Probability that the length of each of the resulting 3 pieces is at least 3cm?

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!

• Pick two random $X_1,X_2$ from $(0,37)$. What are the lenghts of the pieces? – fgp Apr 10 '14 at 20:57
• Well... I've drawn a line, from 0 to 37 and picked to random points on it (with X2 > X1). It looks like the first piece is length of X1, the second is length of X2 - X1, and the final piece being of 37 - X2 length. Is that at all correct? – Shell Apr 10 '14 at 21:09
• Yup. Now you have you to figure out in which cases some of the resulting pieces are less than 3cm. – fgp Apr 10 '14 at 22:51
• Okay... but how would you go about doing that? I know it probably has to do with calculating a double integral, but I don't understand what the ends of those integrals would be? Are they just the values mentioned above? Where do you account for the 3cm? – Shell Apr 10 '14 at 22:56
• Before you start integrating, you need to understand the situation. So far, there are no integrals in sight - just find the cases where no piece is less than 3cm. What conditions does that impose on $X_1,X_2$? – fgp Apr 10 '14 at 22:58

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 $P = \frac{28^2}{37^2}$. 