A) \[\frac{4}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{3}{2}\]
D) \[\frac{5}{12}\]
Correct Answer: A
Solution :
\[P(\overline{E}\cap \overline{F})\,=P(\overline{E})\,.P(\overline{F})=\frac{1}{12}\,=\] \[P(\overline{E}\cap \overline{F})\,=P(\overline{E})\,.P(\overline{F})=\frac{1}{2}\] \[\Rightarrow \,\,(1-P(E))\,(1-P(F)\,=\frac{1}{2}\] \[=1-x-y+xy=\frac{1}{2}\] \[1-x-y+\frac{1}{12}\,=\frac{1}{2}\] \[1-x-y=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\] \[\] \[x+\frac{1}{12x}\,=\frac{7}{12}\] \[\frac{12{{x}^{2}}+1}{12x}\,=\frac{7}{12}\] \[12{{x}^{2}}-7x+1=0\] \[12{{x}^{2}}-4x-3x+1=0\] \[4x(3x-1)-1(3x-1)=0\] \[x=\frac{1}{3},\,\,x=\frac{1}{4}\] \[y=\frac{1}{4},\,y=\frac{1}{3}\] \[\therefore \,\,\frac{x}{y}\,=\frac{1/3}{1/4}=\frac{4}{3}\,\,or\,\,\frac{1/4}{1/3}=\frac{3}{4}\]
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