Probability

Event and Probability

Experiment: Process that results in one or many observations.

Outcome: Observations of an experiment.

Sample Space (S): Set of all possible outcomes.

Event(s): Any subset of the sample space.

An event is a statement about the outcome.
Ex: If we have an outcome that makes the statement true, we say “Event E occurred”

Example: Experiment: Toss a coin once.
S = { Heads, Tails }

Example: Experiment: Toss a coin twice.
S = { HH, HT, TH, TT }
E = { Heads on the first flip }
This is true for { HH, HT }

Set Operators

Complement: A^c or \bar{A}

All outcomes in the sample space that are not in A.

Union: A \cup B

All outcomes that are in A or B (inclusive).

Intersection: A \cap B

All outcomes that are in both A and B.

Disjoint Events: A \cap B = \emptyset

Events A and B are mutually exclusive (they can’t occur together)
- Ex: A and A^c are always disjoint.

Example: Set operators

Q: Given

S = \{ 1, 2, 3, 4, 5, 6 \}

A = \{ 2, 4, 6 \}

B = \{ 3, 6 \}

Find:

A^c
B^c
A \cup B
A \cap B

A^c: { 1, 3, 5 }
B^c: { 1, 2, 4, 5 }
A \cup B: { 2, 3, 4, 6 }
A \cap B { 6 }

Related Notes: Operations on Sets - CS1300
Note: Probability can be represented with Venn diagrams.

Probability Rules

\boxed{ P(E) = \frac{\text{\# of outcomes in $E$ occured}}{\text{\# of outcomes in $S$ occured}} } \\~\\ \small\textit{Given that all outcomes in $S$ are equally likely}

More Rules:

0 \le P(E) \le 1
- Probabilities are always between 0—1.
P(S) = 1
- Probability of sample space must be one.
P(A^c) = 1 - P(A)
- Complement is 1 - probability.
P(A \cup B) = P(A) + P(B) - P(A \cap B)
- We subtract the intersection to avoid double-counting.

Disjoint Events v.s. Independent Events

Let A, B, and C be events.

A and B are disjoint (mutually exclusive) if and only if P(A \cap B) = 0

If A and B are disjoint, P(A \cup B) = P(A) + P(B)
If A, B, and C are disjoint, P(A \cup B \cup C) = P(A) + P(B) + P(C)
A and B are independent if and only if P(A \cap B) = P(A) P(B)
- aka: The occurrence of A doesn’t affect B

Example: Disjoint v.s. Independent

There’s a box containing 1 white ball and 3 black balls.

A = { The first ball is white } B = { The second ball is white }

Exp 1: Select two balls at random in a row with replacement.

Exp 2: Select two balls at random in a row without replacement.

Exp 1:
- Is not disjoint (A and B can happen).
- Is independent (the result of the first pick has no effect on the second)
Exp 2:
- Is disjoint (A and B cannot both happen)
- Is dependent (A affects the occurrence of B)

Marginal & Joint Probability

Marginal Probability: Probability of a single event without considering any other events.

Joint Probability: Probability of two events occurring simultaneously.

Note: Both probabilities are out of n.

Example: Marginal probability

Ex: Select one employee at random out of 100 employees.

	Smoke	Not Smoke
Male	45	15	60
Female	10	30	40
	55	45	100

P(Male) = 60/100 = 0.6
P(Smoker) = 55/100 = 0.55
P(Male \cup Smoker) = 45/100 = 0.45

Conditional Probability

Conditional Probability: P(A|B): Probability of A given B

Probability that event A will occur under the condition that the event B already occurred.

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

Example: Marginal probability

Q: Construct the probability table from the following information

There are 30 students in class, 10 students are female and 12 students are from LA county. There are 3 female students from LA county.

	Female	Male
LA County
Other Counties

	Female
LA County	3	12
Other Counties
	10	30

	Female	Male
LA County	3		12
Other Counties			18
	10	20	30

	Female	Male
LA County	3	9	12
Other Counties	7	11	18
	10	20	30

Q: One student will be selected at random, what is the probability that the selected student is:

Male?
From LA County?
Male and from LA County.
If a male is selected, what is the probability that he is from LA county.

