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I am trying to pick two points from $[0, 1]$ at random and with uniform probability. Let the result be the pair $(X, Y)$ in the square $[0, 1] \times [0, 1]$. Suppose that the distribution of $(X, Y)$ in the square is uniform, that is, the probability of $(X, Y)$ being in a particular region in the square is the area of that region.

I am trying to find probabilities such as these: $Prob(0 \leq X \leq 1/3)$

$Prob(X \leq Y )$

$Prob(X < Y )$

I am not sure where to start on this. I would appreciate any help in the right direction.

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    $\begingroup$ Area of the part of the square where the event happens, divided by $1$. $\endgroup$ – André Nicolas Feb 3 '16 at 1:46
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In general, these problems can be interpreted in terms of portions of the area of a square. To see this, let's write out the probabilities formally. We could write these as double-integrals.

Your first problem: $$P(X\leq\frac{1}{3})=\int_{0}^{1}\int_{0}^{\frac{1}{3}}f\left(x,y\right)dx\space dy$$

Since these are independent random variables, we can split $f(x,y)$ into $f(x)f(Y)$.

$$P(X\leq\frac{1}{3})=\int_{0}^{1}\left(\int_{0}^{\frac{1}{3}}f\left(x\right)dx\right)f\left(y\right)dy$$

But here, $f(x)=1$ and $f(y)=1$

$$P(X\leq\frac{1}{3})=\int_{0}^{1}\left(\int_{0}^{\frac{1}{3}}dx\right)dy$$

This is just an expression for the area of the left third of a square. It has area $\frac{1}{3}$.

$$P(X\leq\frac{1}{3})=\frac{1}{3}$$

Of course, since these are independent, we could have really just ignored $Y$ altogether in that one. $P(X\leq\frac{1}{3})=F(\frac{1}{3})=\frac{1}{3}$. But, your second problem is not so trivial. We can write:

$$P(X\leq Y)=\int_{0}^{1}\left(\int_{0}^{y}dx\right)dy$$

This is an expression for the area of a square cut in half along a diagonal. It has area $\frac{1}{2}$

$$P(X\leq Y)\frac{1}{2}$$

I'll leave the last one to you. (Hint: the area of a line is zero).

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  • $\begingroup$ It's probably better to just write $f(x,y)$ rather than $f(x)$ times $f(y)$, since you could have a true joint distribution in certain problems. (Here you don't, though.) $\endgroup$ – Ian Feb 3 '16 at 2:05
  • $\begingroup$ Good point. I'll add a note. $\endgroup$ – CommonerG Feb 3 '16 at 2:06
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Since $(X,Y)$ is uniformly distributed within the unit square, the probability of point $(X,Y)$ being in some region $A$ is exactly the area of the region of $A$ within the unit square.

$$\mathsf P\big((X,Y)\in A\big) = {\lvert A~\cap~[0;1]{\times}[0;1]\rvert}$$

Just determine what that area is.


$\mathsf P(0{<} X{\leq} 1/3) = 1/3$ as exactly one third of the square has an $X$ value nevermore than $1/3$.

$\mathsf P(X{\leq}Y) = 1/2$ as half of the points in the square has an $X$ value nevermore than their $Y$ value.

Also $\mathsf P(X{<}Y) = 1/2$ as half of the points in the square has an $X$ value less than their $Y$ value.

PS: Note that $\mathsf P(X{=}Y) = 0$ as a line has no area; it is almost impossible to randomly select (a uniformly distributed) point that is exactly on that line.

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