91Ó°ÊÓ

It is given that X and Y are independent Poisson variables. It is also given \(Var\left( X \right) + Var\left( Y \right) = 5 \ldots \left( 1 \right)\)

Let the parameters of X and Y be\(\lambda \,\,and\,\,\mu .\)

Therefore,

\(X \sim P\left( \lambda \right)\,\,and\,\,Y \sim P\left( \mu \right)\)

Let us define a new variable, W=X+Y.

The sum of the independent Poisson with parameters\(\lambda \,\,and\,\,\mu .\)is a Poisson with parameter\(\lambda \,\, + \,\,\mu .\)

\(W \sim P\left( {\lambda + \mu } \right)\)

By the property of a Poisson distribution, the variance of the variable is equal to its parameter value.

Therefore,

\(\begin{array}{l}Var\left( X \right) = \lambda \\Var\left( Y \right) = \mu \end{array}\)

By using (1)

\(\begin{array}{l}Var\left( X \right) + Var\left( Y \right) = 5\\ \Rightarrow \lambda + \mu = 5\\ \Rightarrow Var\left( W \right)\end{array}\)

Therefore, \(W \sim P\left( 5 \right)\)

\(\begin{array}{l} = \Pr \left( {W < 2} \right)\\ = \Pr \left( {W = 0} \right) + \Pr \left( {W = 1} \right)\\ = \frac{{{e^{ - 5}}{5^0}}}{{0!}} + \frac{{{e^{ - 5}}{5^1}}}{{1!}}\\ = 6{e^{ - 5}}\\ = 0.0404\end{array}\)

Hence the answer is\({\bf{0}}{\bf{.0404}}\)