Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 177884 by DAVONG last updated on 10/Oct/22

Answered by mr W last updated on 10/Oct/22

$$6blackandnredballsinbag.$$$$totake3ballstherearetotally$$$$C_3^{n+6}ways.$$$$totake1blackand2redballsthere$$$$are{C}_1^6{C}_2^{n}ways.$$$$P_n=\frac{C_1^6{C}_2^n}{C_3^{n+6}}$$$$=\frac{6\times n(n-1)\times 6}{2\times (n+6)(n+5)(n+4)}$$$$=\frac{18n(n-1)}{(n+6)(n+5)(n+4)}$$$$P_{n}ismaximumwhenn=11or12$$$$P_{n,max}=\frac{18\times 11\times 10}{17\times 16\times 15}=\frac{33}{68}$$$$\Rightarrow \left(D\right)$$

Commented by mr W last updated on 10/Oct/22

$${it}^{\prime }snoteasytoseewhen{P}_{n}is$$$$maximum.ijustusedthefunction$$$$f\left(x\right)=\frac{18x(x-1)}{(x+6)(x+5)(x+4)}andfound$$$$itsmaximumatx\approx 11.5.sincen$$$$isinteger,weget{P}_{n}ismaximum$$$$whenn=11or12.$$

Commented by Acem last updated on 10/Oct/22

$$Whatdoestheamount{P}_{n}referto?$$

Commented by Acem last updated on 10/Oct/22

$$Doesitindicattheprobabilityofthisevent$$$$C_1^6\times C_2^{n}occuring?$$

Commented by mr W last updated on 10/Oct/22

$$yes.asthequestionsays:P_{n}isthe$$$$probabilityoftaking1blackballand$$$$2redballs.$$

Commented by Acem last updated on 11/Oct/22

$$Yes!$$$$f\left(n\right)=\frac{18n^2-18n}{n^3+15n^2+74n+120}={\mathit{P}}_{\mathit{n}}$$$$f^{{}^{\prime }}\left(n\right)=\frac{-18n^4+36n^3+1602n^2+4320n-2160}{{(n^3+15n^2+74n+120)}^2}$$$$f^{{}^{\prime }}\left(n\right)=0$$$$-18n^4+36n^3+1602n^2+4320n-2160=0$$$$n_1\cong -4.38,n_2\cong -5.54,n_3\cong 0.43Refuses$$$${\mathit{n}}_4\cong 11.5$$$$\{\begin{array}{l}n=11\Rightarrow P_{11,max}=0.48529411764706\\ n=12\Rightarrow P_{12,max}=0.48529411764706\end{array}$$$$$$$$\mathit{N}\mathit{o}\mathit{t}\mathit{e}:Igot{n}_{1-4}fromGraphcalculator$$$$\mathit{H}\mathit{o}\mathit{w}\mathit{c}\mathit{a}\mathit{n}\mathit{i}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{a}\mathit{n}\mathit{e}\mathit{q}\mathit{u}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}{\mathit{a}\mathit{x}}^4+....$$$$$$$$$$

Commented by mr W last updated on 11/Oct/22

$$usuallyitisaquickwaytodetermine$$$$P_{n,max}usinggraphcalculator.$$$$butreallysolving{f}^{\prime }\left(n\right)=0isnota$$$$goodway,evenimpossiblelikeabove.$$$$besides,inanexaminationitisnot$$$$allowedtousegraphcalculator.$$$$$$$$thebestandyeteasyandexactway$$$$isthefollowing:$$$$weonlyneedtofindthe{n}_{max}such$$$$that{P}_n⩾P_{n-1},i.e.$$$$\frac{18n(n-1)}{(n+6)(n+5)(n+4)}⩾\frac{18(n-1)(n-2)}{(n+5)(n+4)(n+3)}$$$$\Rightarrow \frac{n}{n+6}⩾\frac{n-2}{n+3}$$$$n^2+3n⩾n^2+4n-12$$$$n⩽12$$$$thatmeanstilln=12,P_n⩾P_{n-1}$$$$anduponn=13,P_n<P_{n-1}.$$$$so{P}_{12}isthemaximum.$$$$P_{12}=\frac{18\times 12\times 11}{18\times 17\times 16}=\frac{33}{68}$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by Acem last updated on 11/Oct/22

