Menu Close

please-help-me-solve-these-two-questions-1-A-magician-cuts-a-rope-into-two-parts-at-a-point-selected-at-random-what-is-the-probability-that-the-length-of-the-longer-rope-is-at-least-8-times-




Question Number 50299 by OTCHRRE ABDULLAI last updated on 15/Dec/18
please help me solve these two  questions   1. A magician cuts a rope into two     parts at a point selected at   random. what is the probability that  the length of  the longer  rope is at least   8 times the length of the shorter   rope.  2. find the sum of co−efficient of   thebinomial  expansion  of    (4x−1)^(16)
$${please}\:{help}\:{me}\:{solve}\:{these}\:{two} \\ $$$${questions}\: \\ $$$$\mathrm{1}.\:{A}\:{magician}\:{cuts}\:{a}\:{rope}\:{into}\:{two}\: \\ $$$$\:\:{parts}\:{at}\:{a}\:{point}\:{selected}\:{at}\: \\ $$$${random}.\:{what}\:{is}\:{the}\:{probability}\:{that} \\ $$$${the}\:{length}\:{of}\:\:{the}\:{longer}\:\:{rope}\:{is}\:{at}\:{least}\: \\ $$$$\mathrm{8}\:{times}\:{the}\:{length}\:{of}\:{the}\:{shorter}\: \\ $$$${rope}. \\ $$$$\mathrm{2}.\:{find}\:{the}\:{sum}\:{of}\:{co}−{efficient}\:{of}\: \\ $$$${thebinomial}\:\:{expansion}\:\:{of}\:\: \\ $$$$\left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{16}} \\ $$$$ \\ $$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18
please sir i had the probability to be   1−(5/9)=0.444   using an idea from your diagram  please check for me sir
$${please}\:{sir}\:{i}\:{had}\:{the}\:{probability}\:{to}\:{be}\: \\ $$$$\mathrm{1}−\frac{\mathrm{5}}{\mathrm{9}}=\mathrm{0}.\mathrm{444}\: \\ $$$${using}\:{an}\:{idea}\:{from}\:{your}\:{diagram} \\ $$$${please}\:{check}\:{for}\:{me}\:{sir} \\ $$
Commented by Abdo msup. last updated on 15/Dec/18
we have (4x−1)^(16)  =Σ_(k=0) ^(16)   C_(16) ^k (4x)^k (−1)^(16−k)   =Σ_(k=0) ^(16)   C_(16) ^k  (−4)^k  x^k  ⇒sum of coefficient is  =Σ_(k=0) ^(16)  C_(16) ^k (−4)^k  =(−4+1)^(16)  =3^(16)  .
$${we}\:{have}\:\left(\mathrm{4}{x}−\mathrm{1}\right)^{\mathrm{16}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{16}} \:\:{C}_{\mathrm{16}} ^{{k}} \left(\mathrm{4}{x}\right)^{{k}} \left(−\mathrm{1}\right)^{\mathrm{16}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{16}} \:\:{C}_{\mathrm{16}} ^{{k}} \:\left(−\mathrm{4}\right)^{{k}} \:{x}^{{k}} \:\Rightarrow{sum}\:{of}\:{coefficient}\:{is} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{16}} \:{C}_{\mathrm{16}} ^{{k}} \left(−\mathrm{4}\right)^{{k}} \:=\left(−\mathrm{4}+\mathrm{1}\right)^{\mathrm{16}} \:=\mathrm{3}^{\mathrm{16}} \:. \\ $$
Commented by mr W last updated on 16/Dec/18
to Abdullai sir:  now you can also solve if the question  is “what is the probability that the  length of the longer rope is at least  6 times but at most 9 times of the  shorter rope”. can you give it a try?
$${to}\:{Abdullai}\:{sir}: \\ $$$${now}\:{you}\:{can}\:{also}\:{solve}\:{if}\:{the}\:{question} \\ $$$${is}\:“{what}\:{is}\:{the}\:{probability}\:{that}\:{the} \\ $$$${length}\:{of}\:{the}\:{longer}\:{rope}\:{is}\:{at}\:{least} \\ $$$$\mathrm{6}\:{times}\:{but}\:{at}\:{most}\:\mathrm{9}\:{times}\:{of}\:{the} \\ $$$${shorter}\:{rope}''.\:{can}\:{you}\:{give}\:{it}\:{a}\:{try}? \\ $$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18
Thank you very much my boss  God bless you
$${Thank}\:{you}\:{very}\:{much}\:{my}\:{boss} \\ $$$${God}\:{bless}\:{you} \\ $$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18
i will sir
$${i}\:{will}\:{sir} \\ $$
Commented by mr W last updated on 16/Dec/18
no sir! try to draw a disgram which  shows the areas where the cut can be  placed and then calculate the sum of  these areas. the fraction of these areas  is the result.  the result should be (3/(35)), i think.
