738 lines
24 KiB
TeX
738 lines
24 KiB
TeX
\documentclass[a4paper]{article}
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\usepackage{fancyhdr}
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\usepackage{extramarks}
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\usepackage{amsmath}
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\usepackage{amsthm}
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\usepackage{amsfonts}
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\usepackage{tikz}
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\usepackage[plain]{algorithm}
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\usepackage{algpseudocode}
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\usepackage{enumitem}
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\graphicspath{ {./images/} }
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\usetikzlibrary{automata,positioning}
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%
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% Basic Document Settings
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%
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\topmargin=-0.45in
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\evensidemargin=0in
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\oddsidemargin=0in
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\textwidth=6.5in
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\textheight=9.0in
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\headsep=0.25in
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\linespread{1.1}
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\pagestyle{fancy}
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\lhead{\hmwkAuthorName}
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\chead{\hmwkClass\ : \hmwkTitle}
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\rhead{\firstxmark}
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\lfoot{\lastxmark}
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\cfoot{\thepage}
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\renewcommand\headrulewidth{0.4pt}
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\renewcommand\footrulewidth{0.4pt}
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\setlength\parindent{0pt}
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%
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% Create Problem Sections
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%
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\newcommand{\enterProblemHeader}[1]{
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\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
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\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
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}
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\newcommand{\exitProblemHeader}[1]{
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\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
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\stepcounter{#1}
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\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
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}
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\setcounter{secnumdepth}{0}
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\newcounter{partCounter}
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\newcounter{homeworkProblemCounter}
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\setcounter{homeworkProblemCounter}{1}
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\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
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\newenvironment{homeworkProblem}[1][-1]{
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\ifnum#1>0
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\setcounter{homeworkProblemCounter}{#1}
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\fi
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\section{Question \arabic{homeworkProblemCounter}}
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\setcounter{partCounter}{1}
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\enterProblemHeader{homeworkProblemCounter}
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}{
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\exitProblemHeader{homeworkProblemCounter}
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}
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\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
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\newcommand{\hmwkTitle}{Probability Practice Exercises}
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\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
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\newcommand{\hmwkStudentID}{\textbf{2200624}}
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\title{
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\vspace{2in}
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\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
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\vspace{3in}
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}
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\author{\hmwkAuthorName \\ \hmwkStudentID}
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\date{\today}
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\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
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%
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% Various Helper Commands
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%
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% Useful for algorithms
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\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
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% For derivatives
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\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
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% For partial derivatives
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\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
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% Integral dx
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\newcommand{\dx}{\mathrm{d}x}
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% Alias for the Solution section header
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\newcommand{\solution}{\textbf{\large My Solution\\}}
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% Probability commands: Expectation, Variance, Covariance, Bias
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\newcommand{\E}{\mathrm{E}}
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\newcommand{\Var}{\mathrm{Var}}
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\newcommand{\Cov}{\mathrm{Cov}}
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\newcommand{\Bias}{\mathrm{Bias}}
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\begin{document}
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\maketitle
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\pagebreak
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\begin{homeworkProblem}
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A jar contains four coins: a nickel (5 cents), a dime (10 cents), a quarter (25 cents), and a half-dollar (50 cents).
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Three coins are randomly selected without replacement from the jar.
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\begin{enumerate}[label=(\alph*)]
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\item List all the possible outcomes in sample space S.
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\item What is the probabilty that the selection will contain the half-dollar?
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\item What is the probability that the total amount drawn will equal 65 cents or less?
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\item Which interpretations of probability are you applying?
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\end{enumerate}
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\solution
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\part
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$$S = \{(5,10,25), (5,25,50), (10, 25,50), (5,10,50)\}$$
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\part
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$$P(\text{half dollar}) = \frac{3}{4}$$
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\part
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Sum of the outcomes:
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$$S = \{40, 80, 85, 65\}$$
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$$P(\text{total} \leq 65) = \frac{2}{4} = \frac{1}{2}$$
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\part
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Classical interpretation.
