X is a poisson variate with parameter $\mathrm{\lambda }$. Then match the following.

A) P(X= 0)                        1) ${\mathrm{e}}^{-\mathrm{\lambda }}.\sinh \mathrm{\lambda }$

B) P(X≥1)                         2) ${\mathrm{e}}^{-\mathrm{\lambda }}.\cosh \mathrm{\lambda }$

C) P(X= 1, 3, 5,.....)          3) ${\mathrm{e}}^{-\mathrm{\lambda }}.\tanh \mathrm{\lambda }$

D) P(X= 0, 2, 4, .....)         4) $1-{\mathrm{e}}^{-\mathrm{\lambda }}$

                                           5) ${\mathrm{e}}^{-\mathrm{\lambda }}$

The correct matching is

Given X follows Poisson Variate with Parameters $\text{'}\mathrm{\lambda }\text{'}$

A)   $\mathrm{P}(\mathrm{X}=0)=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^0}{0!}={\mathrm{e}}^{-\mathrm{\lambda }}$

B)   $\mathrm{P}(\mathrm{X}\ge 1)=1-\mathrm{P}(\mathrm{X}=0)$

                       $=1-{\mathrm{e}}^{-\mathrm{\lambda }}$

C)   $\mathrm{P}(\mathrm{X}=1, 3, 5, ......)$

        $$\begin{gathered} =\mathrm{P}(0, \mathrm{\lambda })+\mathrm{P}(2, \mathrm{\lambda })+\mathrm{P}(4, \mathrm{\lambda })+....... \\ =\left[\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^0}{0!}+\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^2}{2!}+\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^4}{4!}+........\right] \\ ={\mathrm{e}}^{-\mathrm{\lambda }}\left[1+\frac{{\mathrm{\lambda }}^2}{2}+\frac{{\mathrm{\lambda }}^4}{4!}+......\right] \\ ={\mathrm{e}}^{-\mathrm{\lambda }} \left[\frac{{\mathrm{e}}^{\mathrm{\lambda }}+{\mathrm{e}}^{-\mathrm{\lambda }}}{2}\right] \\ ={\mathrm{e}}^{-\mathrm{\lambda }} \cos \mathrm{h} \mathrm{\lambda } \end{gathered}$$

        $\therefore$ P(sum of terms in odd places)

          $={\mathrm{e}}^{-\mathrm{\lambda }} \cos \mathrm{h} \mathrm{\lambda }$

D)   $\mathrm{P}(\mathrm{x}=0, 2, 4, .......)$

       Similarly P (sum of terms in even places)

         $$\begin{gathered} ={\mathrm{e}}^{-\mathrm{\lambda }}(\mathrm{\lambda }+\frac{{\mathrm{\lambda }}^3}{3!}+\frac{{\mathrm{\lambda }}^5}{5!}+............) \\ ={\mathrm{e}}^{-\mathrm{\lambda }} \left[\frac{{\mathrm{e}}^{\mathrm{\lambda }}-{\mathrm{e}}^{-\mathrm{\lambda }}}{2}\right] \\ ={\mathrm{e}}^{-\mathrm{\lambda }} \sin \mathrm{h} \mathrm{\lambda } \end{gathered}$$