at 1:11
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    $\begingroup$ that's part of it, but you have to also include the entire rectangle $B<\frac 12$. The hyperbolic arc $C=\frac 1{2B}$ intersects your square at $(.5,1)$ and $(1,.5)$ but everything to the left of the line $B=.5$ is in the desired region. $\endgroup$
    – lulu
    Sep 12 '17 at 1:55
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    $\begingroup$ Just to stress: your region only covers $B≥.5$. You are missing a lot of solutions! $\endgroup$
    – lulu
    Sep 12 '17 at 1:56
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$BC < \tfrac 12$ is a hyperbola.   You are after the area under the curve that is also inside the unit square.   Easiest found by noting the curve intersects the edge of the square at cartessian points $(\tfrac 12,1)$ and $(1,\tfrac 12)$.

So the requisite area is: $\displaystyle\frac 12+ \int_{1/2}^1\int_0^{1/(2c)} \;\mathrm d\, b\;\mathrm d\, c$

Alternatively, the area for the square excluding the "bite" is: $\displaystyle 1-\int_{1/2}^2\int_{1/(2c)}^1\;\mathrm d\, b\;\mathrm d\, c$


$\lvert B-C\rvert < 1/2$ is the region between the lines $C=B+\tfrac 12$ and $C=B-\tfrac 12$.   This region intersects the unit square to form an irregular hexagon: $\operatorname{Hexagon}(0,0)(\tfrac 12,0)(1,\tfrac 12)(1,1)(\tfrac 12,1)(0,\tfrac 12)$.

The area is easiest found by noting the area excluded from the unit square consists of two right triangles; $\triangle(0,\tfrac 12)(\tfrac 12,1)(0,1)$ and $(\tfrac 12,0)(1,0)(1,\tfrac 12)$.

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