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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In a genetics experiment, the researcher mated two Drosophila fruit flies
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and observed the traits of 300 offsprint. The results are shown in the table
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\includegraphics[width=\textwidth]{q2.png}
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One of these offspring is randomly selected and observed for the two genetic traits
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\begin{enumerate}[label=(\alph*)]
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\item What is the probability that the fly has normal eye color and normal wing size?
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\item What is the probability that the fly has vermillion eyes?
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\item What is the probability that the fly has either vermillion eyes or miniature wings, or both?
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\item Which interpretations of probability are you applying?
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\end{enumerate}
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\solution
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\part
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$$P(\text{Normal Eye Color} \cap \text{Normal Wing Size}) = \frac{140}{300} = \frac{7}{15} = 0.46667$$
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\part
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$$P(\text{Vermillion Eyes}) = \frac{3+151}{300} = \frac{77}{150} = 0.513$$
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\part
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$$
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\begin{align}
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P(\text{vermillion eyes } \cup \text{ miniature wings})&= P(\text {vermillion eyes}) + P(\text{miniature wings}) - P(\text{vermillion eyes} \cap \text{miniature wings})\\
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&= \frac{154}{300} + \frac{157}{300} - \frac{151}{300}
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&= \frac{160}{300}
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&= \frac{8}{15}
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\end{align}
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$$
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\part
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Relative frequency interpretation.
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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City residents were surgeyed recently to determine readership of newspapers.
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50\% of the residents read the morning paper, 60\% read the evening paper and
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20\% read both newspapers. Find the probability that a resident selected
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at random reads the morning or evening paper
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\solution
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Given: \\
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$P(\text{morning}) = 0.5$\\
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$P(\text{evening}) = 0.6$\\
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$P(\text{both}) = 0.2 = P(\text{morning } \cap \text{ evening})$\\
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To find: $P(\text{morning or evening})$\\
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$$
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\begin{align}
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P(\text{morning} \cup \text{evening}) &= P(\text{morning}) + P(\text{evening}) - P(\text{morning} \cap \text{evening})$\\
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&= 0.5 + 0.6 - 0.2\\
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&= 0.9
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\end{align}
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$$
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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The experience of a train passenger is that the train is cancelled with probability 0.02
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When it does run, it has probability of 0.9 of being on time.
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\begin{enumerate}[label=(\alph*)]
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\item Find the probability that the train runs on time
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\item Given that the train is not on time, find the probability that it has been cancelled
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\end{enumerate}
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\solution
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Given:\\
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P(\text{cancelled}) = 0.02\\
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P(\text{cancelled}^c) = 0.98\\
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P(\text{on time} | \text{cancelled}^c) = 0.9\\
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\part
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$$P(\text{on time}) = 0.98 \cdot 0.9 = 0.882$$
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\part
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From part (a):
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$$P(\text{on time}^c) = 1 - 0.882 = 0.118$$\\
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$$
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\begin{align}
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P(\text{cancelled} | \text{on time}^c) &= \frac{P(\text{cancelled} \cap \text{on time}^c)}{P(\text{on time}^c)}\\
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&= \frac{0.02\cdot 1}{0.02 \cdot 1 + 0.98 \cdot 0.1}\\
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&= 0.1694915\\
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&\approx 0.1694
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\end{align}
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$$
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Draw the probability tree diagram to visualise.\\
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Note: Once the train is cancelled, there is \textbf{NO WAY} the train can be on time,\\
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Therefore, $P(\text{on time} | \text{cancelled}) = 0$ and $P(\text{on time}^c | \text{cancelled}) = 1$.
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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In a group of students 60\% are female and 40\% are male.
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A third of the female students study french but only a quarter of the male students study french
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\\
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A student is chosen at random from the group.
