Finding $P(x<2)$ given that x follows a poisson distribution and $P(x>2)=0.3$ Let X be a random variable with a Poisson distribution, such that $P(X>2)=0.3$ find $P(X<2)$.
My initial thought when seeing this problem was to use the definition of $P(X=x)$ for the Poisson distribution, however this led to an equation that I couldn’t solve (I believe there are only solutions through numerical methods). My second method is detailed below:
$P(X>2)=1-P(X<2)-P(X=2)=0.3$
Hence, $P(X<2)=0.7-P(X=2$ This led me to a dead-end as I do not know an expression for $P(X=2).$
Any ideas on how to do this problem?
 A: You know
$$
\sum_{k=0}^\infty \frac{\lambda^k\exp(-\lambda)}{k!}=1.
$$
So
$$
Pr(X>2)=\sum_{k=3}^\infty \frac{\lambda^k\exp(-\lambda)}{k!}=1-\frac{\lambda^0\exp(-\lambda)}{0!} - \frac{\lambda^1\exp(-\lambda)}{1!}-\frac{\lambda^2\exp(-\lambda)}{2!}
$$
$$
 = \frac{1}{2}\exp(-\lambda)(-\lambda^2-2\lambda+2\exp(\lambda)-2)=0.3.
$$
WolframAlpha says that this equality holds for $\lambda \approx 1.91378$. Plug that value into the Poisson p.m.f. and you'll get your answer.
A: You want to find to zero of function
$$f(\lambda)=1-e^{-\lambda}\left(1+\lambda+\frac 12 \lambda^2\right)-0.3$$ By inspection or graphing, the solution is "close" to $\lambda=2$.
Let us make a Taylor expansion of $y=f(\lambda)+0.3$ around  $\lambda=2$. This would give
$$y=\left(1-\frac{5}{e^2}\right)+\frac{2 (\lambda -2)}{e^2}-\frac{(\lambda -2)^3}{6
   e^2}+\frac{(\lambda -2)^4}{24 e^2}+O\left((\lambda-2)^6\right)$$ Now, using series reversion
$$\lambda=2+t+\frac{t^3}{12}-\frac{t^4}{48}+\frac{t^5}{48}+O\left(t^6\right)\qquad \text{where} \qquad t=\frac{1}{2} e^2 \left(y+\frac{5}{e^2}-1\right)$$ Make $y=0.3$ to get
$\lambda=1.913775799$ while the exact solution given by Newton method is $1.913775794$
We could also use the simplest $[1,1]$ Padé approximant to get
$$y\sim \frac{\left(1-\frac{5}{e^2}\right)+\left(\frac{1}{4}+\frac{3}{4 e^2}\right) (\lambda
   -2)+\left(\frac{1}{12}+\frac{1}{12 e^2}\right) (\lambda -2)^2 }{1+\frac{\lambda -2}{4}+\frac{1}{12} (\lambda -2)^2}$$ which is a quadratic equation in $(\lambda -2)$ the solution of which being $1.913775803$.
