## Classical Probability

Probability is the mathematical study of measuring uncertainty. Probabilities are classically determined when their numerical values are based upon an enumeration of every possible outcome.

### The Basic Rule

In classical probability, we call the process which generates outcomes a statistical experiment. A list of all possible outcomes of a statistical experiment is called a sample space. We especially desire that the outcomes in our sample space be equally likely.

Caution: A sample space is really a population of outcomes, not a sample of outcomes.

When examined classically, the probability that an event will occur will be equal to the ratio of the number of outcomes producing that event, to the total number of possible outcomes for that experiment (that is, the size of the sample space). More specifically, if $A$ is the name of an event, $f$ is the frequency with which that event occurs in the sample space, and $N$ is the size of the sample space, then:

 $P(A) = \dfrac{f}{N}$

### Rolling One Die

Suppose our statistical experiment involves rolling one die. Since the die has 6 sides, there are six possible outcomes in the sample space. We can write the sample space as the set ${1,2,3,4,5,6}$. We can also create a probability distribution, which is basically a frequency distribution with the frequency column replaced by a column of probabilities. For rolling one die, the frequency distribution is:

 Outcome on the Die Probability 1 $\dfrac16$ 2 $\dfrac16$ 3 $\dfrac16$ 4 $\dfrac16$ 5 $\dfrac16$ 6 $\dfrac16$

We will let $x$ represent the outcome on the die. Then:

• The probability that the outcome will be a 4 is:
$P(x=4)=\dfrac16=0.1667$.

• The probability that the outcome will be more than 4 is:
$P(x > 4) = P(x=5)+P(x=6) = \dfrac26 = 0.3333$.

• The probability that the outcome will be at least 4 is:
$P(x \geq 4) = P(x=4) + P(x=5)+P(x=6) =\dfrac36 = 0.5$.

• The probability that the outcome will be less than 4 is:
$P(x < 4) = P(x=1) + P(x=2) + P(x=3) = \dfrac36 = 0.5$.

• The probability that the outcome will be at most 4 is:
$P(x \leq 4) = P(x=1) + P(x=2) + P(x=3) + P(x=4) = \dfrac46 = 0.6667$.

• The probability that the outcome will not be a 4 is:
$P(x \neq 4) = P(x=1) + P(x=2) + P(x=3) + P(x=5)+P(x=6) = \dfrac56 = 0.8333$.

• The probability that the outcome will be between 2 and 5, inclusive, is:
$P(2 \leq x \leq 5) = P(x=2) + P(x=3) + P(x=4) + P(x=5) = \dfrac46 = 0.6667$.

• The probability that the outcome will be either 4 or 5 is:
$P(x=4 \text{ or } x=5) = P(x=4) + P(x=5) = \dfrac26 = 0.3333$.

• The probability that the outcome will be both 4 and 5 is:
$P(x=4 \text{ and } x=5) = 0$,
because a die roll is either a 4, or it is a 5, but it cannot be both simultaneously.

The basic rule was used in almost all parts of the preceding example. Notice especially how important the "little" words were. Be sure you understand what "at most" and "at least" mean. Also notice the distinction between "and" and "or".

The last part of the example, where we looked for an outcome of both 4 and 5, turned out to have a probability of zero. We call the events "outcome of 4" and "outcome of 5" mutually exclusive events, which means they cannot happen simultaneously. Usually, the outcomes of a sample space, and therefore in a probability distribution, will be constructed so that they are mutually exclusive.

### Rolling Two Dice

Suppose we now roll two dice. The outcomes of this experiment depend on the two separate outcomes of each die, so there are two independent variables. We can display the 36 outcomes in a table.

 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The sample space can help us determine the following probabilities.

• The probability that the first die is a 5 is:
$P(\text{first die is a 5)} = \dfrac{6}{36} = \dfrac16$.
The six outcomes all occurred in the fifth row.

• The probability that the second die is a 5 is:
$P(\text{second die is a 5)} = \dfrac{6}{36} = \dfrac16$.
The six outcomes all occurred in the fifth column.

• The probability that the first die is a 5 is:
$P(\text{first die is not a 5)} = \dfrac{30}{36} = \dfrac56$.
The thirty outcomes occurred in all rows except the fifth row.

• The probability that the sum of the dice is 5 is:
$P(\text{sum is 5)} = \dfrac{4}{36} = \dfrac19$.
The four outcomes all occurred on the diagonal from (4,1) to (1,4).

• The probability both dice are 5 is:
$P(\text{both are 5)} = \dfrac{1}{36}$.
The only outcome occurred at the intersection of the fifth row and fifth column.

• The probability that at least one die is a 5 is:
$P(\text{at least one is a 5)} = \dfrac{11}{36}$.
The eleven outcomes were found by combining the fifth row and fifth column.

• The probability that neither die is a 5 is:
$P(\text{neither is a 5)} = \dfrac{25}{36}$.
The 25 outcomes were found everywhere but the fifth row or fifth column.

### Other Classical Experiments

There are many other types of classical probability problems besides rolling dice. Examples include flipping coins, drawing cards from a deck, guessing on a multiple choice test, selecting jellybeans from a bag, choosing people for a committee, and so on. Any situation that can be analyzed to determine its component parts, with the probabilities computed on the basis of those parts, is fodder for classical probability.