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Probability
└── Homework
    └── W8
        └── Q2.tex

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\documentclass[12pt]{article}

\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage[margin=1in]{geometry}
\usepackage{fancyhdr}
\usepackage{enumerate}
\usepackage[shortlabels]{enumitem}

\pagestyle{fancy}
\fancyhead[l]{Li Yifeng}
\fancyhead[c]{Homework \#8}
\fancyhead[r]{\today}
\fancyfoot[c]{\thepage}
\renewcommand{\headrulewidth}{0.2pt}
\setlength{\headheight}{15pt}

\newcommand{\bE}{\mathbb{E}}
\newcommand{\bP}{\mathbb{P}}

\begin{document}

    \section*{Question 2}

    \noindent During the rifle qualification in the Marine Corps the final ten shots are taken from $500$ yards. You receive one point for a hit and zero for a miss. If each shot is independent and the probability you hit is $p = 0.70$, what is the probability that you hit $7$ or more of your shots?

    \bigskip

    \begin{enumerate}[label={},leftmargin=0in]\item
        \subsection*{Solution}
            Let $X\sim\mathrm{Binomial}(n=10,p=0.70)$. We want

            \[
                \bP(X\ge7)=\sum_{k=7}^{10}\binom{10}{k}(0.70)^k(0.30)^{10-k}\approx0.650
            \]
        \subsection*{Answer}

            \[\boxed{\bP(X\ge7)=\sum_{k=7}^{10}\binom{10}{k}(0.70)^k(0.30)^{10-k}\approx0.650}\]
    \end{enumerate}

\end{document}