100AHW2 - $E$ stands for effect). $ (1) Rule of total...

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\documentclass[11pt]{article} \input{psfig.sty} \usepackage{epsfig} \ \renewcommand{\baselinestretch}{1.1} %\setcounter{page}{0} \thispagestyle{empty} % \setlength{\topmargin}{-0.60in} \setlength{\oddsidemargin}{-0.2in} \setlength{\textwidth}{6.5in} \setlength{\textheight}{9in} \ \def\E{{\rm E}} \def\Var{{\rm Var}} \def\Cov{{\rm Cov}} \ \begin{document} \begin{center} {\huge STAT 100A HWII Due next Wed in class} \end{center} \ \noindent{\bf Problem 1:} If we flip a fair coin $n$ times independently, what is the probability that we observe $k$ heads? $k = 0, 1, . .., n$. Please explain your answer. a \noindent{\bf Problem 2:} Prove the following two identities: \ (1) $P(\bar{A}|B) = 1 - P(A|B)$. ( (1) $P(A \cup B |C) = P(A|C) + P(B|C) - P(A \cap B|C)$. ( \noindent{\bf Problem 3:} Independence. If $P(A|B) = P(A)$, prove \ (1) $P(A \cap B) = P(A) P(B)$. ( (2) $P(B|A) = P(B)$. ( \noindent{\bf Problem 4:} Prove the following two identities ($C$ stands for cause,
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Unformatted text preview: $E$ stands for effect). $ (1) Rule of total probability: $P(E) = P(E|C)P(C) + P(E|\bar{C})P(\bar{C})$. ( (2) Bayes rule: $P(C|E) = P(C)P(E|C)/[P(C)P(E|C) + P(\bar{C})P(E|\bar{C}]$. ( \noindent{\bf Problem 5:} Suppose $1\%$ of the population is inflicted with a particular disease. For a medical test, if a person has the disease, then $95\%$ chance the person will be tested positive. If a person does not have the disease, then $90\%$ chance the person will be tested negative. Using precise notation, calculate c (1) The probability that a randomly selected person will be tested positive. ( (2) If the person is tested positive, what is the chance that he or she has the disease? If the person is tested negative, what is the chance that the person has no disease? n n \end{document}...
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This note was uploaded on 01/29/2010 for the course STATS 100A 262303210 taught by Professor Wu during the Fall '09 term at UCLA.

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100AHW2 - $E$ stands for effect). $ (1) Rule of total...

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