probability
Two points along a straight stick of length 37 cm are randomly selected. The stick is then broken at these two points. Find the
probability that the length of each of the resulting 3 pieces is at least 3cm.
Okay, what little I know so far is that the two points along the stick are uniform(0, 37).
I don't understand where to go from there… Please and thank you!
My approach is maybe more naive than the others posted.
Break the unit interval at $x$ and $y$ where $x < y$. Our lengths are then $x$, $y - x$, and $1 - y$. It's not hard to show that they all have probability $1/3$ of being the shortest. In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$). Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.
I'll take care of the case when $x$ is shortest, that is, $x \leq y - x$ and $x \leq 1 - y$. This is the leftmost triangle. Since we're assuming $x$ is least, we are looking for $$E[x] = \int_0^{1/3} \int_{2x}^{1 - x} 6x \;dy \;dx = 1/9$$
The cases when $y - x$ and $1 - y$ are shortest are similar.
Let $f(x)$ be the probability that we eventually cut off at least $x$ from a stick of length $1$, with $1\ge x\ge1/2$. We can either succeed by immediately cutting off at least $x$, with probability $1-x$, or by leaving $t\ge x$ and then cutting off $x$ from a stick of length $t$. Thus we have
$$ f(x)=1-x+\int_x^1f(x/t)\mathrm dt\;. $$
Substituting $u=x/t$ yields
$$f(x)=1-x+x\int_x^1f(u)/u^2\mathrm du\;.\tag1$$
Then differentiating with respect to $x$ yields
$$f'(x)=-1-f(x)/x+\int_x^1f(u)/u^2\mathrm du\;,$$
and differentiating again yields
$$f''(x)=-f'(x)/x+f(x)/x^2-f(x)/x^2=-f'(x)/x\;,$$
so
$$ \frac{f''(x)}{f'(x)}=-\frac1x $$
and thus
$$ \begin{align} \log f'(x)&=-\log x +c\;,\\ f'(x)&=a/x\;,\\ f(x)&=a\log x+b\;. \end{align} $$
Now $f(1)=0$ yields $b=0$, and then substituting into $(1)$ yields $a=-1$, so $f(x)=-\log x$ and
$$f(1/2)=\log2\approx0.693\;.$$
Best Answer
Since your distributions are uniform, and your probability space is a subset of $\mathbb{R}^2$, you can picture the probabilities directly as areas. The following picture should give you an idea of what's going on. The outer rectangle is your probability space. It has a total area of $37^2$, which amounts of course to a probability of $1$.
Each point represents a pair $(X_1,X_2)$. The orange areas are the points where all pieces have at least length $3$, and have total area $A = A_1 + A_2 = 28^2$. That corresponds to the probability $P = \frac{28^2}{37^2}$.