# 64-points-are-in-a-plane-x-y-x-0-1-2-7-y-0-1-2-7-4-points-are-chosen-at-random-What-is-the-proabability-the-lines-connecting-them-do-not-form-a-square-or-rectangle-

Question Number 2771 by prakash jain last updated on 26/Nov/15
$$\mathrm{64}\:\mathrm{points}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}: \\$$$$\left({x},{y}\right),\:{x}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\},\:{y}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\} \\$$$$\mathrm{4}\:\mathrm{points}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{at}\:\mathrm{random}. \\$$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{proabability}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{connecting} \\$$$$\mathrm{them}\:\mathrm{do}\:\mathrm{not}\:\mathrm{form}\:\mathrm{a}\:\mathrm{square}\:\mathrm{or}\:\mathrm{rectangle}? \\$$
Answered by Rasheed Soomro last updated on 01/Dec/15
$$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{Guide}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{if}}\:\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{am}}\:\boldsymbol{\mathrm{wrong}}. \\$$$$\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{don}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{know}}\:\boldsymbol{\mathrm{much}}\:\boldsymbol{\mathrm{about}}\:\mathcal{PROBABILITY}. \\$$$$\boldsymbol{{Assuming}}\:\boldsymbol{{that}}\:\boldsymbol{{sides}}\:\boldsymbol{{of}}\:\boldsymbol{{rectangle}}\:\boldsymbol{{are}} \\$$$$\boldsymbol{{horizantal}}\:\boldsymbol{{and}}\:\boldsymbol{{vertical}}. \\$$$$\boldsymbol{{Assuming}}\:\boldsymbol{{that}}\:\boldsymbol{{all}}\:\boldsymbol{{vertices}}\:\boldsymbol{{are}}\:\boldsymbol{{different}}\:\boldsymbol{{points}}. \\$$$${Being}\:\overline {{P}_{\mathrm{1}} {P}_{\mathrm{2}} }\:\:{horizantal},\:{P}_{\mathrm{1}} \:{and}\:{P}_{\mathrm{2}\:} \:{have}\:{same}\:{y}−{coordinate}. \\$$$${For}\:{same}\:{reason}\:{P}_{\mathrm{3}} \:{and}\:{P}_{\mathrm{4}} \:{have}\:{same}\:{y}−{coordinate}. \\$$$${Being}\:\overline {{P}_{\mathrm{1}} {P}_{\mathrm{4}} }\:\:{and}\:\overline {{P}_{\mathrm{2}} {P}_{\mathrm{3}} }\:{vertical},\:{P}_{\mathrm{1}} \:{and}\:{P}_{\mathrm{4}} \:{have}\:{same} \\$$$${x}−{coordinate},\:{and}\:{P}_{\mathrm{2}} \:{and}\:{P}_{\mathrm{3}\:} {have}\:{same}\:{x}−{coordinate}. \\$$$${Let} \\$$$${P}_{\mathrm{1}} =\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right),{P}_{\mathrm{2}} =\left({x}_{\mathrm{2}} ,{y}_{\mathrm{1}} \right),{P}_{\mathrm{3}} \left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right)\:\:\:{and}\:\:\:{P}_{\mathrm{4}} =\left({x}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right) \\$$$${where}\:{x}_{{i}} \:\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\},{y}_{{i}} \:\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},…,\mathrm{7}\right\} \\$$$$\underset{−} {{We}\:{first}\:{consider}\:{possibility}\:{of}\:{being}\:{rectangle}/{square}.