Introduction
If the decisions that business managers are required to make were made in an environment of complete and perfect information, then there would be
absolutely no need to learn probability. Such an environment is called deterministic. For better or for worse,
managers in most businesses must function in an environment of incomplete and imperfect information. Managers in the airline or transportation
industries cannot predict the weather perfectly. Managers of taxi companies or trucking and delivery companies cannot predict the number or length of
delays resulting from traffic jams caused by accidents. Retail stores cannot predict the exact number of customers they may serve on a given day.
Universities cannot predict which students, from among those who were granted acceptance, will actually decide to enrol. Human resources managers
cannot predict the exact number or quality of job applicants who will respond to a help wanted ad. Insurance companies cannot predict exactly when or if an
insured event will take place. This environment is called probabilistic and it is the normal functioning environment
for business managers. Consequently, a thorough understandng of probability is essential for a business manager to achieve success.
Probabilities run from straightforward to extremely complex. Some probabilities can be computed by hand. An example is the probability of rolling a 4
with one playing die (a six-sided numbered cube). Other probabilities need a calculator to compute. An example is the probability of getting red at
least 3 times out of 5 flips of a thin two-sided disk that is colored red on one side and blue on the other. The most useful probabilities require the
calculating power of a computer to find. An example is finding the probability of anything whose events follow
a normal distribution.
Your textbook should have a series of tables in its appendix that compute the probabilities of several distributions for you to use in solving your
homework problems. Your professor may require you to use them. A few expensive calculators are able to compute probabilities for the standard normal
distribution. Your professor may require you to purchase a such a calculator for your course. Most spreadsheet programs have an embedded probability
calculator that computes probabilities for a variety of distributions.
We use neither the tables, nor hand-held calculators, nor spreadsheet calculators for this module, though you may if you prefer. We provide you with
a probability calculator written in JavaScript. On each section page and each question page, (though not this page), there is a link to our financial
calculator. It will open a new window or tab in which you can select any of the probability distributions you need for this module.
Empirical Probability
As noted above, some probabilities are generated form complex distribution functions such as the normal probability distribution, whereas other probabilities are generated from what is observed. An example is the number of socks that are observed in a drawer. Say 10 socks are observed in a drawer of which 2 are red, 3 are green, and 5 are blue. The probability of choosing a green sock is the number of green socks observed in the drawer divided by the total number of socks in the drawer. Another example is the number of cars that are exceeding the speed limit. Say over a one week period authorities observe 1,000 cars driving on a stretch of road where the speed limit is 60 kph of which 150 are observed to be driving at a speed above 65 kph. The probability of a car speeding along that stretch of road is the number of speeding cars divided by the total number of cars on that stretch of road. These probabilities are known as empirical probabilities because they are based on empirical observation as opposed to on a mathematical function.
Definitions
The field of probability has terms and definitions that are not intuitive and so require some explanation. We will start with the term
experiment. In probability, an experiment is any action that leads to an outcome in which we have the interest in
computing a probability. The action of rolling a die is an experiment. We are interested in finding the probability of a particular outcome, the value
the die shows on its upper face. The action of drawing a card from an ordinary deck of playing cards is an experiment. We are interested in finding the
probability of a particular outcome, the value shown on the card drawn.
Several experiments may lead to several different outcomes. The set or group of all possible outcomes of an experiment is called the
universe. It is also known as the sample space. We will use the term
universe because we don't want confusion with the word sample which is
used later in a different context. We will use the symbol U to represent the universe. Sets are made up of
elements, the individual items that populate a set. Having defined the universe as a set, the outcomes in it
are elements. The elements of the universe are also called points or
sample points. A universe made up of a countable and indivisible number of elements is known as a
finite universe. An example is the outcomes of rolling one die. That experiment has 6 possible outcomes so its
universe is finite. A universe made up of a countably infinite number of elements or infinitely divisible elements is known as an
infinite universe. An example of a countably infinite universe is the number of atoms in the Milky Way galaxy.
An example of an infinitely divisable universe is the individual weights of 5,000 golf balls. Because the weight of something can be measured to a
degree of precision that extends to an infinite number of places after the decimal point, it is said to be infinitely divisible and can have an infinite number of
values.
An event is defined as a set of elements in the universe; consequently, an event is a subset of the universe.
