Motivating Examples

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & 18\\ \hline \textbf{CWD }^- & 78 & 62\\ \hline \end{array} \]

Can ticks carry Chronic Wasting Disease?

Motivating Examples

\[ \begin{array}{|c|c|c|} \hline & \text{Cancer}^+ & \text{Cancer}^-\\ \hline \text{Daily Smoker} & 484 & 1397 \\ \hline \text{Occaisonal Smoker} & 58 & 127 \\ \hline \text{Never Smoked} & 768 & 1377 \\ \hline \hline \end{array} \]

Do drunk cigarettes count?

Complements

Complements

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & 18\\ \hline \textbf{CWD }^- & 78 & 62\\ \hline \end{array} \]

• What is \(P(\text{Ticks})\)?

• What is \(P(\text{No Ticks})\)?

Complements

\[ \begin{aligned} P(\text{Ticks}) = \frac{42 + 78}{42+ 78 + 18 + 62} \\ \\ = \frac{120}{200} = 0.6 \end{aligned} \]

\[ P(\text{No Ticks}) = 1 - P(\text{Ticks}) = 0.4 \]

Complements

\[ \begin{array}{|c|c|c|} \hline & \text{Cancer}^+ & \text{Cancer}^- \\ \hline \text{Daily Smoker} & 484 & 1397 \\ \hline \text{Occaisonal Smoker} & 58 & 127 \\ \hline \text{Never Smoked} & 768 & 1377 \\ \hline \end{array} \]

• Find \(P(\text{Cancer}^+)\)

• Find \(P(\text{Cancer}^-)\)

Complements

\[ \begin{aligned} P(\text{Cancer}^+) = \frac{484+58+768}{1881+185+2145} \\ \\ = \frac{1310}{4211} = 0.31109 \\ \\ P(\text{Cancer}^-) = 1 - 0.31109 = 0.68891 \end{aligned} \]

Unions

Unions

\[ \begin{array}{|c|c|c|c|} \hline & \text{Cancer}^+ & \text{Cancer}^- & \text{Total}\\ \hline \text{Daily Smoker} & 484 & 1397 & 1881\\ \hline \text{Occaisonal Smoker} & 58 & 127 & 185 \\ \hline \text{Never Smoked} & 768 & 1377 & 2145\\ \hline \text{Total} & 1310 & 2901 & 4211 \\ \hline \end{array} \]

• What is \(P(\text{Never Smoked} \cup \text{Cancer}^-)\)?

Unions

\[ \begin{aligned} \text{Never Smoked} & = 768 + 1377 \\ \\ \text{Cancer}^- & = 1397 + 127 + 1377 \\ \\ P(\text{Never Smoked} \cup \text{Cancer}^-) & = \frac{2145 + 2901 - 1377}{4211} \\ & = \frac{3669}{4211} = 0.8712 \\ \end{aligned} \]

Unions

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & 18\\ \hline \textbf{CWD }^- & 78 & 62\\ \hline \end{array} \]

• Find \(P(\text{CWD}^+ \cup \text{Ticks})\)

Intersections

Intersections

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & \boldsymbol{42} & 18\\ \hline \textbf{CWD }^- & 78 & 62\\ \hline \end{array} \]

Intersections

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & \boldsymbol{18}\\ \hline \textbf{CWD }^- & 78 & 62\\ \hline \end{array} \]

Intersections

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & 18\\ \hline \textbf{CWD }^- & 78 & \boldsymbol{62}\\ \hline \end{array} \]

Intersections

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & 18\\ \hline \textbf{CWD }^- & \boldsymbol{78} & 62\\ \hline \end{array} \]

Mutual Exclusivity

Mutual Exclusivity

Given that \(A\) and \(B\) are mutually exclusive:

\[ \begin{aligned} P(A \cap B) &= 0 \quad \quad \textit{by definition}\\ \\ P(A \cup B) &= P(A) + P(B) - P(A \cap B) \\ \\ & = P(A) + P(B) - 0 \\ \\ & = P(A)+P(B) \end{aligned} \]

Contingency Tables

\[ \begin{array}{|c|c|c|c|} \hline \textbf{Event} & B & B^c & \textbf{Total} \\ \hline A & A\cap B & A \cap B^c & A \\ \hline A^c & A^c \cap B & A^c \cap B^c & A^c \\ \hline \textbf{Total} & B & B^c & S\\ \hline \end{array} \]

Contingency Tables

\[ \begin{array}{|c|c|c|c|} \hline \textbf{Event} & D & D^c & \textbf{Total} \\ \hline A & A\cap D & A \cap D^c & A \\ \hline B & B \cap D & B \cap D^c & B\\ \hline C & C \cap D & C \cap D^c & C\\ \hline \textbf{Total} & D & D^c & S\\ \hline \end{array} \]

