Question-81
Let \(X_{1} ,X_{2} ,X_{3}\) be three i.i.d Poison random variables with mean \(\lambda =3\). Find the probability that exactly one of the \(X_{i} =0\) and the other two variables are greater than \(1\).
\(X_{1} ,X_{2} ,X_{3}\) are i.i.d with \(\text{Poisson}( \lambda )\). Using symmetry, the required probability is:
\[ P( X_{1} =0,X_{2} >1,X_{3} >1) \times 3 \]
Using independence, we have:
\[ \begin{aligned} P( X_{1} =0,X_{2} >1,X_{3} >1) & =P( X_{1} =0)\\ & \times P( X_{2} >1)\\ & \times P( X_{3} >1) \end{aligned} \]
The PDF of a Poisson distribution is \(f_{X}( x) =\frac{e^{-\lambda } \lambda ^{x}}{x!}\). This gives us:
\[ \begin{aligned} P( X_{1} =0) & = e^{-\lambda }\\ & = e^{-3}\\ P( X_{2} >1) & =1-\left( e^{-\lambda } +\lambda e^{-\lambda }\right)\\ & =1-4e^{-3} \end{aligned} \]
The required probability is:
\[ 3e^{-3}\left( 1-4e^{-3}\right)^{2} \]