Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 149458 by mathdanisur last updated on 05/Aug/21

Commented by EDWIN88 last updated on 06/Aug/21

n(S)=C_2 ^7 =((7×6)/(2×1))=21 n(A)= 18 A={(1,2),(1,4),(1,6),(2,3),(2,4), (2,5),(2,6),(3,4),(3,6),(4,5),(4,6), (5,6),(0,1),(0,2),(0,3),(0,4),(0,5) ,(0,6)} then P(A)=((18)/(21)) =(6/7)

$${n}\left({S}\right)={C}_{\mathrm{2}} ^{\mathrm{7}} =\frac{\mathrm{7}×\mathrm{6}}{\mathrm{2}×\mathrm{1}}=\mathrm{21} \\ $$$${n}\left({A}\right)=\:\mathrm{18} \\ $$$${A}=\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{1},\mathrm{4}\right),\left(\mathrm{1},\mathrm{6}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right),\right. \\ $$$$\:\:\:\:\:\left(\mathrm{2},\mathrm{5}\right),\left(\mathrm{2},\mathrm{6}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{3},\mathrm{6}\right),\left(\mathrm{4},\mathrm{5}\right),\left(\mathrm{4},\mathrm{6}\right), \\ $$$$\:\:\:\:\left(\mathrm{5},\mathrm{6}\right),\left(\mathrm{0},\mathrm{1}\right),\left(\mathrm{0},\mathrm{2}\right),\left(\mathrm{0},\mathrm{3}\right),\left(\mathrm{0},\mathrm{4}\right),\left(\mathrm{0},\mathrm{5}\right) \\ $$$$\left.\:\:,\left(\mathrm{0},\mathrm{6}\right)\right\} \\ $$$${then}\:{P}\left({A}\right)=\frac{\mathrm{18}}{\mathrm{21}}\:=\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Commented by EDWIN88 last updated on 06/Aug/21

X_(even) = {0,2,4,6} X_(odd) ={1,3,5} → { ((even×even=even)),((even×odd=even)) :} P(A_(product even) )=((C_2 ^4 +C_1 ^4 ×C_1 ^3 )/C_2 ^7 ) = ((6+4×3)/(21))=((18)/(21))=(6/7)

$${X}_{{even}} =\:\left\{\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6}\right\} \\ $$$${X}_{{odd}} =\left\{\mathrm{1},\mathrm{3},\mathrm{5}\right\}\:\rightarrow\begin{cases}{{even}×{even}={even}}\\{{even}×{odd}={even}}\end{cases} \\ $$$${P}\left({A}_{{product}\:{even}} \right)=\frac{{C}_{\mathrm{2}} ^{\mathrm{4}} +{C}_{\mathrm{1}} ^{\mathrm{4}} ×{C}_{\mathrm{1}} ^{\mathrm{3}} }{{C}_{\mathrm{2}} ^{\mathrm{7}} } \\ $$$$\:\:\:=\:\frac{\mathrm{6}+\mathrm{4}×\mathrm{3}}{\mathrm{21}}=\frac{\mathrm{18}}{\mathrm{21}}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Commented by mathdanisur last updated on 06/Aug/21

Thanlyou Ser

$$\mathrm{Thanlyou}\:\boldsymbol{\mathrm{Ser}} \\ $$

Answered by mr W last updated on 05/Aug/21

p=1−((6+C_2 ^3 )/C_2 ^7 )=1−(9/(21))=(4/7)

$${p}=\mathrm{1}−\frac{\mathrm{6}+{C}_{\mathrm{2}} ^{\mathrm{3}} }{{C}_{\mathrm{2}} ^{\mathrm{7}} }=\mathrm{1}−\frac{\mathrm{9}}{\mathrm{21}}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thank you Ser, but answer: (2/(15))

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{Ser}},\:\mathrm{but}\:\mathrm{answer}:\:\:\frac{\mathrm{2}}{\mathrm{15}} \\ $$

Commented by mr W last updated on 05/Aug/21

i can get the same result in an other way: p=((3×3+3)/(21))=(4/7) other people should check if i′m wrong or you are wrong.

$${i}\:{can}\:{get}\:{the}\:{same}\:{result}\:{in}\:{an}\:{other} \\ $$$${way}: \\ $$$${p}=\frac{\mathrm{3}×\mathrm{3}+\mathrm{3}}{\mathrm{21}}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$ \\ $$$${other}\:{people}\:{should}\:{check}\:{if}\:{i}'{m}\:{wrong} \\ $$$${or}\:{you}\:{are}\:{wrong}. \\ $$

Commented by mathdanisur last updated on 05/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Commented by EDWIN88 last updated on 06/Aug/21

mr W your answer not correct. please check

$${mr}\:{W}\:{your}\:{answer}\:{not}\:{correct}. \\ $$$${please}\:{check} \\ $$

Commented by mr W last updated on 06/Aug/21

i excluded zero. p=1−((6+C_2 ^3 )/C_2 ^7 )=1−(9/(21))=(4/7) if zero is included, then p=1−(C_2 ^3 /C_2 ^7 )=1−(3/(21))=(6/7)

$${i}\:{excluded}\:{zero}. \\ $$$${p}=\mathrm{1}−\frac{\mathrm{6}+{C}_{\mathrm{2}} ^{\mathrm{3}} }{{C}_{\mathrm{2}} ^{\mathrm{7}} }=\mathrm{1}−\frac{\mathrm{9}}{\mathrm{21}}=\frac{\mathrm{4}}{\mathrm{7}} \\ $$$${if}\:{zero}\:{is}\:{included},\:{then} \\ $$$${p}=\mathrm{1}−\frac{{C}_{\mathrm{2}} ^{\mathrm{3}} }{{C}_{\mathrm{2}} ^{\mathrm{7}} }=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{21}}=\frac{\mathrm{6}}{\mathrm{7}} \\ $$

Commented by mathdanisur last updated on 06/Aug/21

Thankyou Ser

$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{Ser}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com