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\begin{enumerate}[label=(\alph*)]
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\item Show that the probability that the chosen student is female and studies french is 0.2
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\item Calculate the probability that the chosen student studies french
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\item Given that the chosen student does study french, calculate the conditional probability that the chosen student is female
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\end{enumerate}
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\solution
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Let f: female\\
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fr: studies french
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\\\\
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Given:
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$P(f) = 0.6$\\
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$P(f^c) = 0.4$\\
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$P(fr) = 0.3$\\
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$P(fr^c) = 0.7$\\
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$P(fr | f) = 0.3$\\
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$P(fr^c | f) = 0.7$\\
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$P(fr | f^c) = 0.25$\\
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$P(fr^c | f^c) = 0.75$\\
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Draw the probability tree to help solve the problem
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\part
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$$
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P(f \cap fr) = P(f) \cdot P(fr | f) = 0.6 \cdot \frac{1}{3} = 0.2
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$$
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\part
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$$P(fr) = 0.2 + (0.4 \cdot 0.25) = 0.2 + 0.1 = 0.3$$
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\part
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$$P(f | fr) = \frac{P(f \cap fr)}{P(fr)} = \frac{0.2}{0.3}$$
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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Of the patients in an emergency unit, 30\% have a sports injury. Pass records
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for this emergency unit suggest that :\\
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A patient with sports injury has a probability of 0.2 of being admitted to hospital\\
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\\
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A patient who doesn’t have a sports injury has a probability of 0.4 of being admitted to
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hospital\\
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\\
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Now, a patient is chosen at random from those in the emergency unit.\\
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\begin{enumerate}[label=(\alph*)]
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\item What is the probability that the chosen patient has a sports injury
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\item What is the probability that the chosen patient has a sports injury and is admitted to hospital
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\item Show that the probability that the chosen patient is admitted to hospital is 0.34
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\item Given that the chosen patient is admitted to hospital, find the conditional probability that the patient has a sports injury
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\end{enumerate}
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\solution
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Given:\\
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$P(\text{sports injury}) = 0.3$\\
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$P(\text{sports injury}^c) = 0.7$\\
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$P(\text{admitted} | \text{sports injury}) = 0.2$\\
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$P(\text{admitted} | \text{sports injury}^c) = 0.4$\\
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\part
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$$P(\text{sports injury}) = 0.3$$
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\part
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$$P(\text{sports injury} \cap \text{admitted}) = P(\text{sports injury}) \cdot P(\text{admitted} | \text{sports injury}) = 0.3 \cdot 0.2 = 0.06$$
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\part
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$$P(\text{admitted}) = P(\text{sports injury} \cap \text{admitted}) + P(\text{sports injury}^c \cap \text{admitted}) = 0.3 \cdot 0.2 + 0.7 \cdot 0.4 = 0.34$$
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\part
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$$P(\text{sports injury} | \text{admitted}) = \frac{P(\text{sports injury} \cap \text{admitted})}{P(\text{admitted})} = \frac{0.06}{0.34} = 0.17647 \approx 0.18$$
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\end{homeworkProblem}
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\pagebreak
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\begin{homeworkProblem}
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Darren and Charles are friends who often go to the cinema together. On such visits there
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is a probability of 0.4 that Darren will buy popcorn.
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The probability that Charles will buy
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popcorn is 0.7 if Darren buys popcorn and 0.35 if he doesn’t.
