Probability of a piece of stick being a certain size?

Suppose Alex has a one meter long wooden stick that Alex breaks randomly in two places such that the location of the breaks are independently distributed uniformly along the length of the wooden stick.

(a) Calculate the probability that one of the pieces will be more than 75 cm.

Is this just $\frac{75}{100}$?

(b) Calculate the probability that the three resulting pieces of the stick can be arranged to form a triangle.

I know the stick cannot form a triangle unless all three pieces are no longer than the sum. But I don't know how to make this into a probability? If $\alpha$ is the first break and $\beta$ is the second break and L is the length of the stick, then $\alpha < \frac{L}{2}$ and $\beta < \frac{L}{2}$. How would I turn this into a probability?

• There are quite a lot of other posts for part b here, e.g. this one. – Parcly Taxel Sep 5 '16 at 1:50

Let $X_1$, $X_2$ be the two uniform variables where the breaks happen.
(a) No it's far from $75/100$. One case where the biggest stick is 75 cm long is when both $X_1$ and $X_2$ occur before the 25% mark. The probability this happens is $\mathbb{P}(X_1<0.25 \cap X_2<0.25)=\mathbb{P}(X_1<1/4)\mathbb{P}(X_2<1/4)= 1/16$. What are the other possibilities of values for $X_1$ and $X_2$ that would give a piece of more than 75cm, and what are their probabilities?
• Ok, so another case could be if $X_1$ comes before 12.5cm and $X_2$ occurs after 87.5cm; the 75cm piece is in the "middle". And the third (and last?), case is when the $X_1$ and $X_2$ occurs after the 75cm mark, so the 75cm is the "first" piece. Am I thinking of this right? Just don't know how you calculated the probability 1/16, but since these are all the same case(?), would the the probability of A just be 3(1/16)? – Deegeeek Sep 5 '16 at 3:05