A) \[\frac{1}{2{{e}^{2}}}\]
B) \[\frac{1}{3{{e}^{2}}}\]
C) \[\frac{2}{3{{e}^{2}}}\]
D) \[\frac{1}{{{e}^{2}}}\]
Correct Answer: C
Solution :
Given, \[P(X=1)=P(X=2)\] \[\frac{{{e}^{-\lambda }}\lambda }{1!}=\frac{{{e}^{-\lambda }}{{\lambda }^{2}}}{2!}\] \[\Rightarrow \] \[\lambda =2\] \[\therefore \] \[P(X=4)=\frac{{{e}^{-2}}{{2}^{4}}}{4!}=\frac{2}{3{{e}^{2}}}\]
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