Suppose that a random variable X follows Poisson distribution. If $P(X=1)=P(X=2)$ then $P(X=5)=$

$\begin {array} {1 1} (1)\;\frac{2}{3}e^{-2} & \quad (2)\;\frac{3}{4}e^{-2} \\ (3)\;\frac{4}{15}e^{-2} & \quad (4)\;\frac{7}{8}e^{-2} \end {array}$

$(3)\;\frac{4}{15}e^{-2}$