How do we represent this event?

My professor did this problem out of our text book and he didn't exactly show us how he did it (he skipped showing the steps in the middle). He got the answer $(1-x)^2$.

If we let $X$ and $Y$ be two independent uniform $(0,1)$ random variables and let $M$ be the minimum of $X$ and $Y$, $0\lt x \lt 1$. How would we represent the event $(M\ge x)$ as the region in the plane and find $P(M\ge x)$ in the area of this region?

How would you go about doing this?

PS I added the homework tag because it could have been a homework problem, however it's not homework. I'm just looking for more people to see my question.

-

The event $M \geq x$ is the intersection of the two events $X \geq x$ and $Y\geq x$. In the plane, this means that the location of $(X,Y)$ must be to the upper-right of $(x,x)$, so $(X,Y)$ must belong to the square $[x,1)\times [x,1)$. This square has area $(1-x)^2$, which is the desired probability.

In other words: \begin{align*} P(M\geq X) &= P(X\geq x\text{ and }Y\geq x)\\ &= P(X\geq x)P(Y\geq x)\text{ since X and Y are independent}\\ &= m([x,1))m([x,1))\text{ since X and Y are uniform (0,1)-random variables}\\ &= (1-x)(1-x)\text{ since the measure of an interval is its length.} \end{align*}

-
so would this mean that $P(M\ge x)=(1-x)^2$? Or is this just to show the area in the plane where $M$ is defined? –  TheHopefulActuary Nov 19 '12 at 19:08
Both -- the area in the plane is the probability of the corresponding event, since $X$ and $Y$ are independent and uniform. I've edited to elaborate. –  Owen Biesel Nov 19 '12 at 22:54
sorry one more question, my professor is being no help to me at all. I am also asked to find the PDF and CDF of the function. I know that the answer for the CDF is $1-(1-x)^2$ and the PDF is $2(1-x)$ but I don't understand how those were found. Wouldn't $P(M\ge x)=(1-x)^2$ be the PDF? –  TheHopefulActuary Nov 20 '12 at 13:27
The CDF is the probability that $M$ is any value up to $x$: $P(M < x) = 1 - P(M\geq x) = 1 - (1-x)^2$. The PDF is the derivative of the CDF: $d(1-(1-x)^2)/dx = 0 - 2(1-x)^1 (-1) = 2(1-x)$. –  Owen Biesel Nov 20 '12 at 21:01

The probability that $M\ge x$ equals the probability that both $X$ and $Y$ are $\ge x$. Since each of these is $1-x$ and they are independent, the result follows.

-