Find the probability of two random real numbers $x$ and $y$ between $0$ and $2$, where $\min(x,y) < 2/3$

Question

Here is a picture of what I did so far.

http://sdrv.ms/HhxIvu

I got a result of $\frac59$, because the total area is $4$, and I'm subtracting the square with side of $\frac43$.

Can anyone confirm that this is the correct answer?

Thank you.

Answer 1

Yes, you are right.

$\left(\frac{4}{3}\right)^2 = \frac{16}{9} $ is the area of empty subsquare. Probability of being in empty subsquare is $\frac{16/9}{4} = \frac{4}{9}$. The probabiliy of being in filled area is equal to $1-\frac{4}{9}$.

So, you are right.

Answer 2

Note that $\min(x,y) < \frac23$ means that the smaller of the two has to be less than $\frac23$. So when one of them is smaller than $\frac23$, the other doesn't even matter! So the only case we're really interested in is when both are $\geq \frac23$, because that is the only condition under which $\min(x,y)<\frac23$ doesn't hold true.

So what's the probability of $\min(x,y) \geq \frac23$? Well, we use independent probability: consider $x$, randomly generated as a real in the interval $[0,2]$. So the chance that $x\geq\frac23$ is $\frac23$. (Check the arithmetic: because we're on the interval $[0,2]$, which can be divided into three lots of $\frac23$.)

It follows that $p(y \geq \frac23)$ is also $\frac23$. So $P(x,y \geq \frac23) = \frac23 \cdot \frac23 = \frac49$. Remember that this is actually the case we don't want to happen, so therefore the probability of $\min(x,y)<\frac23 = 1-\frac49 = \frac59$.