PDF of two watches problem

Question

Consider two battery-powered watches. Let $X_1$ denote the number of minutes past the hour at which the first watch stops and let $X_2$ denote the number of minutes past the hour at which the second watch stops. What is the probability that the larger of $X_1$ and $X_2$ will be between $30$ and $50$?

I was able to follow the solution to this problem until the final answer which is $4/9$. What I don't understand is the problem's pdf which is the following:

$$f(y)=\begin{cases}0&,y\in(-\infty,0)\\\frac{y}{1800}&,y\in[0,60)\\0&,y\in[60,\infty)\end{cases}$$

Why is the value of the pdf $y/1800$ in the interval $[0,60)$? Why isn't it $2/60$?

Thank you in advance!

Answer 1

Let $X_1$ has a pdf of $$f_{x_1}(x)=\begin{cases} \frac1{60}, 0\leq x < 60 \\ 0, \ \text{otherwise} \end{cases}$$

Similar for $X_2$.

Then we are looking for the pdf of $Y=\texttt{max}(X_1,X_2)$, since at least one of the battery-powered watches has to work. We assume that $X_1$ and $X_2$ are independent.

$P(Y\leq y)=\left(\frac{y}{60}\right)^2, 0\leq y < 60$

To obtain the pdf you have to calculate the derivative.

Answer 2

It is tacitly assumed that we have two independent random variables $X_1$, $X_2$, each of them uniformly distributed in $[0,60]$. In order to solve the problem, draw the square $Q:=[0,60]^2$, and find the subset $A\subset Q$ defined by the condition $30\leq\max\{x_1,x_2\}\leq 50$. This $A$ is the union of two trapezoids, and it is easy to see that $$p={{\rm area}(A)\over{\rm area}(Q)}={4\over9}\ .$$