Conditional probability

$P (A | condition)$ = Probability of A given condition

given effectively reduces sample space
P (Q | face card)
- = 4/12

Bayes Theorem

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

$P (A | B) = P (A) \to A and B are independent events$

Extended Bayes' Theorem (applying rule of total probability)

P (B_{j} | A) = \frac{P (A | B_{j}) P (B_{j})}{\sum_{i = 1}^{k} P (A | B_{i}) P (B_{i})}

$P (A \cap B) = P (A) P (B | A)$
$P (A | B) = 1 - P (\bar{A} | B)$
$P (A \cap B) = P (A) P (B | A) = P (B) P (A | B)$
$P (A \cap B) = P (A B)$ (notation)
$P (A B C) = P (A) P (B | A) P (C | A B)$ $= P (A B) P (C | A B)$

$P (A_{1} A_{2} A_{3} \dots)$

If $A_{1}, A_{2}, A_{3}, \dots A_{n}$ are in a sample space such that $P (A_{1}) \neq 0, P (A_{1} A_{2}) \neq 0, \dots P (A_{1} A_{2} \dots A_{n - 1}) \neq 0$ then

$P (A_{1} A_{2} \dots A_{n}) = P (A_{1}) P (A_{2} | A_{1}) P (A_{3} | A_{1} A_{2}) \dots P (A_{n} | A_{1} A_{2} \dots A_{n - 1})$

$P (A B) = P (A) P (B | A) = P (B) P (A | B)$
$P (A) = P (A | B) + P (A | \bar{B})$
$P (A) = P (A | B) P (B) + P (A | C) P (C) + P (A | D) P (D) \dots$

$P (A) = P (A B) + P (A C) + P (A D) \dots$ given context

If $A$ and $B$ are independent $\to (A$ and $\overset{―}{B})$ and $(\overset{―}{A}$ and $B)$ are also independent

P (A_{1} A_{2} \dots A_{k}) = P (A_{1}) P (A_{2}) \dots P (A_{k}) ⟺ A_{1}, A_{2}, \dots A_{k} are independent

Pairwise Independency

3 or more events can be independent when paired against one another (pairwise), but not when all grouped together

Sensitivity

ability of a test to designate an individual with the disease as positive
$P (B | A)$
sensitivity + false negative = $P (B | A) + P (\bar{B} | A) = 1$

Specificity

ability of a test to designate an individual without the disease as negative
$P (\bar{B} | \bar{A})$
specificity + false positive = $P (\bar{B} | \bar{A}) + P (B | \bar{A}) = 1$