Question Number 205394 by Red1ight last updated on 19/Mar/24
$$\mathrm{let}\:{x}\:\mathrm{and}\:{y}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y}\in\mathbb{R} \\ $$$$\mid{x}−{y}\mid\geqslant{d} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{10} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\left.{a}\right)\:{d}=\mathrm{0} \\ $$$$\left.{b}\right)\:{d}=\mathrm{1} \\ $$$$\left.{c}\right)\:{d}=\mathrm{2} \\ $$
Answered by mr W last updated on 20/Mar/24
Commented by mr W last updated on 20/Mar/24
$$\mid{x}−{y}\mid\geqslant{d} \\ $$$$\Rightarrow\begin{cases}{{x}−{y}\geqslant{d}\:\Rightarrow{y}\leqslant{x}−{d}}\\{{or}}\\{{x}−{y}\leqslant−{d}\:\Rightarrow{y}\geqslant{x}+{d}}\end{cases} \\ $$$$\mid{x}−{y}\mid\leqslant{d}\:{means}\:{when}\:{point}\:{lies}\:{in}\: \\ $$$${hatched}\:{zones}. \\ $$$${x}+{y}\leqslant\mathrm{10}\:\Rightarrow{y}\leqslant\mathrm{10}−{x}\:\Rightarrow{shaded}\:{area} \\ $$$${p}=\frac{\left(\mathrm{10}−{d}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{10}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{{d}}{\mathrm{10}}\right)^{\mathrm{2}} \\ $$$${p}_{{d}=\mathrm{0}} =\mathrm{0}.\mathrm{5} \\ $$$${p}_{{d}=\mathrm{1}} =\mathrm{0}.\mathrm{405} \\ $$$${p}_{{d}=\mathrm{2}} =\mathrm{0}.\mathrm{32} \\ $$