In the diagram above $A$ is the centre of the circle and $CB$ is thus the diameter. Point $D$ is an arbitrary point in the circumference.
In $\triangle ACD$ $\angle CDA = \alpha$ since $AC = AD$
Thus
$$2 \cdot \alpha + \theta = 180 ^\circ$$
In $\triangle ADB$ $\angle ADB = \beta$ since $AB = AD$
Thus
$$2 \cdot \beta + (180^\circ - \theta) = 180 ^\circ$$
Add both these together
$$2 \cdot \alpha + 2 \cdot \beta + 180 ^\circ = 360 ^\circ$$
Which we can easily rearrange to show:
$$ \alpha + \beta = 90 ^\circ $$
We have thus proved $\angle CDB$ is a right angle as required.
I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
Best Answer
Let's say we have our triangle $\triangle ABC$ inscribed in a circle, with $\overline{BA}$ its diameter.
It is not too hard to show that the angle labeled $\theta$ in the diagram is equal to $2\angle CAB$. This is sometimes known as the central angle theorem.
This means that, assuming a "random" triangle is chosen by randomly choosing the point $C$ on the top half of the circle, we are randomly choosing an angle $\theta$ between $0^\circ$ and $180^\circ$, and wanting it to come up less than or equal to $120^\circ$. The probability of this would be $\frac{2}{3}$.