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When Darren and Charles visit the cinema together:
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\begin{enumerate}[label=(\alph*)]
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\item Find the probability that both buy popcorn
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\item Show that the probability that neither buys popcorn is 0.39
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\item Find the probability that exactly one of them buys popcorn
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\end{enumerate}
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Sarah sometimes join Darren and Charles on their cinema visits. On these occasions,
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the probability that sarah buys popcorn is 0.55 if both
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of charles and darren buy popcorn and 0.25 if exactly one of darren and charles buys popcorn
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When Darren, Sarah and Charles visit the cinema together:
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\begin{enumerate}[label=(\alph*)]
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\item What is the probability that the three of them buy popcorn
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\item Charles and sarah buy popcorn but Daren Doesn't
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\end{enumerate}
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\solution
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Given:\\
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$P(\text{Darren buys popcorn}) = 0.4$\\
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$P(\text{Charles buys popcorn} | \text{Darren buys popcorn}) = 0.7$\\
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$P(\text{Charles buys popcorn} | \text{Darren doesn't buy popcorn}) = 0.35$\\
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$P(\text{Darren doesn't buy popcorn}) = 0.6$\\
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\part
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$$
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\begin{align}
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& P(\text{Darren buys popcorn} \cap \text{Charles buys popcorn}) \\
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&= P(\text{Darren buys popcorn}) \cdot P(\text{Charles buys popcorn} | \text{Darren buys popcorn})\\
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&= 0.4 \cdot 0.7\\
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&= 0.28
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\end{align}
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$$
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\part
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$$
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\begin{align}
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& P(\text{Darren doesn't buy popcorn} \cap \text{Charles doesn't buy popcorn}) \\
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&= P(\text{Darren doesn't buy popcorn}) \cdot P(\text{Charles doesn't buy popcorn} | \text{Darren doesn't buy popcorn})\\
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&= 0.6 \cdot 0.65\\
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&= 0.39
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\end{align}
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$$
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\part
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$$
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\begin{align}
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& P(\text{Darren buys popcorn} \cap \text{Charles doesn't buy popcorn} \text{ or } \text{Charles buys popcorn} \cap \text{Darren doesn't buy popcorn}) \\
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&= P(\text{Darren buys popcorn}) \cdot P(\text{Charles doesn't buy popcorn} | \text{Darren buys popcorn}) \\
|
||
&+ P(\text{Charles buys popcorn}) \cdot P(\text{Darren doesn't buy popcorn} | \text{Charles buys popcorn})\\
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||
&= 0.4 \cdot 0.3 + 0.6 \cdot 0.35\\
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&= 0.33
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\end{align}
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$$
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\part
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|
||
Given:
|
||
$P(\text{Sarah buys popcorn} | \text{Darren buys popcorn} \cap \text{Charles buys popcorn}) = 0.55$\\
|
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$P(\text{Sarah buys popcorn} | \text{Darren buys popcorn} \cap \text{Charles doesn't buy popcorn}) = 0.25$\\
|
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$P(\text{Sarah buys popcorn} | \text{Darren doesn't buy popcorn} \cap \text{Charles buys popcorn}) = 0.25$\\
|
||
|
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$$
|
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\begin{align}
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||
P(\text{3 buy popcorn}) &= P(\text{Darren buys popcorn} \cap \text{Charles buys popcorn} \cap \text{Sarah buys popcorn}) \\
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||
&= 0.4 \cdot 0.7 \cdot 0.55\\
|
||
&= 0.154
|
||
\end{align}
|
||
$$
|
||
|
||
\part
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||
|
||
$$
|
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\begin{align}
|
||
P(\text{Charles buys popcorn} \cap \text{Sarah buys popcorn} \cap \text{Darren doesn't buy popcorn}) &= 0.6 \cdot 0.35 \cdot 0.25\\
|
||
&= 0.0525
|
||
\end{align}
|
||
$$
|
||
|
||
|
||
\end{homeworkProblem}
|
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\pagebreak
|
||
|
||
\begin{homeworkProblem}
|
||
|
||
|
||
Following a flood, 120 tins were recovered from Lee’s corner shop.
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||
Unfortunately, the water had washed off all the labels.
|
||
Of the tins, 50 contained pet food, 20 contained peas, 35
|
||
contained beans and the rest contained soup.
|
||
|
||
\begin{enumerate}[label=(\alph*)]
|
||
\item Lee selects a tin at random, what is the probability that it contains soup?
|
||
\item Lee selects a tin at random, what is the probability that it doesn’t contain pet food
|
||
\item Lee selects two tins at random (without replacement), what is the probability that both
|
||
contains peas
|
||
\item Lee selects two tins at random (without replacement), what is the probability that one
|
||
contains pet food and the other contains peas
|
||
\item Lee selects 3 tins at random (without replacement). What is the probability that one
|
||
contains pet food, one contains peas and one contains beans.
|
||
\item Find the probability that Lee will have to open more than two tins before he finds one
|
||
that doesn’t contain pet food.