More intuitively, an event can be made up one or more possible outcomes of the experiment. In the case of the six-sided numbered die, all possible
outcomes, the universe, is the numbers 1, 2, 3, 4, 5, and 6. This is written as U = {1, 2, 3, 4, 5, 6}. An event can be any subset of these 6 numbers.
For example, if an event is defined as rolling an even number, it would consist of the numbers 2, 4, and 6. We could write this event, labelled A, as
A = {2, 4, 6}. Should an experiment result in a 2, 4 or 6, then event A will have occurred. Events can contain any number of elements up, from none up
to the number of elements in the universe.
Simple Probability
With the definitions of a universe, an element, and an event, we can discuss the definition of simple probability and compute one in an example. We start
with the probability of an element. Identifying the first element of a universe as e1, the probability of an experiment resulting in an outcome
represented by the first element, written as P({e1}), is whatever probability value we assign to the first element as long as it follows
two fundamental laws of probability. The first fundamental law is the no probability can be negative or greater than 1. The probability we assign to
element 1 must be greater than or equal to zero and less than or equal to 1, written as 0 ≤ P({e1}) ≤ 1.
For the case of the six-sided die, the first element is the number 1, e1 = {1}. Let us assign the probability value of 1/6 to the first
element, P({1}) = 1/6.
The second fundamental law is that the assigned probabilities for each individual element, also called point probabilities, in the universe must sum
to 1. Let us assign the probability value of 1/6 to each of the remaining elements for our die example. Then P({e1}) + P({e2}) + P({e3})
+ P({e4}) + P({e5}) + P({e6}) = P({1}) + ({2}) + P({3}) + P({4}) + P({5}) + P({6}) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6
= 1 satisfies the second law.
Earlier we defined event A as the showing of an even number on the roll of one six-sided die. Event A consisted of 3 elements; the numbers 2, 4, and 6.
How do we find the probability of event A occurring? The probability of an event is computed by summing the point probabilities of the elements in
that event. So P(A) = P({2}) + P({4}) + P({6}) = 1/6 + 1/6 + 1/6 = 0.50. There is a 50% chance of showing an even number upon a roll of one six-sided
die.
An event that has no outcome in it is a null event. The event is said to have an
empty set of elements. As all experiments must have outcomes, it is impossible for an experiment to have a
non-outcome or be null. Null events are impossible and their corresponding probabilities are 0.0. An event that contains all the elements of the universe
is a certain event. It will be the result of an experiment with certainty. The probability of a certain event
is 1.0.
Example 1.
A universe is comprised of integers ranging from the lowest value of -5 through the highest value of +5. The point probabilities of the 11 elements are
P({-5}) = .05, P({-4}) = .05, P({-3}) = .05, P({-2}) = .05, P({-1}) = .10, P({0}) = .10, P({1}) = .10, P({2}) = .10, P({3}) = .10, P({4}) = .15,
P({5}) = .15. You may confirm for yourselves that each point probability is between 0 and 1, and the 11 point probabilities sum to 1.0. The
experiment is the drawing of one plastic chip from a bowl. The chips are coded with integer values from -5 through +5. (The bowl contains 100 chips
with 5 chips having the value -5, 5 chips the value -4, and so on. But that is not important now.) Event W is defined as the drawing a chip with a
negative value. Event X is defined as the drawing of a chip with a positive value. Event Y is defined as the drawing of a chip with a value between
10 and 20. Event Z is defined as the drawing of a chip with an odd value.
What is the probability of event W, P(W)?
The elements of W are the negative chip values. P(W) is the sum of the point probabilities in W. P(W) = P{-5} + P{-4} + P{-3} + P{-2} + P{-1} =
.05 + .05 + .05 + .05 + .10 = .30. There is a 30% chance of drawing a chip with a negative value.
What is the probability of event X, P(X)?
The elements of W are the positive chip values. P(X) is the sum of the point probabilities in X. P(X) = P{1} + P{2} + P{3} + P{4} + P{5} =
.10 + .10 + .10 + .15 + .15 = .60. There is a 60% chance of drawing a chip with a positive value.
What is the probability of event Y, P(Y)?
The elements of Y are the chip values between 10 and 20. P(Y) is the sum of the point probabilities in Y. There are no point probabilities in Y
because its elements are outside the universe. It is a null event. There is a 0% chance of drawing a chip with a value between 10 and 20.
What is the probability of event Z, P(Z)?
