Probability of Satisfying Floor Inequality with Random Real Numbers

[anemone]

Here is this week's POTW:

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Two real numbers $a$ and $b$ are chosen at random. Find the probability that they satisfy \(\displaystyle \left\lfloor{a+b}\right\rfloor \le a+b - \dfrac{1}{4}\).

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[anemone]

Congratulations to kaliprasad for his correct answer!(Cool)

Solution from other:

Spoiler

Let \(\displaystyle a=\left\lfloor{a}\right\rfloor+x\) and \(\displaystyle b=\left\lfloor{b}\right\rfloor+y\) where \(\displaystyle x,y\in [0,\,1)\). So we have

\(\displaystyle \begin{align*}a+b&=\left\lfloor{a+b}\right\rfloor+x+y\\\left\lfloor{a+b}\right\rfloor&=a+b-(x+y)\\x+y&=a+b-\left\lfloor{a+b}\right\rfloor\\\therefore\dfrac{1}{4}&\le x+y \end{align*}\)

\(\displaystyle \implies \dfrac{1}{4}\le x+y <1 \,\,\text{and}\,\,\dfrac{5}{4}\le x+y<2\) subject to \(\displaystyle x,y\in [0,\,1)\)

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The first inequality has the probability of \(\displaystyle 1(1)-\frac{1}{2}\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)-\frac{1}{2}=\frac{15}{32}\).

The second inequality has the probability of \(\displaystyle \frac{1}{2}\left(\frac{3}{4}\right)\left(\frac{3}{4}\right)=\frac{9}{32}\).

The required probability is hence \(\displaystyle \frac{15}{32}+\frac{9}{32}=\frac{3}{4}\).