logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Probability

If X is a Poisson variate such that $ P(2)=9\: P(4)+90\: P(6)$. then mean of X is

$\begin {array} {1 1} (A)\;\pm\: 1 & \quad (B)\;\pm\: 2 \\ (C)\;\pm\: 3 & \quad (D)\;None\: of \: these \end {array}$

 

Download clay6 mobile app

1 Answer

Using the given relation we have
$ \Large\frac{e^{-m}m^2}{2!} = \Large\frac{9e^{-m}m^4}{4!} +\Large\frac{90e^{-m}m^6}{6!}$
$ = m^4+3m^2-4=0$
$ (m^2-1)(m^2+4)=0$
$ m = \pm 1$
$ m^2=-4 $ ( not feasible )
Hence $ m = \pm 1$
Ans : (A)
answered Jan 15, 2014 by thanvigandhi_1
 

Related questions

Ask Question
Tag:MathPhyChemBioOther
...
X