# Topic B4 — Introducing Probability

## Permutations vs. Combinations

Permutations Order Matters

• You have spots to fill (ChoiceSize)
• Draw names from a box (SizePool)
• Names are unique, and you don’t put the names back in the box
• Formula: $\frac{\mathit{SizePool}!}{\mathit{(SizePool-ChoiceSize)}!} = \mathit{SizePool} \times (\mathit{SizePool}-1) \times \dots \times (\mathit{SizePool}-\mathit{ChoiceSize})$
• Simplify by canceling out factorials

Combinations Order Doesn’t Matter

• Intuition: bigger or smaller than permutations? Smaller, because less ways to do it.
• In how many ways can a list have the same elements? $\mathit{ChoiceSize}!$ These are the number of similar lists with different orders!
• Divide permutations by that number

Example

• Pool size 4
• Choice size 2

## Sample space vs. Event

Kind of like population and sample

• Sample space: all the possible outcomes
• Events: a sub set of the sample space

Dice example

• $S={1,2,3,4,5,6}$
• $A=2,3,4$
• $B=4,5,1$
• $A^c = 1,5,6$

## Union, Intersection and Complement

Union:

• Add the two set, don’t repeat the common events
• One or the other occurs
• $P(A \cup B)$

Intersection

• Common between events
• Both occur
• $P(A \cap B)$

Complements

• “The contrary”
• Does not occur

## Unconditional vs. Conditional probability

Unconditional = “Normal”

Conditional on something happening: the prob of the intersection between events over the probability of the condition

$P(A\mid B) = \frac{P(A \cap B)}{P(B)}$

“Occur if the other occurs”

## Dependent vs. Independent events

Independent

• If the conditional prob = the uncoditional (“regular”) one
• $P(A\mid B)=P(A)$
• Occurs with the same probability whether or not the other occurs

Dependent

• $P(A\mid B) \neq P(A)$

## Multiplication

$P(A \cap B) = P(A \mid B) \times P(B) = P(B \mid A) \times P(A)$

Both occur = Prob that A occurs when B occurs, times prob of B occurring

## Total probability rule

An event conditional on two, mutually exclusive, collectively exhaustive events. –> Venn space divided in 2 by $B$ and $B^c$, and $A$ conditional on both

$P(A) = P(A\cap B) + P(A\cap B^c)$

Substituting with the formula above, we get

$P(A) = P(A \mid B) \times P(B) + P(A \mid B^c) \times P(B^c)$

## Bayes’ Theorem

A way to update probabilities: from prior $P(B)$ to updated (i.e. conditional on something happening) $P(B\mid A)$

$P(B\mid A) = \frac{P(A \mid B) \times P(B)}{P(A \mid B) \times P(B) + P(A \mid B^c) \times P(B^c)} = \frac{P(A \cap B)}{P(A)}$

Supplementary Bayes’ theorem material

Great video explanation (and great YouTube channel)