How do you solve for the probability of one outcome from two independent sources of uncertainty? Setup
You have two boxes. Box A has a 50% chance of having the 'desired prize' and Box B has a 25% chance of having the 'desired prize'. You get to keep box A and box B.
Box A and Box B are independent of each other and so it is possible for both of them to have or not have the box. It is possible to not get the desired prize at all, or to receive duplicates of the 'desired prize'.
Question
What are the chances the 'desired prize' is in at least one of your boxes?
 A: The chance of the prize being in at least one of the two boxes means that we have 3 possible scenarios.

*

*Scenario 1: the desired prize is in both boxes
This is simply $0.5$x $0.25$ = $0.125$ as we multiply the probability of the prize being in box A with the probability of it being in box B.

*

*Scenario 2: the desired prize is only in box A
This is now $0.5$ x $(1-0.25)$ = $0.5$x $0.75$ = $0.375$ as we multiply the probability of the prize being in box A with the probability of it not being in box B.

*

*Scenario 3: the desired prize is only in box B
This is, similarly to above, $(1-0.5)$ x $0.25$ = $0.5$x $0.25$ = $0.125$ as we multiply the probability that the prize is not in box A with the probability of it being in box B.
Final Answer
The final step is now just to add up the 3 probabilities as these cover all the situations that you are looking for.
And so our final answer will be $0.125 + 0.375 + 0.125$ $=$ $0.625$
A: As indicated by the comment of Bey, following my answer, the responses here seem to be a case of shooting gallery blues, where the target has moved.

Very interesting opposing answers and comments, all of which I disagree with.  The exception is the interpretation in, and answer of FD_bfa, which I agree with.
This is my alternative approach, given that interpretation:
To me, it seems that the OP's (i.e. original poster's) intention is clear:

*

*Event $E_1$ is the event that there is a prize in Box A. 
$p(E_1) = (1/2).$

*Event $E_2$ is the event that there is a prize in Box B. 
$p(E_2) = (1/4).$

*Events $E_1$ and $E_2$ are independent events.

*The OP is asking (in effect) what is the probability that it is not the case that both events $E_1$ and $E_2$ fail.

Since the $2$ events are independent, and since the probability of failure in the $2$ events $E_1$ and $E_2$ are $(1/2)$ and $(3/4)$ respectively, the probability that both events fail is $(1/2) \times (3/4) = 3/8$.
Therefore, the probability that it is not the case that both events $E_1$ and $E_2$ fail must be
$$1 - (3/8).$$
