Business, as in life, is full of uncertainty. A brief introduction to a few concepts in the world of probability will help with the understanding of other topics such as risk, insurance, marketing research and statistics. The probability of an event happening is expressed as being between zero and one, where zero is never happening and one is definitely happening. First, think of some event where the outcome is uncertain. The probability would be between 0 and 1. Probability is about chances or likely outcomes. Probability is a numerical measure of the likelihood of an event occurring. If we were to randomly select a card from a well shuffled deck of 52 playing cards, we would be just as likely to select a heart as a diamond, spade or club. The probability of selecting a heart is 13/52, or 1/4 or 0.25.


Suppose that a sports figure is the favorite to win a tournament at nine to five (9-5) odds. What does this mean? The odds of an event is the ratio of the probability that the event will not occur to the probability that the event will occur. In our example, the event is “the particular sports figure will win”. The odds of this event are nine to five or 9/5 meaning that it is more likely for the sports figure to lose the tournament than win the tournament. How can we convert odds to probabilities? If the odds of an event are stated as A to B (or A – B or A/B), then the probability of the event is B / (B + A) and in this case 5/14 = 0.3571. The probability that the sports figure will win the tournament is 0.3571 when the odds are 9 to 5.

Understanding Probability

Suppose the probability of winning a prize in a lottery is 1/20, or 5%. In the long term, 5% of the lottery tickets will win a prize and 95% will not. It doesn't mean that if you go out and buy 20 tickets, one of them will win a prize. If you buy many sets of 20 tickets you may find that some sets of tickets have no winners, while other sets have multiple winners. The binomial probability model could be used to calculate the probability of winning at least one prize in a group of tickets. Probabilities often apply to the percentage of a large group. Suppose you know that 40% of the people in your city support a political party, called A. If you randomly select a person, all you can say is that it is not likely that they support party A because the probability is less than 50%.

Rolling Dice Example

With one fair die, the probability of getting any one of the six numerical possibilities is 1/6. What about rolling two die? There are six possible outcomes for the first die and six possible outcomes for the second die. If you multiply the numbers together you get 36 possible outcomes. Suppose we were to look at the sum of the two die and show it in the following chart.

First Die

1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

Sum of 2 Die

1 2 3 4 5 6

Second Die

The probability of getting the sum of 12 can only happen when both die show 6. The probability is 1/36. What is the probability of getting a sum of four? Since this can happen three ways out of a possible 36 ways, the probability is 3/36, or 1/12. The most likely sum is seven. It's probability is 1/6. The probability of getting "doubles" is also 1/6. The probability of getting an even sum is the same as getting an odd sum and is 50%. What if you had three fair die? What's the probability of getting "three of a kind"? Think of rolling the die one at a time. It doesn't matter what the first die's number is. The next two die must turn up the same number as the first die. The probability of that is the product of 1/6 and 1/6 which is 1/36, which is the probability of getting three of a kind. Similarly, the odds of getting five of a kind is 1/6 to the exponent 4, which is 1/1296. The odds of winning a popular lottery can be more than ten million to one. That would be about as likely as rolling a handful of ten fair die with all of the numbers coming up the same. 1/6 to the exponent 9 is about 1/10 million.

Arrangements of a Set

To further understand probability, first we look at sets of items and how they can be arranged. Consider the set A = {a, b, c, d, e, f}.
Note that no two items in the set are the same. How may different ways can we arrange these six elements into a set of three elements where the order of the elements matters? If we begin to make a list of them we have {a, b, c}, {a, b, d}, {a, b, e} and so on. We can think of this as having three spaces to fill from the set A. The first space can be any one of the six elements, so there are six different ways to fill the first space. There are five elements left from the set A, so there are five different ways to fill the second space. There are four different ways to fill the third space.

We intuitively know that if there are n ways to do a first act and for each of those there are r ways to do a second act, there are nr (n multiplied by r) ways to to do the acts successively. Applying this concept to our first example, there are 6 x 5 x 4 = 120 ways to arrange 3 items from 6 items. Expressed another way, there are 120 different 3-arrangements of 6 unique objects.


