Question-17

Suppose \(X \sim Poisson(\lambda)\). Find the value of \(\lambda\) for which \(3 P(X=3) = 2P(X=2) + 4P(X=1)\).

Hint

Apply the PMF formula of Poisson distribution.

Answer

4

Solution

\(3 P(X=3) = 2P(X=2) + 4P(X=1)\)

\(3\times \left(\dfrac{e^{-\lambda} \lambda^3}{3!}\right) = 2\times \left(\dfrac{e^{-\lambda} \lambda^2}{2!}\right) + 4\times\left( \dfrac{e^{-\lambda} \lambda^1}{1!}\right)\)

\(3\times \left(\dfrac{\lambda^3}{3 \times 2!}\right) = 2\times \left(\dfrac{\lambda^2}{2}\right) + 4\times\left(\lambda\right)\)

\(\dfrac{\lambda^2}{2} = \lambda + 4\)

\(\lambda^2 = 2\lambda + 8\)
\(\lambda^2 - 2\lambda - 8 = 0\)
\((\lambda+2)(\lambda-4)=0 \implies \lambda = 4\)