$${no}\:{sir}!\:{try}\:{to}\:{draw}\:{a}\:{disgram}\:{which} \\ $$$${shows}\:{the}\:{areas}\:{where}\:{the}\:{cut}\:{can}\:{be} \\ $$$${placed}\:{and}\:{then}\:{calculate}\:{the}\:{sum}\:{of} \\ $$$${these}\:{areas}.\:{the}\:{fraction}\:{of}\:{these}\:{areas} \\ $$$${is}\:{the}\:{result}. \\ $$$${the}\:{result}\:{should}\:{be}\:\frac{\mathrm{3}}{\mathrm{35}},\:{i}\:{think}. \\ $$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18
ok sir thanks
$${ok}\:{sir}\:{thanks} \\ $$
Answered by mr W last updated on 15/Dec/18
(1)  1−(7/9)=(2/9)≈0.222    (2)  Sum. of all coef.=(4×1−1)^(16) =3^(16)
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{1}−\frac{\mathrm{7}}{\mathrm{9}}=\frac{\mathrm{2}}{\mathrm{9}}\approx\mathrm{0}.\mathrm{222} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${Sum}.\:{of}\:{all}\:{coef}.=\left(\mathrm{4}×\mathrm{1}−\mathrm{1}\right)^{\mathrm{16}} =\mathrm{3}^{\mathrm{16}} \\ $$
Commented by mr W last updated on 15/Dec/18
Commented by OTCHRRE ABDULLAI last updated on 15/Dec/18
This man is great  please can you add explanation to   question 1 for me
$${This}\:{man}\:{is}\:{great} \\ $$$${please}\:{can}\:{you}\:{add}\:{explanation}\:{to}\: \\ $$$${question}\:\mathrm{1}\:{for}\:{me} \\ $$
Commented by mr W last updated on 15/Dec/18
when the cut is in the red area, the  length of the longer part is less than  the 8 times of the shorter part, this  probability is 7/9.  when the cut is in the green areas, the  length of the longer part is at least  8 times of the shorter part, this  probability is 1−7/9=2/9=0.222
$${when}\:{the}\:{cut}\:{is}\:{in}\:{the}\:{red}\:{area},\:{the} \\ $$$${length}\:{of}\:{the}\:{longer}\:{part}\:{is}\:{less}\:{than} \\ $$$${the}\:\mathrm{8}\:{times}\:{of}\:{the}\:{shorter}\:{part},\:{this} \\ $$$${probability}\:{is}\:\mathrm{7}/\mathrm{9}. \\ $$$${when}\:{the}\:{cut}\:{is}\:{in}\:{the}\:{green}\:{areas},\:{the} \\ $$$${length}\:{of}\:{the}\:{longer}\:{part}\:{is}\:{at}\:{least} \\ $$$$\mathrm{8}\:{times}\:{of}\:{the}\:{shorter}\:{part},\:{this} \\ $$$${probability}\:{is}\:\mathrm{1}−\mathrm{7}/\mathrm{9}=\mathrm{2}/\mathrm{9}=\mathrm{0}.\mathrm{222} \\ $$
Commented by Cheyboy last updated on 15/Dec/18
Thank alot Sir I have learn from it.
$$\mathrm{Thank}\:\mathrm{alot}\:\mathrm{Sir}\:\mathrm{I}\:\mathrm{have}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{it}. \\ $$$$ \\ $$
Commented by OTCHRRE ABDULLAI last updated on 15/Dec/18
In fact mr W  you are great God bless you
$${In}\:{fact}\:{mr}\:{W}\:\:{you}\:{are}\:{great}\:{God}\:{bless}\:{you} \\ $$
Commented by mr W last updated on 15/Dec/18
thanks alot sirs!
$${thanks}\:{alot}\:{sirs}! \\ $$
Commented by peter frank last updated on 15/Dec/18
yeah true  he is among of best  solver in this forum  i real appriciate him and GOD bless him
$$\mathrm{yeah}\:\mathrm{true}\:\:\mathrm{he}\:\mathrm{is}\:\mathrm{among}\:\mathrm{of}\:\mathrm{best} \\ $$$$\mathrm{solver}\:\mathrm{in}\:\mathrm{this}\:\mathrm{forum}\:\:\mathrm{i}\:\mathrm{real}\:\mathrm{appriciate}\:\mathrm{him}\:\mathrm{and}\:\mathrm{GOD}\:\mathrm{bless}\:\mathrm{him} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *