---
title: "Sample Spaces, Events, and Probability Rules"
subtitle: "SD03a — Probabilities"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
      - text: "Download R Markdown"
        icon: file-earmark-arrow-down
        href: SD03a-01-sample-spaces-and-probability-rules.Rmd?download=1
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, message = FALSE, warning = TRUE)
```

**Course:** Statistics and Probability for Data Science (DS14270).  
**Source deck:** Samer El Kababji, *SD03a — Probabilities*.

This lesson develops the deck's sample-space notation, probability axioms,
event operations, and addition rules. The practice questions and the extra
computational checks extend the deck's concepts.

## Experiments, outcomes, and sample spaces

An **experiment** is a process with more than one possible outcome. Its
**sample space** $S$ is the set of every possible outcome.

Suppose three welds are inspected in sequence. Write $N$ for nondefective and
$D$ for defective. Each outcome is a string of length three, so

$$
S=\{NNN,NND,NDN,NDD,DNN,DND,DDN,DDD\}.
$$

There are $2^3=8$ outcomes because each of the three positions has two
possibilities.

An **event** is a subset of $S$. The event that all three welds are
nondefective is the simple event $\{NNN\}$. The compound event that exactly
two welds are defective is

$$
A=\{NDD,DND,DDN\}.
$$

```{r weld-sample-space}
weld_grid <- expand.grid(
  weld_1 = c("N", "D"),
  weld_2 = c("N", "D"),
  weld_3 = c("N", "D")
)
weld_outcomes <- apply(weld_grid, 1, paste0, collapse = "")
defect_count <- nchar(gsub("N", "", weld_outcomes))

S <- sort(weld_outcomes)
A <- sort(weld_outcomes[defect_count == 2])

S
A
```

## Assigning probabilities to outcomes

For a finite sample space with outcomes $O_1,\ldots,O_n$, assign probabilities
$p_1,\ldots,p_n$ satisfying

$$
0\leq p_i\leq 1,
\qquad
\sum_{i=1}^{n}p_i=1,
\qquad
P(O_i)=p_i.
$$

For any event $E\subseteq S$, add the probabilities of the outcomes it
contains:

$$
P(E)=\sum_{O_i\in E}P(O_i).
$$

**Worked example — three independent weld results.** If $N$ and $D$ are
equally likely for each inspection, all eight sequences have probability
$1/8$. Therefore

$$
P(A)=P(NDD)+P(DND)+P(DDN)
=\frac18+\frac18+\frac18=\frac38.
$$

```{r weld-event-probability}
outcome_probability <- setNames(rep(1 / length(S), length(S)), S)
P_A <- sum(outcome_probability[A])

c(number_of_outcomes = length(S), P_A = P_A)
stopifnot(isTRUE(all.equal(P_A, 3 / 8)))
```

Equal probabilities are an assumption here. In another experiment, the
outcomes may have different probabilities while still summing to one.

## Complements

The complement $A^c$ contains all outcomes in $S$ that are not in $A$:

$$
A^c=S\setminus A,
\qquad
P(A^c)=1-P(A).
$$

For the weld event,

$$
A^c=\{NNN,NND,NDN,DNN,DDD\},
\qquad
P(A^c)=1-\frac38=\frac58.
$$

```{r weld-complement}
A_complement <- setdiff(S, A)
P_A_complement <- sum(outcome_probability[A_complement])

