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)


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Question Number 50299 by OTCHRRE ABDULLAI last updated on 15/Dec/18

$p l e a s e h e l p m e s o l v e t h e s e t w o$ $q u e s t i o n s$ $1 . A m a g i c i a n c u t s a r o p e i n t o t w o$ $p a r t s a t a p o i n t s e l e c t e d a t$ $r a n d o m . w h a t i s t h e p r o b a b i l i t y t h a t$ $t h e l e n g t h o f t h e l o n g e r r o p e i s a t l e a s t$ $8 t i m e s t h e l e n g t h o f t h e s h o r t e r$ $r o p e .$ $2 . f i n d t h e s u m o f c o - e f f i c i e n t o f$ $t h e b i n o m i a l e x p a n s i o n o f$ ${(4 x - 1)}^{16}$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18

$p l e a s e s i r i h a d t h e p r o b a b i l i t y t o b e$ $1 - \frac{5}{9} = 0.444$ $u s i n g a n i d e a f r o m y o u r d i a g r a m$ $p l e a s e c h e c k f o r m e s i r$
Commented by Abdo msup. last updated on 15/Dec/18

$w e h a v e {(4 x - 1)}^{16} = \sum_{k = 0}^{16} C_{16}^{k} {(4 x)}^{k} {(- 1)}^{16 - k}$ $= \sum_{k = 0}^{16} C_{16}^{k} {(- 4)}^{k} x^{k} \Rightarrow s u m o f c o e f f i c i e n t i s$ $= \sum_{k = 0}^{16} C_{16}^{k} {(- 4)}^{k} = {(- 4 + 1)}^{16} = 3^{16} .$
Commented by mr W last updated on 16/Dec/18

$t o A b d u l l a i s i r :$ $n o w y o u c a n a l s o s o l v e i f t h e q u e s t i o n$ $i s ‘ ‘ w h a t i s t h e p r o b a b i l i t y t h a t t h e$ $l e n g t h o f t h e l o n g e r r o p e i s a t l e a s t$ $6 t i m e s b u t a t m o s t 9 t i m e s o f t h e$ $s h o r t e r {r o p e}^{″} . c a n y o u g i v e i t a t r y ?$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18

$T h a n k y o u v e r y m u c h m y b o s s$ $G o d b l e s s y o u$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18

$i w i l l s i r$
Commented by mr W last updated on 16/Dec/18
$no sir! try to draw a disgram which shows the areas where the cut can be placed and then calculate the sum of these areas. the fraction of these areas is the result. the result should be (3/(35)), i think.$
$n o s i r! t r y t o d r a w a d i s g r a m w h i c h$ $s h o w s t h e a r e a s w h e r e t h e c u t c a n b e$ $p l a c e d a n d t h e n c a l c u l a t e t h e s u m o f$ $t h e s e a r e a s . t h e f r a c t i o n o f t h e s e a r e a s$ $i s t h e r e s u l t .$ $t h e r e s u l t s h o u l d b e \frac{3}{35}, i t h i n k .$
Commented by OTCHRRE ABDULLAI last updated on 16/Dec/18

$o k s i r t h a n k s$
	Answered by mr W last updated on 15/Dec/18

	$(1)$ $1 - \frac{7}{9} = \frac{2}{9} \approx 0.222$ $(2)$ $S u m . o f a l l c o e f . = {(4 \times 1 - 1)}^{16} = 3^{16}$
	Commented by mr W last updated on 15/Dec/18

	Commented by OTCHRRE ABDULLAI last updated on 15/Dec/18

	$T h i s m a n i s g r e a t$ $p l e a s e c a n y o u a d d e x p l a n a t i o n t o$ $q u e s t i o n 1 f o r m e$
	Commented by mr W last updated on 15/Dec/18

	$w h e n t h e c u t i s i n t h e r e d a r e a, t h e$ $l e n g t h o f t h e l o n g e r p a r t i s l e s s t h a n$ $t h e 8 t i m e s o f t h e s h o r t e r p a r t, t h i s$ $p r o b a b i l i t y i s 7 / 9 .$ $w h e n t h e c u t i s i n t h e g r e e n a r e a s, t h e$ $l e n g t h o f t h e l o n g e r p a r t i s a t l e a s t$ $8 t i m e s o f t h e s h o r t e r p a r t, t h i s$ $p r o b a b i l i t y i s 1 - 7 / 9 = 2 / 9 = 0.222$
	Commented by Cheyboy last updated on 15/Dec/18

	$Thank alot Sir I have learn from it .$
	Commented by OTCHRRE ABDULLAI last updated on 15/Dec/18

	$I n f a c t m r W y o u a r e g r e a t G o d b l e s s y o u$
	Commented by mr W last updated on 15/Dec/18

	$t h a n k s a l o t s i r s!$
	Commented by peter frank last updated on 15/Dec/18

	$yeah true he is among of best$ $solver in this forum i real appriciate him and GOD bless him$
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