Very basic probability problem Firstly, apologies if this question is too trivial or just plain inappropriate for this site...
I want to catch a ball. I have exactly two chances to catch the ball. The ball is thrown in such a way that my chance of catching the ball on the first throw is $55\%$. The next time, the ball is thrown faster, so my chance of catching that throw is only $25\%$.
I only need to catch the ball once. Before the first ball is thrown, what's my overall chance of success?
(My assumption is that this is a simple average (so $40\%$), but I wonder if I'm making a mistake.)

When succeeding means doing something at least once, failing means doing something zero times, and often the latter is easier to compute. In this case, failing the first time has a probability 0.45, and failing the second 0.75; assuming these events are independent, the chance of failing both times is $$0.45 \cdot 0.75 = \frac{27}{80},$$ and so your chance of succeeding is $$1 - \frac{27}{80} = \frac{53}{80} = 0.6625.$$
