Question

A : the sum of the times in odd places in a P.D is $e^{-\lambda }$ cosh $\lambda$
R : cosh $\lambda =\frac{{\lambda }^1}{1!}+\frac{{\lambda }^3}{3!}+\frac{{\lambda }^5}{5!}+......$

• A. Both A and R are true and R is the correct
  explanation of A
• B. Both A and R are true but R is not correct
  explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is C A is true but R is false

Considering all odd places in Poisson's Distribution, we get
$e^{-\lambda }+\frac{e^{-\lambda }.{\lambda }^2}{2!}+\frac{e^{-\lambda }.{\lambda }^4}{4!}....$
$=e^{-\lambda }[1+\frac{{\lambda }^2}{2!}+\frac{{\lambda }^4}{4!}+....]$
$=e^{-\lambda }\left[\frac{e^{-\lambda }+e^{\lambda }}{2}\right]$
$=e^{-\lambda }\left(\cosh \lambda \right)$.
Hence reason is false.