The elements of Z are the odd chip values. P(Z) is the sum of the point probabilities in Z. P(Z) = P{-5} + P{-3} + P{-1} + P{1} + P{3} + P{5} =
.05 + .05 + .10 + .10 + .10 + .15 = .55. There is a 55% chance of drawing a chip with an odd value.
Sample Questions about Empirical Probability.
Counting Techniques
As you saw above, to compute the simple probability of an event, one needs to know the total number of elements in the universe and the point
probabilities of the elements that are contained in the event as it is defined. When each element has the same probability of occurring, then it
suffices to know the number of elements that are contained in the event. In the above examples, we tallied the relevant elements to compute
probability. That was possible because the number of elements was small. In situations where there are millions of elements, the tallying technique
is not feasible. Some other methed is needed to count the number of elements in a universe and in an event. Three of those methods are discussed in
this section.
A key aspect of determining which counting method to use depends on a concept called replacement. When drawing an
element from a universe in an experiment, if that element is available to be drawn a second or third time in that same experiment,
then the experiment is taking place with replacement. Replacement means that all elements in the universe are
available for each and every draw in the same experiment. Examples are the alphabet when drawing letters to spell
ones name, or the 10 digits when drawing digits to compose a telephone number. After writing the first letter of our name, we 'put back' that letter
into the 26-letter pool that is the alphabet making it available to be used again in our name. The name 'Bobby' can use the letter 'b' 3 times because
it was replaced after each time it was drawn. The fictious phone number 555-7000 uses the digits '5' and '0' 3 times each because they were replaced
after each time they were drawn. If an element in the universe is only available to be drawn ONCE in the same experiment,
so that it is NOT available for any subsequent draw in that same experiment, then the experiment is taking place
without replacement. Without replacement means that an element in the universe is available for only ONE draw in an
experiment. An example is selecting players for a team in a player draft. Once a player has been selected by a team in a draft, that player is NOT available to be
selected again by that or any other team in that same draft.
First Counting Principle
The first counting principle is for experiments with replacement. If an experiment consists of one draw with j
possible outcomes and a second draw with k possible outcomes, then the universe contains jk
elements. An example is the experiment of rolling one six-sided die twice. There are 6 possible outcome on the
first roll and another 6 possible outcomes on the second roll. The total outcomes in the universe is 6 * 6 = 36. Another example is an experiment
consisting of drawing one of 8 pairs of shoes and drawing one of 12 suits. The outcome is the combination of a pair shoes and a suit. There are 8 * 12
ways of choosing a pair of shoes and a suit yielding 96 total outcomes in the universe.
The application of this counting technique to probability is best demonstrated using popular government lotteries. One lottery involves choosing a 3
digit number (Pick 3) and another involves choosing a 4 digit number (Pick 4). The lottery prize is won by the outcome that matches a randomly chosen
winning number. How many elements are in the universe for the Pick 3 lottery? The Pick 4 lottery? The Pick 3 lottery is an experiment in composed of
3 draws. Each draw has 10 possible outcomes, the 10 digits from 0 through 9. The universe for the Pick 3 lottery would then be all the possible ways
to choose from 10 digits 3 times and that is 103 or 1,000 outcomes. Similarly, there are 104 or 10,000 outcomes for the Pick 4 lottery.
How does this relate to probability? Since the wining number is randomly chosen (We will define that term precisely in the module on statistics.),
each element in the universe has the same probability of being the winning number. The point probability of each element is the same at 1/1,000 for
the Pick 3 lottery and 1/10,000 for the Pick 4 lottery. As there is only 1 winning number for this experiment, the probability of winning the lottery
is 1 over the total number of outcomes in the universe or the point probability of each element. That means the probability of winning the Pick 3 is
0.1% and the Pick 4 is 0.01%.
Example 2.
A bank lets its customers choose a personal identification number (PIN) comprised of 4 characters that can be numerical digits or letters. What is the
probability of someone correctly guessing a bank customer's PIN on the first attempt?
The experiment is to make 4 draws with each draw having 10 + 26 = 36 possible outcomes. There are 364, or 1,679,616 elements in the universe.
There is only one correct outcome, so the probability of guessing the correct PIN on the first attempt is 1,679,616-1 or 6.0 x 10-5%.
