Suppose that cell death occurs according to a Poisson process with rate lambda = 15 per day. Calculate the probability that after 10 days have died least 160 cells. I am very confused, I dont know to solve this. I know that:
$P[N(t)=n]=\frac{(\lambda t)^n*e^-\lambda t}{n!}$
But I want to know
$P[N(10)<160]$
Maybe if
$P(T<=t)=1-e^{-\lambda t}$ then $P(T<=160)=1-e^{-15(160)}$???
In ten days, the expected number of deaths is (how many)? That is the $\lambda$ in your Poisson distribution, and you want the chance that $N \ge 160$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The probability of $N$ death cells in 10 days is given by: \begin{align} &\sum_{n_{1} = 0}^{\infty}\sum_{n_{2} = 0}^{\infty}\ldots\sum_{n_{10} = 0}^{\infty} {\pars{\lambda t}^{n_{1}}\expo{-\lambda t} \over n_{1}!} {\pars{\lambda t}^{n_{2}}\expo{-\lambda t} \over n_{2}!}\ldots {\pars{\lambda t}^{n_{10}}\expo{-\lambda t} \over n_{10}!}\delta_{n_{1} + n_{2} + \cdots + n_{10},N} \\[3mm]&=\expo{-10\lambda t}\sum_{n_{1} = 0}^{\infty}\sum_{n_{2} = 0}^{\infty}\ldots\sum_{n_{10} = 0}^{\infty} {\pars{\lambda t}^{n_{1}} \over n_{1}!} {\pars{\lambda t}^{n_{2}} \over n_{2}!}\ldots {\pars{\lambda t}^{n_{10}} \over n_{10}!} \int_{\verts{z} = 1}{1 \over z^{N + 1 - n_{1} - n_{2} - \cdots - n_{10}}}\, {\dd z \over 2\pi\ic} \\[3mm]&=\expo{-10\lambda t} \int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{1 \over z^{N + 1}} \bracks{\,\sum_{n = 0}^{\infty}{\pars{\lambda t z}^{n} \over n!}}^{10} =\expo{-10\lambda t} \int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\,{\expo{10\lambda tz} \over z^{N + 1}} \\[3mm]&=\expo{-10\lambda t} \sum_{\ell = 0}^{\infty}\ \overbrace{\int_{\verts{z} = 1}{\dd z \over 2\pi\ic}\, {z^{\ell} \over z^{N + 1}}}^{\ds{\delta_{\ell,N}}}\ {\pars{10\lambda t}^{\ell} \over \ell!}= {\pars{10\lambda t}^{N}\expo{-10\lambda t} \over N!} =\color{#00f}{\large{\pars{15/16}^{N} \over N!}\,\expo{-15/16}} \end{align}
Your question answer is given by: \begin{align} \expo{-15/16}\sum_{N = 160}^{\infty}{\pars{15/16}^{N} \over N!}\, &= 1 - \expo{-15/16}\sum_{N = 0}^{159}{\pars{15/16}^{N} \over N!} =\color{#00f}{\large 1 - {\Gamma\pars{160,15/16} \over 159!}} \\[3mm]&\approx 2.1001 \times 10^{-15} \end{align} where $\ds{\Gamma\pars{a,x}}$ is the Incomplete Gamma Function.