20/30
12/30
9/30
9/20

Conditional Probability and Independent Events

If A and B are independent, then P(A|B) = P(A) and P(B|A) = P(B), furthermore:

\boxed{ \text{If A and B are independent: } P(A) = P(A|B) = P(A | B^c ) }

Multiplication Rule

\boxed{ P(A \cap B) = P(A|B) P(B) \lor P(B|A) P(A) }

Example: Multiplication rule

Q: There are 10 students and 6 of them are women.

We take a sample of two students (w/out replacement), find the probability that both students are women.

\begin{aligned} P(W_1 \cap W_2) &= P(W_1) P(W_2 | W_1) \\ &= \frac{6}{10} \times \frac{5}{9} \\ &= \frac{1}{3} \end{aligned}

Q: Find the probability that the first student is a woman and the second student is a man.

\begin{aligned} P(W_1 \cap M_2) &= P(W_1) P(M_2 | W_1) \\ &= \frac{6}{10} \times \frac{4}{9} \\ &= \frac{4}{15} \end{aligned}

Example: Multiplication rule

Q: Let P(A)=0.7 and P(B|A)=0.4, find P(A \cap B)

\begin{aligned} P(A \cap B) &= P(B|A)P(A) \\ &= 0.4 \times 0.7 \\ &= 0.28 \end{aligned}

Example: Multiplication rule

Q: 70% of a population are male. 40% of males smoke.

When a person is chosen at random from this population, find the probability that the person is a male smoke.

Let A = Male and B = Smoker

We have been told that P(A)=0.7 and P(B|A)=0.4, so:

P(A \cap B) = P(B|A) P(A) = 0.4 \times 0.7 = 0.28

Rule of Total Probability

Rule of Total Probability:

Helps you find complicated marginal probabilities.

\boxed{ \text{Rule of Total Probability: } P(A) = P(A|B)P(B) + P(A|B^c) P(B^c) }

Why?

\begin{aligned} P(A) &= P{ (A \cap B) \cup (A \cap B^c) } \\ &= P(A \cap B) + P(A \cap B^c) \textit{(disjoint)} \\ &= P(A|B)P(B) + P(A|B^c) P(B^c) \textit{(multiplication rule)} \end{aligned}

Example: Rule of total probability

Q: There are 10 students and 6 of them are women.

Suppose we take a sample of two students (w/o replacement). Find the probability that the second student is a woman.

We will consider two cases:

First student is W_1
First student is M_1

\begin{aligned} P(W_2) &= P(W_1 \cap W_2) + P(W_1^c \cap W_2) \\ &= P(W_2 | W_1) P(W_1) + P(W_2 | W_1^c) P(W_1^c) \\ &= P(W_2 | W_1) P(W_1) + P(W_2 | M_1) P(M_1) \\ &= \frac{5}{9} \times \frac{6}{10} + \frac{6}{9} \times \frac{4}{10} \\ &= \frac{3}{5} \end{aligned}

Example: Rule of total probability

Q: There is an urn containing three white balls and four black. One ball will be drawn at random and put back into the urn with another of a kind. Another ball will be drawn at random after adding another ball of the same color into the urn.

What is the probability the second ball is black?

We will consider two cases:

First ball is white (W_1).
First ball is black (B_1).

\begin{aligned} P(B_2) &= P(B_1 \cap B_2) + P(W_1 \cap B_2) \\ &= P(B_2 | B_1) P(B_1) + P(B_2 | W_1) P(W_1) \\ &= \frac{5}{8} \times \frac{4}{7} + \frac{4}{8} \times \frac{3}{7} \\ &= \frac{4}{7} \end{aligned}

Bayes Rule

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

Helps you find complicated conditional probabilities.

Example: Bayes Rule

\begin{aligned} P(W_1 | W_2) &= \frac{P(W_1 \cap W_2)}{P(W_2)} \\ &= \frac{P(W_2|W_1) P(W_1)}{P(W_2|W_1)P(W_1) + P(W_2|M_1)P(M_1)} \\ &= \frac{ \frac{5}{9} \times \frac{6}{10} }{ \frac{3}{5} } \\ &= \frac{5}{9} \end{aligned}