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+ \section{Probability}
+
+ \[ \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) \]
+ \[ \Pr(A \cup B) = 0 \tag{mutually exclusive} \]
+
\section{Conditional probability}
\[ \Pr(A|B) = \frac{\Pr(A \cap B)}{\Pr(B)} \quad \text{where } \Pr(B) \ne 0 \]
\[ \Pr(A) = \Pr(A|B) \cdot \Pr(B) + \Pr(A|B^{\prime}) \cdot \Pr(B^{\prime}) \tag{law of total probability} \]
+
+ \[ \Pr(A \cap B) = \Pr(A|B) \times \Pr(B) \tag{multiplication theorem} \]
For independent events:
\begin{itemize}
- \item \(\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)\)
+ \item \(\Pr(A \cap B) = \Pr(A) \times \Pr(B)\)
\item \(\Pr(A|B) = \Pr(A)\)
\item \(\Pr(B|A) = \Pr(B)\)
\end{itemize}
+ \subsection{Discrete random distributions}
+
+ Any experiment or activity involving chance will have a probability associated with each result or \textit{outcome}. If the outcomes have a reference to \textbf{discrete numeric values} (outcomes that can be counted), and the result is unknown, then the activity is a \textit{discrete random probability distribution}.
+
+ \subsubsection{Discrete probability distributions}
+
+ If an activity has outcomes whose probability values are all positive and less than one ($\implies 0 \le p(x) \le 1$), and for which the sum of all outcome probabilities is unity ($\implies \sum p(x) = 1$), then it is called a \textit{probability distribution} or \textit{probability mass} function.
+
+ \begin{itemize}
+ \item \textbf{Probability distribution graph} - a series of points on a cartesian axis representing results of outcomes. $\Pr(X=x)$ is on $y$-axis, $x$ is on $x$ axis.
+ \item \textbf{Mean $\mu$} or \textbf{expected value} \(E(X)\) - measure of central tendency. Also known as \textit{balance point}. Centre of a symmetrical distribution.
+ \begin{align*}
+ \overline{x} = \mu = E(X) &= \frac{\Sigma(xf)}{\Sigma(f)} \\
+ &= \sum_{i=1}^n (x_i \cdot P(X=x_i)) \\
+ &= \int_{-\infty}^{\infty} x\cdot f(x) \> dx \quad \text{(for pdf } f \text{)}
+ &= \sum_{-\infty}^{\infty}
+ \end{align*}
+ \item \textbf{Mode} - most popular value (has highest probability of \(X\) values). Multiple modes can exist if \(>1 \> X\) value have equal-highest probability. Number must exist in distribution.
+ \item \textbf{Median \(m\)} - the value of \(x\) such that \(\Pr(X \le m) = \Pr(X \ge m) = 0.5\). If \(m > 0.5\), then value of \(X\) that is reached is the median of \(X\). If \(m = 0.5 = 0.5\), then \(m\) is halfway between this value and the next.
+ \[ m = X \> \text{such that} \> \int_{-\infty}^{m} f(x) dx = 0.5 \]
+ \item \textbf{Variance $\sigma^2$} - measure of spread of data around the mean. Not the same magnitude as the original data. For distribution \(x_1 \mapsto p_1, x_2 \mapsto p_2, \dots, x_n \mapsto p_n\):
+ \begin{align*}
+ \sigma^2=\operatorname{Var}(x) &= \sum_{i=1}^n p_i (x_i-\mu)^2 \\
+ &= \sum (x-\mu)^2 \times \Pr(X=x) \\
+ &= \sum x^2 \times p(x) - \mu^2
+ \end{align*}
+ \item \textbf{Standard deviation $\sigma$} - measure of spread in the original magnitude of the data. Found by taking square root of the variance: $\sigma =\operatorname{sd}(X)=\sqrt{\operatorname{Var}(X)}$
+ \end{itemize}
+
+ \subsubsection{Expectation theorems}
+
+ \begin{align*}
+ E(aX \pm b) &= aE(X) \pm b \\
+ E(z) &= z \\
+ E(X+Y) &= E(X) + E(Y) \\
+ E(X)^n &= \Sigma x^n \cdot p(x) \\
+ &\ne [E(X)]^2
+ \end{align*}
+
+
+ \section{Binomial Theorem}
+
+ \begin{align*}
+ (x+y)^n &= {n \choose 0} x^n y^0 + {n \choose 1} x^{n-1}y^1 + {n \choose 2} x^{n-2}y^2 + \dots + {n \choose n-1}x^1 y^{n-1} + {n \choose n} x^0 y^n \\
+ &= \sum_{k=0}^n {n \choose k} x^{n-k} y^k \\
+ &= \sum_{k=0}^n {n \choose k} x^k y^{n-k}
+ \end{align*}
+
+ \begin{enumerate}
+ \item powers of \(x\) decrease \(n \rightarrow 0\)
+ \item powers of \(y\) increase \(0 \rightarrow n\)
+ \item coefficients are given by \(n\)th row of Pascal's Triangle where \(n=0\) has one term
+ \item Number of terms in \((x+a)^n\) expanded \& simplified is \(n+1\)
+ \end{enumerate}
+
+ Combinations: \(^n\text{C}_r = {N\choose k}\) (binomial coefficient)
+ \begin{itemize}
+ \item Arrangements \({n \choose k} = \frac{n!}{(n-r)}\)
+ \item Combinations \({n \choose k} = \frac{n!}{r!(n-r)!}\)
+ \item Note \({n \choose k} = {n \choose k-1}\)
+ \end{itemize}
+
+ \subsubsection{Pascal's Triangle}
+
+ \begin{tabular}{>{$}l<{$\hspace{12pt}}*{13}{c}}
+ n=\cr0&&&&&&&1&&&&&&\\
+ 1&&&&&&1&&1&&&&&\\
+ 2&&&&&1&&2&&1&&&&\\
+ 3&&&&1&&3&&3&&1&&&\\
+ 4&&&1&&4&&6&&4&&1&&\\
+ 5&&1&&5&&10&&10&&5&&1&\\
+ 6&1&&6&&15&&20&&15&&6&&1
+ \end{tabular}
+
\end{document}