A) \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{5}=1\]
B) \[\frac{{{x}^{2}}}{5}-\frac{{{y}^{2}}}{4}=1\]
C) \[\frac{{{x}^{2}}}{2}-\frac{{{y}^{2}}}{3}=1\]
D) \[\frac{{{x}^{2}}}{3}-\frac{{{y}^{2}}}{2}=1\]
Correct Answer: A
Solution :
Let\[P(A)=x\]and\[P(B)=y\] Since, A and B are independent events. Therefore, \[P(\overline{A}\cap B)=\frac{2}{15}\] \[\Rightarrow \] \[P(\overline{A}\cap B)=\frac{2}{15}\] \[\Rightarrow \] \[P(\overline{A}).P(B)=\frac{2}{15}\] \[\Rightarrow \] \[[1-P(A)]P(B)=\frac{2}{15}\] \[\Rightarrow \] \[(1-x)y=\frac{2}{15}\] Also, \[P(A\cap \overline{B})=\frac{1}{6}\] \[P(A).P(\overline{B})=\frac{1}{6}\] \[\Rightarrow \] \[P(A).[1-P(B)]=\frac{1}{6}\] \[\Rightarrow \] \[x(1-y)=\frac{1}{6}\] \[\Rightarrow \] \[x-xy=\frac{1}{6}\] Subtracting Eq. (i) from Eq. (ii), we get \[x-y=\frac{1}{6}-\frac{2}{15}\] \[\Rightarrow \] \[x-y=\frac{1}{30}\] \[\Rightarrow \] \[x=\frac{1}{30}+y\] Substituting this value of\[x\]in Eq. (i), we get \[y-\left( \frac{1}{30}+y \right)y=\frac{2}{15}\] \[\Rightarrow \] \[y-\frac{y}{30}-{{y}^{2}}=\frac{2}{15}\] \[\Rightarrow \] \[{{y}^{2}}-\frac{29}{30}y+\frac{2}{15}=0\] \[\Rightarrow \] \[30{{y}^{2}}-29y+4=0\] \[\Rightarrow \] \[(6y-1)(5y-4)=0\] \[\Rightarrow \] \[y=\frac{1}{6}\]or\[y=\frac{4}{5}\]
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