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Find the probability that the two positive real numbers $x$ and $y$ satisfy the following condition:

$$x+y> {\frac 12}$$ if it is given that $x< {\frac 12}$ , $y< {\frac 12}$

NOTE:Although it is quite obvious, it is also given that $y<1$ and $x<1$.Also, $x$ and $y$ are variables that vary from $0$ to $1$.
PS:The answer is not $1$. The given answer is $\frac14$ but I wish to know the solution. The problem I am facing is that I don't know how to find $P\left((y<\frac{1}{2})\cup (\frac{1}{2}x\right))$. I think I can use inclusion-exclusion but it doesn't work

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    This is a sure event: pick $x=y=1/3$. Hence, such numbers surely exist, the probability is $1$. – 2017-02-02
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    At the moment the answer is 1, since if x and y are both a third, it satisfies the condition, and x 1/3 exists - I didn't copy honest! – 2017-02-02
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    You had better write out the problem exactly as you found it, Luci. Then, you had better make some effort to solve it on your own, and let us know where you get stuck. – 2017-02-02
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    @GerryMyerson. I am truly sorry for the mistake, I wrote the problem exactly as I found it and have edited the problem. Kindly, revisit the problem. Deepest apologies. – 2017-02-02
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    It seems a kind of conditional probability of two uniform random variables between zero and $1$, it seem you are asking for $$\Pr[X+Y>1/2|Y<1/2,X<1/2]$$ – 2017-02-02

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Since $X$ and $Y$ both are less than $\frac{1}{2}$ hence draw this figure .it will be a square of area $\frac{1}{8}$. now if you want $x+y>\frac{1}{2}$ just see that if $x>\frac{1}{4}$ and if $y>\frac{1}{4}$ only then can the result be greater than $\frac{1}{2}.$

$P(x+y>\frac{1}{2})= \frac{Favourable Area}{Total Area}=\frac{\frac{1}{2}.\frac{1}{4}.\frac{1}{4}}{\frac{1}{2}\frac{1}{2}\frac{1}{2}}.$

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    This is nonsense. First, where is this "square of area $1/8$"? Second, it is certainly not necessary that both $x$ and $y$ exceed $1/4$, e.g., you could have $x=.4$, $y=.2$. – 2017-02-02