|
||
\end{enumerate}
|
||
|
||
\solution
|
||
|
||
\part
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||
|
||
Soup left: 120 - 50 - 20 - 35 = 15\\
|
||
|
||
$$P(\text{soup}) = \frac{15}{120} = 0.125$$
|
||
|
||
\part
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||
|
||
$$P(\text{pet food}^c) = \frac{120-50}{120} = 0.58333$$
|
||
|
||
\part
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||
|
||
$$P(\text{peas} \cap \text{peas}) = \frac{20}{120} \cdot \frac{19}{119} = 0.00254$$
|
||
|
||
\part
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||
|
||
$$P(\text{pet food} \cap \text{peas}) = \frac{50}{120} \cdot \frac{20}{119} \cdot 2 = 0.140056$$
|
||
|
||
\part
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||
|
||
$$P(\text{pet food} \cap \text{peas} \cap \text{beans}) = \frac{50}{120} \cdot \frac{20}{119} \cdot \frac{35}{118} \cdot 3! = 0.12462612164 \approx 0.125$$
|
||
|
||
\part
|
||
|
||
What are the chances that it's just
|
||
|
||
$$\frac{50}{120} \cdot \frac{49}{119} = 0.1715686 \approx 0.172$$
|
||
|
||
|
||
|
||
\end{homeworkProblem}
|
||
\pagebreak
|
||
|
||
\begin{homeworkProblem}
|
||
A school employs 75 teachers. The following table summarizs their length of service at the school, classified by gender.
|
||
|
||
\begin{tabular}{|c|c|c|c|}
|
||
\hline
|
||
& $<$ 3 years & 3 years to 8 years & $>$ 8 years\\
|
||
\hline
|
||
Female & 12 & 20 & 13\\
|
||
\hline
|
||
Male & 8 & 15 & 7\\
|
||
\hline
|
||
\end{tabular}
|
||
\\
|
||
\begin{enumerate}[label=(\alph*)]
|
||
\item Find the probability that a randomly selected teacher is a female
|
||
\item Find the probability that a randomly selected teacher is female given that the teacher has more than 8 years service
|
||
\item Find the probability that a randomly selected teacher is female given that the teacher has less than 3 years service
|
||
\item State, giving a reason, whether or not the event of selecting a female teacher is independent of the event of selecting a teacher with less than 3 years service
|
||
\end{enumerate}
|
||
|
||
\solution
|
||
|
||
\part
|
||
|
||
$$P(\text{female}) = \frac{12+20+13}{75} = \frac{45}{75} = 0.6$$
|
||
|
||
\part
|
||
|
||
$$P(\text{female} | >\text{8 years}) = \frac{P(\text{female} \cap >\text{8 years})}{P(>\text{8 years})} = \frac{\frac{13}{75}}{\frac{20}{75}} = 0.65$$
|
||
|
||
\part
|
||
|
||
$$P(\text{female}|< \text{3 years}) = \frac{P(\text{female} \cap < \text{3 years})}{P(< \text{3 years})} = \frac{\frac{12}{75}}{\frac{20}{75}} = 0.6$$
|
||
|
||
\part
|
||
|
||
Test for independence: $$P(\text{female}| < \text{3 years}) = P(\text{female})$$\\
|
||
Since both are 0.6, the events are independent.
|
||
|
||
\end{homeworkProblem}
|
||
\pagebreak
|
||
|
||
\begin{homeworkProblem}
|
||
|
||
A group of students bought, in total, 25 items of clothing at two shops: Mango and
|
||
Iora. The following table shows how many tops, jeans and sweaters were
|
||
bought at each of the two shops:\\
|
||
\begin{tabular}{|c|c|c|c|}
|
||
\hline
|
||
& Top & Jeans & Sweaters\\
|
||
\hline
|
||
Mango & 3 & 7 & 5\\
|
||
\hline
|
||
Iora & 2 & 5 & 3\\
|
||
\hline
|
||
\end{tabular}
|
||
\\
|
||
One item of clothing is chosen at random from these 25 items:\\
|
||
\begin{enumerate}[label=(\alph*)]
|
||
\item What is the probability that the chosen item is a top?
|
||
\item What is the probability that the chosen item was bought from Iora?
|
||
\item What is the probability that the chosen item is a top and was bought from Iora?
|
||
\item State with a reason whether the events (chosen item is a top) and (chosen item was bought from iora) are independent.
