Probability Notes Math 309

Sample spaces, events, axioms Math 309 Chapter 1

Some Definitions Experiment - means of making an observation Sample Space (S) - set of all outcomes of an experiment listed in a mutually exclusive and exhaustive manner Event - subset of a sample space Simple Event - an event which can only happen in one way; )

Since events are sets, we need to understand the basic set operations Intersection (A  B) - everything in A and B Union (A  B) - everything in A or B or both Complement (A C ) - everything not in A Difference A – B = A  B C – everything in A that is not in B

You should be able to sketch Venn diagrams to describe the intersections, unions, & complements of sets. Note that these set operations obey the commutative, associative, and distributive laws

DeMorgan’s Laws (A  B) C = (A C  B C ) (A  B) C = (A C  B C ) Convince yourself that these are reasonable with Venn diagrams!

Another definition - A and B are mutually exclusive iff A  B = 

Axioms of Probability (these are FACT, no proof needed!) Let A represent an event, S the sample space, P(S) = 1 For pairwise mutually exclusive events, the probability of their union is the sum of their respective probabilities, i.e. P(A 1  A 2 ...  A n ...) = P(A 1 )+P(A 2 )+... +P(A n ) +...

Theorems (You should be able to prove these using the axioms and definitions.) Thm The probability of the empty set is zero. Thm 1.2 – Let {A 1, A 2,...,A n } be a mutually exclusive set of events. Then P(A 1  A 2 ...  A n ) = P(A 1 ) + P(A 2 ) P(A n )

From Axiom 3, it can be shown that: (Prop. 1*) Let {A 1, A 2,...,A n } be a mutually exclusive set of events. Then P(A 1  A 2 ...  A n ) = P(A 1 ) + P(A 2 ) P(A n )

Let A and B be mutually exclusive, our last theorem with n = 2 gives: P(A  B) = P(A) +P(B) Letting B = A C along with axioms 1 & 2 gives: 0 <= P(A)<= 1. (Can you show this?)

Let A and B be mutually exclusive, our last theorem with n = 2 gives: P(A  B) = P(A) +P(B)

More Theorems Let A and B be any two events. Thm P(A C ) = 1 - P(A) Thm 1.5- If A  B, then P(B-A) = P(B  A C ) = P(B) – P(A) – Corollary: If A is a subset of B, then P(A) <= P(B). Thm 1.6 P(A  B) = P(A) + P(B) - P(A  B) Thm 1.7 P(A) = P(A  B) + P(A  B C )

More Theorems (Propositions) Let A and B be any two events. Prop P(A C ) = 1 - P(A) Prop If A is a subset of B, then P(A) <= P(B) Prop P(A  B) = P(A) + P(B) - P(A  B) Prop. 2* - P(A) = P(A  B) + P(A  B C )

Unions get complicated if events are not mutually exclusive! P(A  B  C) = P(A) + P(B) + P(C) - P(A  B) - P(A  C) - P(B  C) + P(A  B  C) A C B

Sample Spaces with Equally Likely Outcomes In an experiment where all sample points are equally likely, one can find the probability of an event by counting two sets.

Combinatorial Methods Math 309 Chapter 1

Combinatorics Basic Principle of Counting –(a.k.a. Multiplication Principle) Permutations –Permutations with indistinguishable objects Combinations

Basic Counting Principle If experiment 1 has m outcomes and experiment 2 has n outcomes, then there are mn outcomes for both experiments. The principle can be generalized for r experiments. The number of outcomes of r experiments is the product of the number of outcomes of each experiment.

We define experiment as a means of making an observation (e.g. flip a coin, choose a color). Each experiment could be making a choice from a different set.

Permutations # of arrangements of one set, order matters application of the basic counting principle where we return to the same set for the next selection P(n,r) = n!/(n-r)!

Permutations with Indistinguishable Objects Order the objects as if they were distinguishable Then “divide out” those arrangements that look identical.

Combinations the number of selections, order doesn’t matter – C(n,r) = n!/[(n-r)!r!] the number of arrangements can be counted by selecting the objects and then ordering them –i.e. P(n,r) = C(n,r)*r!

Observations about Combinations C(n, r) = C(n, n-r) C(n, n) = C(n, 0) = 1 C(n, 1) = n = C(n, n-1) C(n, 2) = n(n-1)/2

Combining Counting Techniques If we are careful with language, –when we say “AND”, we multiply –“AND”  multiplication  intersection –when we say “OR”, we add –“OR”  addition  union

Conditional Probability and Independence Math 309 Chapter 3

Conditional Probability P(A|B) P(A|B) is read, “the probability of A given B” B is known to occur. P(A|B) = P(A  B) / P(B), if P(B) > 0 i.e. the conditional probability is the probability that both occur divided by what is given occurs

The multiplication rule and  intersection  multiply P(A  B) = P(A)*P(B|A) = P(B)*P(A|B) (Note that this is an algebraic manipulation of the formula for conditional probability.) Intersections get more complicated when there are more events, e.g. P(A  B  C  D) = P(A)* P(B|A)*P(C|A  B)*P(D|A  B  C)

Independent Events A and B are independent if any of the following are true: –P(A  B) = P(A)*P(B) –P(A|B) = P(A) –P(B|A) = P(B) You need to check probabilities to determine if events are independent. If A, B, C, & D are pairwise independent, –P (A  B  C  D) = P(A)*P(B)*P(C)*P(D)