Second Counting Principle
The second counting principle is for experiments without replacement. Say an experiment consists of three draws and the first draw has
j possible outcomes. The second draw in the same experiment will have only j-1
possible outcomes because one outcome was used for the first draw and is not available for subsequent draws due to the absence of replacement. The
third draw in the same experiment will have j-2 possible outcomes because a second outcome was used up in the
second draw making 2 outcomes unavailable for subsequent draws. Following the first counting principle, the universe contains j * (j-1) * (j-2)
elements. An example is the experiment of selecting the first place, second place, and third place finishers in a horse race. Say there are 8 horses
starting in the race. That provides for 8 possible outcomes for the first place draw. The second place draw will have 7 possible outcomes, and 6 for
the third place draw. The total outcomes in the universe is 8 * 7 * 6 = 336.
This second counting principle involves multiplying a number by itself minus 1, then by itself minus 2, and then again by itself minus 3, etc. The name
of this mathematical process is factorial, and its symbol is !. An
example of the factorial operator is 5!. It is 5 * 4 * 3 * 2 * 1 = 120. Almost all scientific calculators have a factorial key. An unintuitive law
regarding factorial is that 0! = 1. You should note that in the horse race example, it matters which horse comes in first, second, or third. Switching
the first and second place horses yields a different outcome. This is the 'order matters' situation. A short-cut
method of computing the number of outcomes in the universe in this situation is called permutation, meaning
ordered arrangement. The symbol and formula for permutation is nPr = n! / (n - r)!,
where r is the number of items to be arranged and n is the number of items to select from for arrangement. In the horse race example, we select 3 items
to arrange (first, second, and third places) from the 8 horses participating in the race. 8P3 = 8! / (8 - 3)! = 8 * 7 * 6 * 5 * 4
* 3 * 2 * 1 / 5 * 4 * 3 * 2 * 1 = 8 * 7 * 6 = 336.
How does permutation relate to probability? If the horse race is fair and all the horses have equal ability, then each of the 336 different ways of
arranging the 8 horses in first, second, and third places, elements in the universe, will have the same probability of occurring. The point probability
of each element in the 8-horse race experiment is 1/336 or 0.30%. As there is only 1 winning arrangement for this experiment, the probability of
winning a bet that exactly matches the first, second, and third place horses, in order, is 0.30%.
Example 3.
A giant slalom race has 12 skiers participating. What is the probability of identifying and ranking, in one attempt, the skiers with the 4 fastest times?
The race is fair and all the skiers have equal ability.
The experiment is to choose 4 skiers out of 12 in the correct order of fastest times. The universe has 12P4 = 12! / (12 - 4)! = 11,880
possible outcomes. There is only one correct outcome, so the probability of guessing the correct 4 skiers on the first attempt is 11,880-1 or 8.4 x 10-3%.
Third Counting Principle
The third counting principle is also for experiments without replacement. However, the outcomes do not depend on the order of the draws. In the horse
race scenario, the outcome would be the selection of the horses placing in the top 3 without regard for which horse came in first, second, or third. This
'order does not matter' situation is called combination, meaning
unordered arrangement. The symbol and formula for combination is nCr = n! / [r! * (n - r)!],
where r is the number of items to be arranged and n is the number of items to select from for arrangement. In the horse race example above, we select
3 items to arrange in any order from the 8 horses participating in the race. 8C3 = 8! / [3! * (8 - 3)!] = 8 * 7 * 6 * 5 * 4
* 3 * 2 * 1 / 3 * 2 * 1 * 5 * 4 * 3 * 2 * 1 = 8 * 7 = 56.
How does combination relate to probability? If the horse race is fair and all the horses have equal ability, then each of the 56 different ways of
arranging the 8 horses in first three places, elements in the universe, will have the same probability of occurring. The point probability
of each element in the 8-horse race experiment is 1/56 or 1.79%. As there is only 1 winning outcome for this experiment, the probability of
winning a bet that matches the first three placing horses is 1.79%.
Example 4.
A government lottery consists of selecting 5 numbers, without replacement, out of 30 numbered 1 through 30. Winning the lottery consists of
matching all 5 numbers, in any order. What is the probability of winning this lottery?
The experiment is to choose and match 5 numbers out of 30 in any order. The universe has 30C5 = 30! / [5! * (30 - 5)!] = 142,506
possible outcomes. There is only one correct outcome, so the probability of matching the correct 5 numbers on one attempt is 142,506-1
or 7.0 x 10-4%.
Sample Questions about Counting Techniques.
Return to module description page.