|
||
\item Given that the chosen item is not a top, find the conditional probability that it was
|
||
\end{enumerate}
|
||
\solution
|
||
|
||
\begin{tabular}{|c|c|c|c|c|}
|
||
\hline
|
||
& Top & Jeans & Sweaters & Total\\
|
||
\hline
|
||
Mango & 3 & 7 & 5 & 15\\
|
||
\hline
|
||
Iora & 2 & 5 & 3 & 10\\
|
||
\hline
|
||
Total & 5 & 12 & 8 & 25\\
|
||
\hline
|
||
\end{tabular}
|
||
\\\\
|
||
\part
|
||
|
||
$$P(\text{top}) = \frac{5}{25} = 0.2$$
|
||
|
||
\part
|
||
|
||
$$P(\text{Iora}) = \frac{10}{25} = 0.4$$
|
||
|
||
\part
|
||
|
||
$$P(\text{top} \cap \text{Iora}) = \frac{2}{25} = 0.08$$
|
||
|
||
\part
|
||
|
||
$$P(\text{top} | \text{Iora}) = \frac{P(\text{top} \cap \text{Iora})}{P(\text{Iora})} = \frac{0.08}{0.4} = 0.2$$
|
||
|
||
Test for independence: $$P(\text{top}) = P(\text{top} | \text{Iora})$$\\
|
||
Since both are 0.2, the events are independent.
|
||
\\\\
|
||
\part
|
||
|
||
$$P(\text{Iora} | \text{top}^c) = \frac{P(\text{Iora} \cap \text{top}^c)}{P(\text{top}^c)} = \frac{0.4 \cdot 0.8}{0.8} = 0.4$$
|
||
|
||
|
||
|
||
\end{homeworkProblem}
|
||
\pagebreak
|
||
|
||
\begin{homeworkProblem}
|
||
A volunteer for the Health Literacy Center was investigating the attitudes of students towards
|
||
smoking on campus. A random sample of 730 students from all four grade levels was taken.
|
||
Each student was given the statement "Smoking is dangerous for your health" and asked
|
||
whether they agreed, had no opinion or disagreed. The following contingency table summarizes the results.
|
||
|
||
\begin{enumerate}[label=(\alph*)]
|
||
\item Complete the following table knowing that
|
||
\begin{enumerate}
|
||
\item 387 Male students participated
|
||
\item 277 students agreed and 97 students had no opinion
|
||
\item 154 from those who agreed were female
|
||
\item Only 36 from the males who participated had no opinion
|
||
\end{enumerate}
|
||
\item Among the participants, what is the probability that a student agreed P (A)?
|
||
\item Show whether the events of students agreeing(A) and being a female student (F) are independent
|
||
\item What is the probability that a student is a female given that the student had no opinion $P(F|N)$? What is the probability that a student disagreed, given that the student is Male $P(D|M)$?
|
||
\item Use Bayes' Theorem to calculate the probability $P(M|A)$
|
||
\end{enumerate}
|
||
|
||
\solution\\
|
||
|
||
\part
|
||
|
||
\begin{tabular}{|c|c|c|c|c|}
|
||
\hline
|
||
& Agree (A) & No Opinion (N) & Disagree (D) & Totals\\
|
||
\hline
|
||
Male (M) & 123 & 36 & 228 & 387 \\
|
||
\hline
|
||
Female (F) & 154 &61 &128 &343 \\
|
||
\hline
|
||
Totals &277 & 97 & 356 &730 \\
|
||
\hline
|
||
\end{tabular}
|
||
\\\\
|
||
\part
|
||
|
||
$$P(A) = \frac{277}{730} = 0.38$$
|
||
|
||
\part
|
||
|
||
$$P(A|F) = \frac{P(A \cap F)}{P(F)} = \frac{0.21}{0.47} = 0.45$$\\
|
||
Test for independence: $$P(A) = P(A|F)$$\\
|
||
Since $P(A) = 0.38$ and $P(A|F) = 0.45$, the events are not independent.
