MHB Probability of Satisfying Floor Inequality with Random Real Numbers

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The discussion revolves around a probability problem involving two randomly chosen real numbers, a and b, and the condition that their sum satisfies the inequality \left\lfloor{a+b}\right\rfloor \le a+b - \dfrac{1}{4}. Participants engage in solving the problem and sharing their approaches. The correct solution was provided by a user named kaliprasad, who received congratulations for their contribution. The thread also emphasizes the importance of following the guidelines for participating in the Problem of the Week (POTW). Overall, the focus is on understanding and solving the specified mathematical inequality.
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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 $$\left\lfloor{a+b}\right\rfloor \le a+b - \dfrac{1}{4}$$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to kaliprasad for his correct answer!(Cool)

Solution from other:
Let $$a=\left\lfloor{a}\right\rfloor+x$$ and $$b=\left\lfloor{b}\right\rfloor+y$$ where $$x,y\in [0,\,1)$$. So we have

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

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

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

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

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