Question-149458

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

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

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

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

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

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

Commented by EDWIN88 last updated on 06/Aug/21

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

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

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

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