Question:
Joan tries to solve a really hard problem once each day.  She has a 1/4 probability of solving it each day.  What is the probability that she will solve it before her sixth try?

Answer:
We must find the probability that Joan can solve it at any time before the sixth try, so it is the sum of the probabilities that she will solve it on her first, second, third, fourth, and fifth tries.  We could evaluate all those cases, but seeing all that casework, we wonder if it will be easier to find the probability that she fails to solve it before 6 tries, and subtract the result from 1.

In order for her to fail to solve it before her sixth try, she must fail 5 times.  The probability of failure on each try is $1 - \frac{1}{4} = \frac{3}{4}$, so the probability that she fails on each of her first 5 tries is $\left(\frac{3}{4}\right)^5 = \frac{243}{1024}$.  Therefore, the probability that she succeeds before her sixth try is \[1-\frac{243}{1024} = \boxed{\frac{781}{1024}}.\]