A) 7/25
B) 18/25
C) 13/25
D) 19/25
E) 16/25
Correct Answer: E
Solution :
\[\therefore \] \[P(X=0)=kp(X=1)=2k{{\left( \frac{1}{5} \right)}^{1}}\] \[p(X=2)=3k{{\left( \frac{1}{5} \right)}^{1}},....\] Since, \[P(X=0)+P(X=1)+P(X=2)\] \[+.....=1\] \[\begin{align} & \underline{\begin{align} & \therefore \,\,\,\,\,\,\,\,\,k+2k\left( \frac{1}{5} \right)+3k{{\left( \frac{1}{5} \right)}^{2}}+.....=1 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{k}{5}+2k{{\left( \frac{1}{5} \right)}^{2}}+......=\frac{1}{5} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,- \\ \end{align}} \\ & \,\,\,\,\,\,\,\,\,\,\,k+k\frac{1}{5}+k{{\left( \frac{1}{5} \right)}^{2}}+.....=\frac{4}{5} \\ \end{align}\] \[\Rightarrow \] \[\frac{k}{1-\frac{1}{5}}=\frac{4}{5}\] \[\Rightarrow \] \[k=\frac{16}{25}\] \[\therefore \] \[P(X=0)=\frac{16}{25}(0+1){{\left( \frac{1}{5} \right)}^{0}}\]
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