Question

For a poission distribution variable $X$ is such that $P(X=2)=9P(X=4)+90P(X=6)$ the mean is

• A. $2$
• B. $3$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

For $P.D.P(X=r)=\frac{e^{-\lambda }{\lambda }^r}{r!},r=0,1,2,\cdots$

$\therefore \frac{e^{-\lambda }{\lambda }^2}{2!}=\frac{9e^{-\lambda }{\lambda }^4}{4!}+90\frac{e^{-\lambda }{\lambda }^6}{6!}$ (given)

$\Rightarrow \frac{{\lambda }^2}{2}=\frac{9}{24}{\lambda }^4+\frac{90}{720}{\lambda }^6$

$\Rightarrow {\lambda }^4+3{\lambda }^2-4=0$

$\Rightarrow ({\lambda }^2+4)({\lambda }^2-1)=0=>\lambda =\pm 1$

$\Rightarrow \lambda =1$ as $\lambda >0$ and $({\lambda }^2+4=0$ impossible$)$

$\therefore$ mean$=\lambda =1$