- Probability is the branch of mathematics
quantifies uncertainty.
- Probabilites are measured on a scale from zero to one
where events with no chance of occuring have probability zero
and events that are certain to occur have probability one.
- Probability may be thought of as the long-run relative frequency
with which a particular event would occur
if the same random experiment could be repeated indefinitely.
This viewpoint is the
**frequentist**interpretation of probability. - Another school of thought views probability
as a measure of the strength of belief.
This explicitly subjective viewpoint of probability is called
**Bayesian**. - The set of all possible outcomes in a random experiment
is called the
**outcome space**. - An
**event**is a subset of the outcome space. - If we assume that all of the outcomes are equally likely,
then the probability of an event is the number of outcomes in the event
divided by the number of outcomes in the outcome space.
- For example, in rolling a single die,
the outcome space is the set {1,2,3,4,5,6}.
The event "a 1 is rolled" has probability 1/6 while the event "an even number is rolled" has probability 3/6.

- Two events are
**mutually exclusive**if they have no outcomes in common. In other words, if one event occurs, the other cannot possibly occur. The events "a 1 is rolled" and "an even number are rolled" are a pair of mutually exclusive events. **Addition Principle:**If A and B are mutually exclusive events, then P(A or B) = 0.Here, "A or B" is the event that either A occurs or B occurs or both occur. It does not mean that exactly one of A or B occur.

- Two events are
**independent**if the outcome of one does not affect the probability of the outcome of the other. **Multiplication Principle:**If A and B are independent events, then P(A and B) = P(A) * P(B).Here, "A and B" is the event of outcomes in both A and B. The symbol "*" means multiplication.

**Inclusion-Exclusion:**If A and B are not mutually exclusive, there is another formula for the probability of their union. This formula is always correct.P(A or B) = P(A) + P(B) - P(A and B)

In calculating P(A), we count all outcomes in A. In calculating P(B), we count all outcomes in B. We have overcounted all outcomes in both A and B, so get the correct probability by subtracting these off.

- Counting (enumeration) problems often involve these types of calculations.
**Factorials**. The number of ways to arrange k labeled items in is k! = k*(k-1)*(k-2)* ... * 1 if k is a positive integer.Because there is exactly one way to arrange zero items in order, 0!=1.

**Permutations**. The number of ways to select r items from k labeled items in order is_{k}P_{r}= k*(k-1)*(k-2)* ... *(k-r+1) if k and r are positive integers and r is not greater than k.There are r factors in this product.

Notice that

_{k}P_{k}= k!.Also, in general,

_{k}P_{r}= k! / (k-r)!.**Combinations:**. The number of ways to choose r items from k labeled items without regard to order is_{k}C_{r}= k! / (r! * (k-r)!).Notice that

_{k}C_{r}=_{k}P_{r}/ r!. There are r! ways to place the r chosen items in order after they have been selected.

**Example**. If a box has ten balls, five black and five white and three balls are drawn out at random without replacement, what is the probability of drawing out three black balls?**Solution 1:**There are_{10}P_{3}= 10*9*8 = 720 ways to choose three of the ten balls in order. There are_{5}P_{3}= 5*4*3 = 60 ways to choose three of the five black balls in order. Thus, the probability is 60/720 = 1/12.**Solution 2:**There are_{10}C_{3}= 10*9*8/(3*2*1) = 120 ways to choose three of the ten balls without regard to order. There are_{5}C_{3}= 5*4*3/(3*2*1) = 10 ways to choose three of the five black balls without regard to order. Thus, the probability is 10/120 = 1/12.

Last modified: January 18, 2001