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.)
 A: 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 $$1 - \frac{27}{80} = \frac{53}{80} = 0.6625.$$
A: The probability of catching the ball on the first throw is $0.55$, the probability of catching the ball on the second throw is $0.45\cdot0.25$ (the ball wasn't caught on the first throw, but it was caught on the second throw). So the probability of catching the ball is then equal to
$$
0.55+0.45\cdot0.25=0.6625.
$$
A: Ok, it is easier to calculate the probability that you don't get it.
So probability that you don't get it the first time multiplied by that you don't get the second time ( I guess you assume that the two probabilities are independent).
A: Play $10000$ times.
You will catch $55\%$ of the balls, i.e. $5500$ on the first throws, and miss $4500$ of them.
Among the $4500$, you will catch $25\%$, i.e. $1125$ on the second throw.
Total $6625$.