|
||
|
||
\part
|
||
|
||
$$P(F|N) = \frac{P(F \cap N)}{P(N)} = \frac{\frac{61}{730}}{\frac{97}{730}} = 0.63$$
|
||
|
||
$$P(D|M) = \frac{P(D \cap M)}{P(M)} = \frac{\frac{228}{730}}{\frac{387}{730}} = 0.59$$
|
||
|
||
\part
|
||
|
||
$$P(M|A) = \frac{P(M \cap A)}{P(A)} = \frac{\frac{123}{730}}{\frac{277}{730}} = 0.44$$
|
||
|
||
\end{homeworkProblem}
|
||
\pagebreak
|
||
|
||
\begin{homeworkProblem}
|
||
Letter A is sent by first-class post and has a probability of 0.9 of being delivered next day. Letter B is sent by second-class post and has a probability of only 0.3 if being delivered the next day.
|
||
\begin{enumerate}[label=(\alph*)]
|
||
\item Find the probability that both letters are delivered the next day
|
||
\item Find the probability that neither letter is delivered the next day
|
||
\item Find the probability that at least one of the letters is delivered the next day
|
||
\item Given that at least one of the letters is delivered the next day, find the probability that letter A is delivered the next day
|
||
\end{enumerate}
|
||
|
||
\solution\\
|
||
|
||
Draw the Probability tree diagram to solve the problem.
|
||
|
||
\part
|
||
|
||
$$P(A \cap B) = P(A) \cdot P(B) = 0.9 \cdot 0.3 = 0.27$$
|
||
|
||
\part
|
||
|
||
$$P(A^c \cap B^c) = P(A^c) \cdot P(B^c) = 0.1 \cdot 0.7 = 0.07$$
|
||
|
||
\part
|
||
|
||
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.9 + 0.3 - 0.27 = 0.93$$
|
||
OR, Using a tree diagram:\\
|
||
$$P(A \cup B) = 0.9 \cdot 0.3 + 0.9 \cdot 0.7 + 0.1 \cdot 0.3 = 0.93$$
|
||
|
||
\part
|
||
|
||
$$P(a|\text{at least 1 letter}) = \frac{P(a \cap \text{at least 1 letter})}{P(\text{at least 1 letter})} = \frac{0.9 \cdot 0.3 + 0.9 \cdot 0.7}{0.93} = 0.9677419 \approx 0.9677$$
|
||
|
||
\end{homeworkProblem}
|
||
\pagebreak
|
||
|
||
\begin{homeworkProblem}
|
||
\textbf{ANSWERS}
|
||
\begin{enumerate}
|
||
\item (b) $\frac{3}{4}$, $c = \frac{1}{2}$, classical interpretation
|
||
\item $\frac{7}{15}$, $\frac{77}{150}$, $\frac{8}{15}$, Relative frequency interpretation
|
||
\item 0.9
|
||
\item 0.882, 0.1694
|
||
\item 0.2,0.3, 0.2/0.3
|
||
\item 0.3, 0.06, 0.34, 0.17647
|
||
\item 0.28, 0.39, 0.33,0.154, 0.0525
|
||
\item 0.125, 0.0583, 0.0266, 0.140, 0.125, 0.172
|
||
\item 0.6, 0.65, 0.6, independent
|
||
\item $\frac{1}{5}$, $\frac{2}{5}$, $\frac{2}{25}$, independent, $\frac{2}{5}$
|
||
\item B)P(A) = $\frac{377}{730}$, C) not Independent, D) $\frac{61}{97}, \frac{228}{387}$, E) 0.44404
|
||
\item A) 0.27, B)0.07, C)0.93, D)0.9677
|
||
\end{enumerate}
|
||
|
||
11a)
|
||
|
||
\begin{tabular}{|c|c|c|c|c|}
|
||
\hline
|
||
& Agree (A) & No Opinion (N) & Disagree (D) & Totals\\
|
||
\hline
|
||
Male (M) & 123 & 36 & 228 & 387 \\
|
||
\hline
|
||
Female (F) & 154 &61 &128 &343 \\
|
||
\hline
|
||
Totals &277 & 97 & 356 &730 \\
|
||
\hline
|
||
\end{tabular}
|
||
\end{homeworkProblem}
|
||
|
||
\end{document}
|