The probability mass function (p.m.f.) f(x) of a discrete random variable X is a function that satisfies the following properties:

(a) f(x) > 0, if x ε S

(b) Σf(x) = 1, for x ε S

(c) P(X ε A) = Σf(x), for x ε A where A is a proper subset of S 

Given a random experiment with an outcome space S, a function X that assigns one and only one real number X(s)=x to each element in S is called a random variable.

The space of X is the set of real numbers {x: X(s)=x, s ε S}, where s ε S means that s belongs to the set S.

mathematical expectation

(expected value)

If f(x) is the p.m.f. of the random variable X of the discrete type with space S, and if the summation Σu(x)f(x) for x ε S exists, then the sum is called the mathematical expectation or expected value of the function u(X), and it is denoted by E[u(X)]. That is

 E[u(X)] = Σu(x)f(x) for x ε S

properties of mathematical expectation

When it exists, the mathematical expectation E satisfies the following properties: 

(a) If c is a constant, then E(c)=c.

(b) If c is a constant and u is a function, then

E[cu(X)]=cE[u(X)].

(c) If c₁ and c₂ are constants and u₁ and u₂ are functions, then E[c₁u₁(X) + c₂u₂(X)] = c₁E[u₁(X)] + c₂E[u₂(X)] 

(A ∪ B)' = A' ∩ B'

(A ∩ B)' = A' ∪ B'

A ∪ B = B ∪ A

A ∩ B = B ∩ A 

 (A ∪ B) ∪ C = A ∪ (B ∪ C)

(A ∩ B) ∩ C = A ∩ (B ∩ C) 

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) 

1. A Bernoulli (success-failure) experiment is performed n times.

2. The trials are independent.

3. The probability of success on each trial is a constant p; the probability of failure is q = 1 - p.

4. The random variable X equals the number of successes in the n trials.

μ = np

σ² = np(1-p) or σ² = npq

σ = √(np(1-p)) or σ = √(npq)

The conditional probability of an event A, given that event B has occurred, is defined by

P(A|B) = P(A∩B)/P(B)

provided that P(B) > 0.

The probability that two events, A and B, both occur is given by the multiplication rule:

P(A ∩ B) = P(A)P(B|A)

or by

P(B ∩ A) = P(B)P(A|B).

addition rule for two events

(Hogg and Tanis theorem 1.2-5)

The probability that either (or both) of two events, A and B, occur is given by the addition rule:

P(A U B) = P(A) + P(B) - P(A ∩ B).

proof hint 1: Rewrite A U B as the union of mutually exclusive events: A U (A' ∩ B).

proof hint 2: Do that trick again: B = (A ∩ B) U (A' ∩ B)

Probability is a real-valued set function P that assigns, to each event A in the sample space S, a number P(A), called the probability of the event A, such that the following properties are satisfied:

(a) P(A) ≥ 0,

(b) P(S) = 1, 

(c) If A₁, A₂, A₃, ... are events and A_i ∩ A_j = Ø, i ≠ j, then

P(A₁ U A₂ U ... U A_k) = P(A₁) + P(A₂) + ... + P(A_k)

for each positive integer k, and

P(A₁ U A₂ U A₃ U ...) = P(A₁) + P(A₂) + P(A₃) + ...

for an infinite, but countable, number of events.

probability of the complement event A

(Hogg and Tanis theorem 1.2-1)

For each event A, P(A) = 1 - P(A')

Proof hint: S = A + A'

probability of Ø

(Hogg and Tanis theorem 1.2-2)

P(Ø) = 0

Proof hint: Let A = Ø then A' = S, then apply probability of inverse theorem

relationship between P(A) and P(B) if A is a subset of B

(Hogg and Tanis theorem 1.2-3)

If events A and B are such that A is a subset of B, then P(A) ≤ P(B) 

Proof hints: start with B = A U (B ∩ A'). Remember that for any event A, P(A) ≥ 0.

upper limit of probability of event A

(Hogg and Tanis theorem 1.2-4)

P(A) ≤ 1

Proof hint: A is a subset of S

addition rule for three events

(Hogg and Tanis theorem 1.2-6)

The probability that at leat one of three events, A, B or C, occur is given by the addition rule:

P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B)- P(A ∩ C) -  P(B ∩ C) + P(A ∩ B ∩ C)

proof hint: A U B U C = A U (B U C)

Events A and B are independent if and only if P(A ∩ B) = P(A)P(B). Otherwise A and B are called dependent events.

(Note that this is equivalent to saying P(B|A) = P(B).)

independence of events and complements

(Hoggand Tanis theorem 1.5-1)

If A and B are independent events, then the following pairs of events are also independent:

(a) A and B',

(b) A' and B,

(c) A' and B'

proof hint for (a): P(B' |A) =  1 - P(B|A)