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Question Number 18945 by Tinkutara last updated on 01/Aug/17
$$\mathrm{A}\:\mathrm{point}\:'\mathrm{A}'\:\mathrm{is}\:\mathrm{randomly}\:\mathrm{chosen}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{of}\:\mathrm{side}\:\mathrm{length}\:\mathrm{1}\:\mathrm{unit}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{probability}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{A}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{does}\:\mathrm{not}\:\mathrm{exceed}\:\mathrm{x}. \\ $$
Commented by dioph last updated on 02/Aug/17
$$\mathrm{Area}\:\mathrm{of}\:\mathrm{square}\:{S}:\:{A}_{{S}} \:=\:\mathrm{1} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{circle}\:{C}\:\mathrm{with}\:\mathrm{radius}\:{x}:\:{A}_{{C}} \:=\:\:\pi{x}^{\mathrm{2}} \\ $$$$\mathrm{say}\:\mathrm{point}\:\mathrm{is}\:{P}: \\ $$$${p}\left({P}\:\in\:{C}\:\mid\:{P}\:\in\:{S}\right)\:=\:\frac{{p}\left({P}\:\in\:{C}\:\cap\:{S}\right)}{{p}\left({P}\:\in\:{S}\right)} \\ $$$$=\:\begin{cases}{\frac{{A}_{{C}} }{{A}_{{S}} }\:=\:\pi{x}^{\mathrm{2}} ,\:\mathrm{if}\:{x}\:\leqslant\:\mathrm{0}.\mathrm{5}}\\{\frac{{A}_{{C}\:\cap\:{S}} }{{A}_{{S}} }\:=\:...,\:\mathrm{if}\:\mathrm{0}.\mathrm{5}\:<\:{x}\:\leqslant\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\frac{{A}_{{S}} }{{A}_{{S}} }\:=\:\mathrm{1},\:\mathrm{if}\:{x}\:>\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{working}\:\mathrm{on}\:\mathrm{second}\:\mathrm{case} \\ $$