A_complement
c(P_A = P_A, P_A_complement = P_A_complement,
  total = P_A + P_A_complement)
```

The complement rule is especially useful for probabilities such as "at least
one," whose complement is often a single simpler event.

## Intersection and union

For events $A$ and $B$:

$$
A\cap B=\{\text{outcomes in both }A\text{ and }B\},
$$

$$
A\cup B=\{\text{outcomes in }A\text{ or }B\text{ or both}\}.
$$

Thus, $P(A\cap B)$ is the probability that both events occur, while
$P(A\cup B)$ is the probability that at least one occurs.

The addition rule prevents outcomes in the intersection from being counted
twice:

$$
P(A\cup B)=P(A)+P(B)-P(A\cap B).
$$

If $A\cap B=\varnothing$, the events are **mutually exclusive** or
**disjoint**, and the rule reduces to

$$
P(A\cup B)=P(A)+P(B).
$$

## Worked example: rolling two dice

Roll two fair six-sided dice. Let

$$
A=\{\text{the sum is }6\},
\qquad
B=\{\text{at least one die is }6\}.
$$

There are 36 equally likely ordered outcomes. Event $A$ has five outcomes:

$$
(1,5),(2,4),(3,3),(4,2),(5,1).
$$

Event $B$ has $6+6-1=11$ outcomes; $(6,6)$ is subtracted once because it was
counted twice. No outcome can have sum 6 and contain a die showing 6, so
$A\cap B=\varnothing$. Therefore

$$
P(A\cup B)=\frac5{36}+\frac{11}{36}
=\frac{16}{36}=\frac49.
$$

```{r two-dice-example}
dice <- expand.grid(die_1 = 1:6, die_2 = 1:6)
A_dice <- with(dice, die_1 + die_2 == 6)
B_dice <- with(dice, die_1 == 6 | die_2 == 6)

probability <- function(event) mean(event)

c(
  P_A = probability(A_dice),
  P_B = probability(B_dice),
  P_intersection = probability(A_dice & B_dice),
  P_union = probability(A_dice | B_dice)
)
stopifnot(isTRUE(all.equal(probability(A_dice | B_dice), 4 / 9)))
```

## Probability axioms and consequences

Probability is built from three axioms:

1. $P(E)\geq0$ for every event $E$.
2. $P(S)=1$.
3. For pairwise disjoint events $E_1,E_2,\ldots$,

$$
P\!\left(\bigcup_{i=1}^{\infty}E_i\right)
=\sum_{i=1}^{\infty}P(E_i).
$$

These axioms imply useful propositions:

$$
P(\varnothing)=0,
\qquad
P(E)+P(E^c)=1,
$$

and, for three events,

$$
\begin{aligned}
P(A\cup B\cup C)
={}&P(A)+P(B)+P(C)\\
&-P(A\cap B)-P(A\cap C)-P(B\cap C)\\
&+P(A\cap B\cap C).
\end{aligned}
$$

The final triple-intersection term is added back because subtracting all three
pairwise intersections removes outcomes shared by all three too many times.

## Practice questions

1. Toss two fair coins. Write the sample space and the event that exactly one
   head occurs. Find its probability.
2. In the weld experiment, find the probability of at least one defective
   weld under the equal-probability assumption.
3. Roll one fair die. Let $C=\{1,2,3\}$ and $D=\{3,4,5\}$. Find
   $P(C\cap D)$ and $P(C\cup D)$.
4. Are $C$ and $D$ in Question 3 mutually exclusive? Explain using their
   intersection.
5. For two fair dice, let $E$ be "the first die is even" and $F$ be "the sum
   is even." Compute $P(E\cup F)$ using the addition rule.

::: {.callout-note collapse="true"}
### Check your answers

1. $S=\{HH,HT,TH,TT\}$, the event is $\{HT,TH\}$, and its probability is
   $2/4=1/2$.
2. The complement is $\{NNN\}$, so the probability is
   $1-P(NNN)=1-1/8=7/8$.
3. $C\cap D=\{3\}$, so $P(C\cap D)=1/6$. Also,
   $C\cup D=\{1,2,3,4,5\}$, so $P(C\cup D)=5/6$.
4. No. Their intersection contains the outcome 3.
5. $P(E)=1/2$, $P(F)=1/2$, and $P(E\cap F)=1/4$. Hence
   $P(E\cup F)=1/2+1/2-1/4=3/4$.
:::