Question Number 33724 by mondodotto@gmail.com last updated on 22/Apr/18
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{three}}\:\boldsymbol{\mathrm{persons}}\:\boldsymbol{\mathrm{selected}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{random}} \\ $$$$\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{stopped}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{street}},\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{probability}}\:\boldsymbol{\mathrm{that}} \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{were}}\:\boldsymbol{\mathrm{born}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{friday}}? \\ $$$$\left(\boldsymbol{\mathrm{ii}}\right)\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{were}}\:\boldsymbol{\mathrm{born}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{friday}}\:\boldsymbol{\mathrm{and}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{others}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{thursday}}? \\ $$$$\left(\boldsymbol{\mathrm{iii}}\right)\boldsymbol{\mathrm{none}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{born}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{friday}} \\ $$$$ \\ $$
Commented by mondodotto@gmail.com last updated on 22/Apr/18
$$\mathrm{it}\:\mathrm{is}\:\mathrm{urgent}\:\mathrm{please}\:\mathrm{help}\:\mathrm{this} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Apr/18
$$\mathrm{52}\:{weeks}\:=\mathrm{1}{year}\:{so}\:\mathrm{52}\:{frida} \\ $$$$\left({i}\right)\mathrm{52}{C}_{\mathrm{3}} \:/\mathrm{365}{C}_{\mathrm{3}} \: \\ $$$$\left({ii}\right)\left(\mathrm{52}{C}_{\mathrm{2}} ×\mathrm{52}{C}_{\mathrm{1}} \:\right)/\mathrm{365}{C}_{\mathrm{3}\:} \\ $$$$\left({iio}\right)\mathrm{1}−\left(\mathrm{52}{C}_{\mathrm{3}} /\mathrm{365}{C}_{\mathrm{3}} \:\right) \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 22/Apr/18
$$\mathrm{week}\:\mathrm{has}\:\mathrm{7}\:\mathrm{days} \\ $$$$\left(\mathrm{i}\right)\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{343}} \\ $$$$\left(\mathrm{ii}\right)\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{3}} }×\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}=\frac{\mathrm{3}}{\mathrm{343}} \\ $$$$\left(\mathrm{iii}\right)\:\left(\frac{\mathrm{6}}{\mathrm{7}}\right)^{\mathrm{3}} =\frac{\mathrm{216}}{\mathrm{343}} \\ $$