} \\$$$$\boldsymbol{{Possibility}}\:\boldsymbol{{of}}\:\boldsymbol{{cboosing}}\:\boldsymbol{{P}}_{\mathrm{1}} ,\boldsymbol{{P}}_{\mathrm{2}} ,\boldsymbol{{P}}_{\mathrm{3}} \:\boldsymbol{{and}}\:\boldsymbol{{P}}_{\mathrm{4}} \: \\$$$$\boldsymbol{{is}}\:\boldsymbol{{Possibility}}\:\boldsymbol{{of}}\:\boldsymbol{{cboosing}}\:\boldsymbol{{their}}\:\boldsymbol{{coordinates}} \\$$$$\boldsymbol{{x}}_{\mathrm{1}} ,\boldsymbol{{y}}_{\mathrm{1}} ,\boldsymbol{{x}}_{\mathrm{2}} \:\:\boldsymbol{{and}}\:\boldsymbol{{y}}_{\mathrm{2}} \\$$$$\mathcal{C}{hoosing}\:{x}_{\mathrm{1}} \:{has}\:\frac{\mathrm{1}}{\mathrm{8}}\:{possibility}\:{and}\:{y}_{\mathrm{1}} \:{has}\:\frac{\mathrm{1}}{\mathrm{8}}\:{possibility}. \\$$$${Hence}\:{choosing}\:{P}_{\mathrm{1}} \:{has}\:\frac{\mathrm{1}}{\mathrm{64}}\:{possibility}. \\$$$${Choosing}\:{x}_{\mathrm{2}} \:\left[\neq{x}_{\mathrm{1}} \right]\:{after}\:{choosing}\:{x}_{\mathrm{1}} \:{has}\:\frac{\mathrm{1}}{\mathrm{7}}\:{possibility}. \\$$$${Choosing}\:{x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{has}\:\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{7}}\:=\frac{\mathrm{1}}{\mathrm{448}}\:{possibility}. \\$$$${Choosing}\:{y}_{\mathrm{2}\:} {after}\:{choosing}\:{y}_{\mathrm{1}} \:{has}\:\frac{\mathrm{1}}{\mathrm{7}}\:{possibility}.\: \\$$$${Choosing}\:{all}\:{the}\:{coordinates}\:{x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{x}_{\mathrm{2}} \:{and}\:\:{y}_{\mathrm{2}} \:\:{in}\:{succesion} \\$$$${has}\:\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{7}}×\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{3136}}\:{possibility}\:{of}\:\underset{−} {{being}}\:{rectangle}. \\$$$${Possibility}\:{of}\:\underset{−} {{not}\:{being}}\:{rectangle}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3136}}=\frac{\mathrm{3135}}{\mathrm{3136}} \\$$
Commented by prakash jain last updated on 04/Dec/15
$$\mathrm{May}\:\mathrm{the}\:\mathrm{question}\:\mathrm{was}\:\mathrm{not}\:\mathrm{clear}\:\mathrm{but}\:\mathrm{the}\:\mathrm{coordiantes} \\$$$$\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:\mathrm{are}\:\left({x},{y}\right)\:\mathrm{are}\:\left(\mathrm{0},\mathrm{0}\right)..\left(\mathrm{0},\mathrm{7}\right) \\$$$$\left(\mathrm{1},\mathrm{0}\right)\:\mathrm{to}\:\left(\mathrm{1},\mathrm{7}\right) \\$$$$\mathrm{so}\:\mathrm{you}\:\mathrm{know}\:\mathrm{when}\:\mathrm{two}\:\mathrm{lines}\:\mathrm{are}\:\bot^{{r}} . \\$$
Commented by Rasheed Soomro last updated on 30/Nov/15
$$\bullet\mathcal{I}\:{understood}\:{from}\:\:{your}\:{question}:\:'{the}\:{square}\:{array} \\$$$${of}\:\mathrm{64}\:{points}\:{of}\:{order}\:\mathrm{8}×\mathrm{8}\:{from}\:\left(\mathrm{0},\mathrm{0}\right)\:{to}\:\left(\mathrm{7},\mathrm{7}\right)'.\:{Am}\:{I}\:{wrong}? \\$$$$\bullet{Are}\:{the}\:{sides}\:{of}\:{rectangle}\:{horizantal}\:{and}\:{vertical}\:{only}? \\$$
Commented by Rasheed Soomro last updated on 03/Dec/15
$${What}\:{is}\:{the}\:{meaning}\:{of}\:'\:{two}\:\boldsymbol{{points}}\:{are}\:\bot^{{r}} \:' \\$$
Commented by prakash jain last updated on 04/Dec/15
$$\bot^{{r}} =\mathrm{perpendicular}.\:{A}\mathrm{ctually}\:\mathrm{I}\:\mathrm{meant}\:\mathrm{lines}. \\$$$$\mathrm{I}\:\mathrm{was}\:\mathrm{travelling}\:\mathrm{for}\:\mathrm{last}\:\mathrm{few}\:\mathrm{days}\:\mathrm{returned} \\$$$$\mathrm{only}\:\mathrm{today}. \\$$
Commented by Rasheed Soomro last updated on 04/Dec/15
$$\mathcal{T}{h}\mathcal{A}{nk}\mathcal{S}! \\$$