$$The{P}_{n\mathrm{\_}max}⩾P_{n-1}isa\mathit{g}\mathit{e}\mathit{n}\mathit{i}\mathit{u}\mathit{s}\mathit{i}\mathit{d}\mathit{e}\mathit{a}!!$$$$BigThanks$$$$$$$$\mathit{N}\mathit{o}\mathit{t}\mathit{e}:Inordertoavoidthestudentfrom$$$$calculatingthecasesinwhichnless$$$$than12,inparallelwith{P}_{n\mathrm{\_}max}⩾P_{n-1}$$$$weaddbesideanotherlogicalinequality:$$$${\mathit{P}}_{\mathit{n}\mathbf{\_}\mathit{m}\mathit{a}\mathit{x}}⩾{\mathit{P}}_{\mathit{n}+1}\mathrm{\&}{P}_{n\mathrm{\_}max}⩾P_{n-1}$$$$$$$$\frac{18n(n-1)}{(n+6)(n+5)(n+4)}⩾\frac{18(n+1).n}{(n+7)(n+6)(n+5)}$$$$$$$$\frac{n-1}{n+4}⩾\frac{n+1}{n+7}\Rightarrow n⩾11$$$$$$$$Thenwehaven⩽12\mathrm{\&}n⩾11i.e.$$$$$$$$n_1=11,n_2=12bothachievetobe{P}_{11or12\mathrm{\_}max}$$$$P_{11or12-max}=\frac{33}{68}$$$$$$$$Mygreetings$$

Commented by Acem last updated on 11/Oct/22

$$Itisnecessarytodothat$$$$Becausefromonlythisn⩽12$$$$thestudentmaythinkthatallvaluesofn$$$$lessthan12i.e.(12,11,10...,2)redballs$$$$achievetobe{P}_{n⩽12\mathrm{\_}max}$$

Commented by mr W last updated on 11/Oct/22

$$thenwecandothis$$$$P_n<P_{n+1}$$$$\frac{18n(n-1)}{(n+6)(n+5)(n+4)}<\frac{18(n-1)(n-2)}{(n+5)(n+4)(n+3)}$$$$\frac{n}{n+6}<\frac{n-2}{n+3}$$$$\Rightarrow n>12$$$$i.e.uponn=13,P_n<P_{n+1}$$$$\Rightarrow P_{12}ismaximum.$$$$we{don}^{\prime }tneedtoknowwhether{P}_{11}$$$$isalsomaximum.$$

Commented by Acem last updated on 12/Oct/22

$$$$$$Infactthe\frac{18(n-1)(n-2)}{(n+5)(n+4)(n+3)}isnot{P}_{n+1}$$$$Also,whenwewanttosetnwhichmake{P}_{max}$$$$itmustbe{P}_{n,max}greaterthantermbefore$$$$andafteriti.e.$$$$P_{n\mathrm{\_}max}⩾P_{n+1}\mathrm{\&}{P}_{n\mathrm{\_}max}⩾P_{n-1}$$$$Finallyisaythatn⩽12isnotenough$$$$because12redballsmakes{P}_{max}$$$$andalso11redballsdoes$$$$$$$$Inadditiontothat,then⩽12makesthe$$$$studentillusionthatallvaluesless$$$$than12make{P}_{max}$$$$$$

Commented by mr W last updated on 12/Oct/22

$$asisaidwhenweonlyneedtoget{P}_{n,max}$$$${it}^{\prime }senoughtoknowthat{P}_{12}the$$$$maximum.we{don}^{\prime }tneedtocare$$$$whether{P}_{11}isalsomaximum.$$$$from$$$$P_1⩽P_2⩽P_3⩽....⩽P_{11}⩽P_{12}>P_{13}>P_{14}>...$$$$wecanbesurethat{P}_{12}isthemaximum.$$$$itisnotsaidthatonly{P}_{12}isthe$$$$maximum,butthisisforthe$$$$questionnotimportant.$$

Commented by Acem last updated on 13/Oct/22

$$$$$${It}^{\prime }sokandgood$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com