Conditional Probability

Conditional Probability

\[ P(A \cap B) = \frac{P(A \cap B)}{P(S)} \]

\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

Conditional Probability

\[ \begin{array}{|c|c|c|c|} \hline & \text{Cancer}^+ & \text{Cancer}^- & \text{Total}\\ \hline \text{Daily Smoker} & 484 & 1397 & 1881\\ \hline \text{Occaisonal Smoker} & 58 & 127 & 185 \\ \hline \text{Never Smoked} & 768 & 1377 & 2145\\ \hline \text{Total} & 1310 & 2901 & 4211 \\ \hline \end{array} \]

• Given that you have cancer, what is the probability that it’s attributed to occasional smoking?

Conditional Probability

\[ P(\text{Occasional Smoker} | \text{Cancer}^+) \\ \]

\[ \begin{aligned} & = \frac{P(\text{Occasional Smoker} \cap \text{Cancer}^+ )}{P(\text{Cancer}^+)} \\ \\ & = \frac{58}{1310} = 0.044 \end{aligned} \]

Conditional Probability

\[ \begin{array}{|c|c|c|c|} \hline & \text{Cancer}^+ & \text{Cancer}^- & \text{Total}\\ \hline \text{Daily Smoker} & 484 & 1397 & 1881\\ \hline \text{Occaisonal Smoker} & 58 & 127 & 185 \\ \hline \text{Never Smoked} & 768 & 1377 & 2145\\ \hline \text{Total} & 1310 & 2901 & 4211 \\ \hline \end{array} \]

• Find \(P( \text{Daily Smoker} | \text{Cancer}^+)\)

Conditional Probability

\[ P( \text{Daily Smoker} | \text{Cancer}^+) \]

\[ \begin{aligned} & = \frac{P( \text{Daily Smoker} \cap \text{Cancer}^+)}{P(\text{Cancer}^+)} \\ \\ & = \frac{484}{1310} = 0.3694 \end{aligned} \]

Conditional Probability

\[ \begin{array}{|c|c|c|c|} \hline & \text{Cancer}^+ & \text{Cancer}^- & \text{Total}\\ \hline \text{Daily Smoker} & 484 & 1397 & 1881\\ \hline \text{Occaisonal Smoker} & 58 & 127 & 185 \\ \hline \text{Never Smoked} & 768 & 1377 & 2145\\ \hline \text{Total} & 1310 & 2901 & 4211 \\ \hline \end{array} \]

• Do drunk cigarettes count?

Conditional Probability

\[ P(\text{Never Smoked} | \text{Cancer}^+ ) \]

\[ \begin{aligned} & = \frac{P(\text{Never Smoked} \cap \text{Cancer}^+ )}{P(\text{Cancer}^+)} \\ \\ & = \frac{768}{1310} = 0.5862 \end{aligned} \]

Conditional Probability

\[ \begin{array}{|c|c|c|} \hline & \textbf{Ticks} & \textbf{No Ticks}\\ \hline \textbf{CWD }^+ & 42 & 18\\ \hline \textbf{CWD }^- & 78 & 62\\ \hline \end{array} \]

• Can ticks carry Chronic Wasting Disease?

Conditional Probability

\[ \begin{aligned} P(\text{Ticks} | \text{CWD}^+) = \frac{42}{60} = 0.7 \\ \\ P(\text{Ticks} | \text{CWD}^-) = \frac{78}{140} = 0.557\\ \\ P(\text{CWD}^+ | \text{Ticks}) = \frac{42}{120} = 0.35 \\ \\ P(\text{CWD}^- | \text{Ticks}) = \frac{78}{120} = 0.65 \\ \end{aligned} \]

Independence

By definition, if \(A\) and \(B\) are independent:

\[ P(A|B) = P(A) \]

\[ P(B|A) = P(B) \]

Independence

Assume \(A\) and \(B\) are independent:

\[ \begin{aligned} P(A|B) = \frac{P(A \cap B)}{P(B)} = P(A) \\ \\ \frac{P(A \cap B)}{P(B)} = P(A)\\ \\ P(A \cap B) = P(A)P(B) \\ \end{aligned} \]

Independence

What’s the probability that flipping a coin 10 times results in 10 heads?

\[ P(H) = 0.5 \]

\[ P(T) = 0.5 \]

Independence

\[ P(H_1) \times P(H_2) \times \ ... \times P(H_{10}) \]

\[ 0.5^{10} = 0.000977 \]

Independence

For independent and equi-probable events, the probability of event \(A\) occurring \(n\) many times in sequence :

\[ \prod_{i=1}^n P(A_i) = P(A_1) \times P(A_2) \times \ ... \times P(A_n) = P(A)^n \]