Answer:
Clearly, \[P(X=0)+P(X=1)+P(X=2)=1\] \[\Rightarrow \] \[p+p+P(X=2)=1\] \[\Rightarrow \] \[P(X=2)=1-2p\] So, the probability distribution of X is as given below
\[\therefore \] \[E(X)=0\times p\,+1\times p+2(1-2p)=2-3p\] and \[E({{X}^{2}})={{0}^{2}}\times p\,+{{1}^{2}}\times p+{{2}^{2}}(1-2p)=4-7p\] It is given that, \[E({{X}^{2}})=E(X)\] \[\Rightarrow \] \[4-7p=2-3p\] \[\Rightarrow \] \[4p=2\] \[\Rightarrow \] \[p=\frac{1}{2}\] \[{{\mathbf{x}}_{\mathbf{i}}}\] 0 1 2 \[{{\mathbf{p}}_{\mathbf{i}}}\] p p \[(1-2p)\]
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