A permutation is an ordered arrangement of a set of objects. Given the three objects a, b, c, there are 6 different ways to
arrange them: abc, acb, bac, bca, cab, cba. In this example we are finding the number of permutations of size 3 in a set of 3.
If we are selecting from n unique objects to make r-arrangements of objects in order, then we can define this as a permutation.
Permutation notation can be written in three different ways such as:

The number of r-arrangements of n objects is:

The number of 3-arrangements of 7 unique objects is:

The number of ways of arranging the letters in the word math if the first letter must be a t is the same as arranging the letters mah which is:

The number of ways of arranging all the elements of set D = {a, b, c, d, e} where the letters b and d must be together can be approached first by thinking of the letters b and d as one object. However, for each of these arrangements, the letters b and d can be interchanged without altering the position of the other letters. If we let the number of arrangements of all letters in the set D, with b and d together, be represented by x, then the solution is as follows:


The factorial notation is useful when discussing probability. The factorial notation uses and exclamation mark. Five factorial can be written as 5! and is equal to 5x4x3x2x1 = 120. Five factorial is the product of all natural numbers from 1 to 5 inclusive. Similarly, four factorial is the product of all natural numbers from 1 to 4 inclusive. Zero factorial is defined as being equal to one. One factorial is also equal to one. Many calculators have a factorial function button that displays an exclamation mark that makes
calculating factorials easier. The following formula can be used when calculating permutations.

Arrangements With Like Elements

How many 5-digit numbers can be made with the set of numbers A = {1, 3, 4, 7, 8} ? The answer is 5! = 120. How many 5 digit numbers can be made from this set A = {1, 1, 2, 2, 3} ? The answer is less than 120 because some of the elements in the set are the same. Interchanging the 1’s with each other would not produce a different result. To do this we must imagine some means of distinguishing between like objects. For example, if  the two ones were different, there would be two ways to arrange them. There would be two ways to arrange the twos also.

The number of n-arrangements of n objects, if n1are alike of one kind, n2 alike of a second, … nr alike of an r
, is:

How many ways can you arrange the letters in the word probability? The letters b and i both occur twice and there are 11 letters in total.

Subsets of a Set

A combination is an unordered grouping of a set of objects. Given the three objects a, b, c, there are 3 different groups that can be formed by choosing 2 objects from 3 objects: ab, ac and bc. Consider a set of elements A = {a, b, c, d}. One subset of A is {a, b, c}. We can call this subset a 3-element subset, or simply call it a 3-subset. We can say that we have found a 3-subset of a given set of five elements. When we select our three elements it is important to know if the order matters. If we simply wanted to choose three people out of five and wanted to know how many ways we could do this where the order of the people in the subset doesn’t matter, we would be referring to combinations, not permutations (previous discussion).

To calculate the number of r-element subsets of a set of n elements (order of the subset doesn’t matter) use the following formula.

This formula can also be written in the form.

Although it will not be shown here, the following equation can be proven.

Find the number of 6-subsets of a set of 49 elements.

Eight people are in a meeting. Each shakes hands with each of the others. How many handshakes occur?

If seven points are printed on a piece of paper and no three points are collinear (form a straight line
when connected), what’s the maximum number of line
segments that can be drawn to join pairs of points?

How many diagonals has a polygon of n sides? The way to approach this is to think of the total number of line segments connecting all points and then
subtracting n from that to subtract the outer edges, leaving you with the number of diagonals.

To better understand this formula, like all formulas, it helps to use examples. The example of a triangle is a good place to start. It has 3 sides but no diagonals.
A square has 4 sides and 2 diagonals. For the triangle, if you substitute 3 into
the formula, the result should be zero. If you substitute 4 into the formula, that answer should be 2.

A committee of 9 people have formed for a project. Two will be assigned for task A, three will be assigned to do task B and the remaining four will do task C. How many different ways can the 9 people be assigned?

Sample Space
Suppose that we will observe some process or experiment in which the outcome is not known in advance. For example, suppose we plan to roll two dice and we’re interested in the sum of the two numbers appearing on the top faces. Before we can talk about probabilities of various sums, say 4 or 7, we have to understand what outcomes are possible in this experiment. If we roll two dice, each die could show 1, 2, 3, 4, 5, 6. So the sum of the two faces could be any whole number from 2 to 12. We call this set of possible outcomes in the random experiment the sample space . Here the sample space can be written as {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

Probability Rules:

Any probability assigned must be a nonnegative number. The probability of the sample space (the collection of all possible outcomes) is equal to 1. If you have two outcomes that can’t happen at the same time, then the probability that either outcome occurs is the sum of the probabilities of the individual outcomes.

Relative Frequencies

Suppose you wanted to figure out the probability that the next person that enters your store will purchase something. To do this you will to collect data and based on that data find the percentage of the time that the event, identified as “purchase something" occurred. The total number of “observations” would be the total number of people that entered the store for a given period of time. The more data that you gather, the more reliable your results will be. The percentage that you calculate is the relative frequency of the event. These relative frequencies are only estimates that are based on finite samples of data.

In sports, the probability of an athlete scoring a point or points on an attempt is based on relative frequencies. Sports announcers will quote past statistics to relate the probability that the athlete will score. The outcome of the attempt is defined as either “score” or “not score”.

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