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Probability Theory
![The set of all possible outcomes is called the \textbf{sample space}.\\ If each outcome has an equal chance, then each outcome is equally likely.\\ An event consists of one or more outcomes. $$P(E)=\dfrac{\text { number of outcomes matching } E}{\text { number of outcomes in the sample space }}$$ The probability of an event can be written as a fraction, decimal or percentage.\\ An experiment is a situation involving chance that leads to outcomes.\\ A trial is one run of the experiment.\\ An outcome is the result of an experiment.\\ In a random experiment, every possible outcome has the same chance of occurring.\\ \begin{center} \begin{tikzpicture} \node[below, red] at (0,-0.6cm) {\scriptsize impossible}; \node[below, red] at (8,-0.6cm) {\scriptsize certain}; \foreach \y/\label in {0/no way,1.3/ not likely,4/even chance,6.8/almost definitely} \draw[line width=1pt,latex-] (\y,4pt)--(\y,1cm) node [anchor=south] {\scriptsize \label}; \draw[line width=1pt,latex-] (8,4pt)--(8,0.6cm) node [anchor=south] {\scriptsize must happen}; \draw[line width=1pt] (0,0)--(8,0); \foreach \x/\label in {0/0,4/\frac{1}{2},8/1} \draw[line width=1pt] (\x,0)--(\x,-4pt) node [below,red] {\(\label\)}; \fill (0,0) circle (2pt); \fill (8,0) circle (2pt); \end{tikzpicture} %\includegraphics[width=0.6\textwidth]{1038e232-c766-4e0d-a1a6-6d0bb97a2c21} \end{center} \begin{multicols}{2} \textbf{Example 1}\\ A bag contains 4 blue, 3 red and 2 white counters, what is the probability of:\\ \textbf{i)} Selecting a red counter?\\ \textbf{ii)} Not selecting a white counter?\\ \textbf{iii)} Selecting a blue, red or white counter?\\ \textbf{iv)} selecting a green counter?\\ \textbf{Example 1 solution}\\ $\begin{aligned}[t] \textbf{i)} &\text { Total number of counters }=4+3+2=9\\ &\begin{aligned} P(\text {red}) &=\frac{3}{9} \\ &=\frac{1}{3} \end{aligned} \end{aligned}$\\ $\begin{aligned} \textbf{ii) } \quad P(\text { not white }) & =P(\text { red on blue }) \\ & =\frac{3+4}{9}=\frac{7}{9} \end{aligned}$\\ $\begin{aligned} \textbf{iii) } \quad P(\text { blue, red or white }) & =\frac{4+3+2}{9}=1 \end{aligned}$\\ $\begin{aligned} \textbf{iv) } \quad P(\text { green }) & =\frac{0}{9} \\ & =0 \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ From a standard die (numbered \((1\) to \(6\) ), what is the probability of:\\ \textbf{i)} rolling a 4?\\ \textbf{ii)} rolling a prime number?\\ \textbf{iii)} not rolling an even number?\\ \textbf{Example 2 solution}\\ $\begin{aligned} \textbf{i) } \quad &\text{There are 6 numbers}\\ &\therefore P(4)=\frac{1}{6} \end{aligned}$\\ \textbf{ii)} The prime numbers ave 2,3 and 5 \\ $\begin{aligned} \therefore P(\text { prime }) & =\frac{3}{6} \\ & =\frac{1}{2} \end{aligned}$\\ \textbf{iii)} The odd numbers are \(1,3,5\) \\ $\begin{aligned} \therefore P(\text { not even }) & =\frac{3}{6} \\ & =\frac{1}{2} \end{aligned}$\\ \end{multicols}](/media/0lzjhlnw/9242.png)
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