Question Number 207752 by efronzo1 last updated on 25/May/24 $$\:\mathrm{Two}\:\mathrm{ships}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{berth}\: \\ $$$$\:\mathrm{in}\:\mathrm{a}\:\mathrm{port}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{the}\: \\ $$$$\:\mathrm{arrival}\:\mathrm{times}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{ships}\: \\ $$$$\:\mathrm{are}\:\mathrm{independent}\:\mathrm{and}\:\mathrm{have}\:\mathrm{the}\: \\ $$$$\:\mathrm{same}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{docking}\: \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{Sunday}\:\left(\mathrm{00}.\mathrm{00}−\mathrm{24}.\mathrm{00}\right) \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{ship} \\ $$$$\:\mathrm{is}\:\mathrm{2}\:\mathrm{hours}\:\mathrm{and}\:\mathrm{the}\:\mathrm{berth}\:\mathrm{time} \\…
Question Number 207751 by efronzo1 last updated on 25/May/24 $$\:\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{a}\:\mathrm{balanced}\:\mathrm{6}−\mathrm{sided}\: \\ $$$$\:\mathrm{dice}\:\mathrm{originally}\:\mathrm{had}\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\:\mathrm{and}\:\mathrm{7}. \\ $$$$\:\mathrm{The}\:\mathrm{dice}\:\mathrm{wre}\:\mathrm{thrown}\:\mathrm{once}\:\mathrm{and}\: \\ $$$$\:\mathrm{the}\:\mathrm{result}\:\mathrm{was}\:\mathrm{observed}.\:\mathrm{If}\:\mathrm{an}\: \\ $$$$\mathrm{odd}\:\mathrm{numbers}\:\mathrm{appears},\:\mathrm{than}\:\mathrm{the}\: \\ $$$$\:\mathrm{number}\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{with}\:\mathrm{the}\: \\ $$$$\:\mathrm{number}\:\mathrm{8}.\:\mathrm{However},\:\mathrm{if}\:\mathrm{an}\:\mathrm{even}\: \\ $$$$\:\mathrm{number}\:\mathrm{appears}\:,\:\mathrm{the}\:\mathrm{number} \\…
Question Number 207665 by efronzo1 last updated on 22/May/24 $$\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{0}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{1}}\end{pmatrix}\:+\frac{\mathrm{1}}{\mathrm{3}}\begin{pmatrix}{\mathrm{20}}\\{\:\:\mathrm{2}}\end{pmatrix}\:+…+\frac{\mathrm{1}}{\mathrm{21}}\:\begin{pmatrix}{\mathrm{20}}\\{\mathrm{20}}\end{pmatrix}\:=? \\ $$ Answered by Tinku Tara last updated on 22/May/24 $$\left(\mathrm{1}+{x}\right)^{\mathrm{20}} =\underset{{n}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\begin{pmatrix}{\mathrm{20}}\\{{n}}\end{pmatrix}{x}^{{n}} \\…
Question Number 207387 by sniper237 last updated on 13/May/24 $${Let}\:\:{cardE}={n}\:,\:{and}\:\:{the}\:{set}\:{of}\:{parts} \\ $$$${S}=\left\{\left({A},{B}\right)\in{P}\left({E}\right)×{P}\left({E}\right)\:/\:\:{A}\cap{B}=\varnothing\right\} \\ $$$${Show}\:{that}\:\:{cardS}=\:\mathrm{3}^{{n}} \\ $$ Answered by Berbere last updated on 13/May/24 $${if}\:{card}\left({A}\right)={k};{E}={A}\cup\overset{−} {{A}}\:\:{the}\:{number}\:{of}\:{subset}\:{of}\:{card}={k}…
Question Number 207374 by sniper237 last updated on 13/May/24 $${Show}\:{that}\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$ Answered by mr W last updated on…
Question Number 206514 by cortano21 last updated on 17/Apr/24 $$\:{In}\:{this}\:{covid}\:−\mathrm{19}\:{pandemic},\:{it}\:{is}\: \\ $$$$\:{known}\:{that}\:{are}\:\mathrm{5},\mathrm{667},\mathrm{355}\:{confirmed}\: \\ $$$$\:{cases}\:{out}\:{of}\:\mathrm{273},\mathrm{500},\mathrm{000}\:{in}\:{X}\:{country} \\ $$$$\:{population}\:{based}\:{WHO}.\: \\ $$$$\:{One}\:{of}\:{the}\:{equipment}\:{to}\:{test}\:{the} \\ $$$$\:{covid}−\mathrm{19}\:{is}\:{GeNose}\:{C}\mathrm{19}−{S}\:{developed} \\ $$$$\:{by}\:{UGM}.\:{GeNose}\:{C}\mathrm{19}−{S}\:{is}\:{a}\:{rapid} \\ $$$$\:{screening}\:{equipment}\:{for}\:{Sars}−{CoV}\mathrm{2} \\…
Question Number 205409 by Red1ight last updated on 20/Mar/24 $$\mathrm{let}\:{x},\:{y},\:{z}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y},{z}\in\mathbb{R} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\left.{a}\right)\:\mathrm{all}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{satisfied} \\ $$$$\mid{x}−{y}\mid\geqslant\mathrm{2} \\ $$$$\mid{x}−{z}\mid\geqslant\mathrm{2} \\ $$$$\mid{y}−{z}\mid\geqslant\mathrm{2} \\ $$$$\left.{b}\right)\:\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\…
Question Number 205394 by Red1ight last updated on 19/Mar/24 $$\mathrm{let}\:{x}\:\mathrm{and}\:{y}\:\mathrm{be}\:\mathrm{random}\:\mathrm{numbers}\:\mathrm{from}\:\mathrm{0}\:\mathrm{to}\:\mathrm{10} \\ $$$$\mathrm{where}\:{x},{y}\in\mathbb{R} \\ $$$$\mid{x}−{y}\mid\geqslant{d} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{10} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\left.{a}\right)\:{d}=\mathrm{0} \\ $$$$\left.{b}\right)\:{d}=\mathrm{1} \\ $$$$\left.{c}\right)\:{d}=\mathrm{2} \\…
Question Number 205230 by louisyanni last updated on 13/Mar/24 $$ \\ $$ Commented by Rasheed.Sindhi last updated on 13/Mar/24 $$\mathcal{R}{eserved}\:{space}\:{for}\:{a}\:{question}? \\ $$ Commented by Frix…
Question Number 201972 by Lekhraj last updated on 17/Dec/23 Answered by mr W last updated on 18/Dec/23 $$\boldsymbol{\Phi}_{\leqslant\boldsymbol{{x}}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\boldsymbol{{erf}}\left(\frac{\boldsymbol{{x}}−\boldsymbol{\mu}}{\:\sqrt{\mathrm{2}}\:\boldsymbol{\sigma}}\right)\right] \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{erf}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}}{\:\sqrt{\boldsymbol{\pi}}}\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{e}}^{−\boldsymbol{{t}}^{\mathrm{2}} } \boldsymbol{{dt}}…