Business Statistics Unit 4

Probability Concepts and Applications

In this unit we will begin our discussion of inferential statistics by looking at probabilities.  When we study probabilities we are simply looking at determining the chance that something will occur in the future.

INTRODUCTION

By now the expert card players who are reading this are asking many questions:  how many card players are there?  Are the other cards face down so that we can't see them or are they face up?  Has the deck been recently reshuffled?  For our purposes we shall just assume that you are playing blackjack by yourself.  So if you just received a queen of hearts, what is the probability of receiving an ace of any suite?

This is a simple calculation.  We had 52 cards to start with and we shall assume the deck was shuffled properly so that all of the cards are at random within the deck.  One card was drawn and it was a queen of hearts.  There are now 51 cards left.  How many aces to we have in a deck?  Four.  Therefore we know that we have 4 chances within the remaining 51 cards to get an ace.  4/51 = .0784.  We have a 7.84% chance of getting an ace for our next card.

PROBABILITY TERMS

A probability is simply the chance, expressed as a value between 0 and 1, including 0 and 1, of an event occurring.  Probabilities are usually stated as a decimal amount (like .10, .50, etc.), but they can also be stated as a fraction (1/4, 3/5, etc.) and as a percentage (10%, 45%, etc.).

Probabilities are based upon an experiment.  An experiment is a process that leads to an outcome that cannot be determined with absolute certainty.  An outcome is simply the result of an experiment.  Outcomes are classified into events.  An event is a subset or collection of outcomes from an experiment. 

The possible outcomes we could have from our experiment is either a head or a tail on each flip.  The event we are looking for is the number of heads we get out of ten flips.  How many possible events are there?  Well we could get anywhere from one to ten heads, so we have ten possible events.

CLASSICAL PROBABILITY - Single Events

Sometimes we already know the possible outcomes of a probability experiment.  For example if we flip a coin we know that a head or a tail are the possible outcomes.  Since each outcome has an equal chance, and there are two possible outcomes, each outcome has a probability of 1/2 or .50.  The probability of getting a head is .50 and the probability of getting a tail is .50.

Another example involves dice.  A die has six sides and each side has dots numbering from one to six.  Each side of a die has an equal probability of occurring.  So if you roll a die, what are your chances of any one side occurring?  Since the die has six sides, and each side has an equal probability of occurring, then the probability of any one side occurring is 1/6 or .1667.

When we can determine in advance what the probability of an event occurring is, it is called a classical probability.  Therefore whenever we are trying to determine the probability of an event (E) occurring, and we are working with a classical probability, the following formula applies:

Formula Box 4.1.  Classical Probability.

The probability (P) of an event (E) occurring is:

P(E) = f / n

Where f = the number of favorable outcomes and n = total number of possible outcomes.

If we apply the formula to our dice example and we are trying to determine the probability of rolling a two with one die, then by using the formula we get:

P(E) = 1/6 = .1667.

How did we get this answer?  We know that there are six possible outcomes with any die, and we are looking for just one possible outcome (a two).  Therefore to determine the probability, we take the number of favorable outcomes (1) and divide it by the total number of possible outcomes (6).

What if we were looking for more than one possible outcome with our die?  Lets assume we were interested in the probability of getting an odd number on a single roll?  How many odd numbers are there on a die?  A one, three, and a five.  We have three possible favorable outcomes. The total number of possible outcomes is six.  Therefore the P(E) = 3 / 6 = .50.

Complementary Events

A complementary event is the probability of a particular event not occurring.  What does this mean?  Let's look at an example using a deck of cards.  Suppose we shuffle a deck of cards and are looking for the probability of drawing an ace on the first draw.  Since there are four aces in the deck, the probability of getting an ace is 4/52 or .0769.  The complementary event is the probability of not getting an ace on the first draw.  To find the probability for any complementary event, simply take the probability for an event and subtract it from1.  In our example here, the probability for the event is .0769.  Therefore the probability for the complementary event is equal to 1 - .0769 = .9231.

Another example of using complementary events.  Let's say a parking lot contains 30 cars.  We look at each car and record its color.  There are 7 red cars in the lot.  The probability of having a red car in the lot is 7/30 or .2333.  What is the complementary event?  Its is the probability of not having a red car.  Therefore we take 1 - .2333 = .7667.  The probability for the complementary event is .7667.

Relative Frequency Probability - Single Events

The second type of probability we are going to study is called relative frequency or empirical probability.  Relative frequency (empirical) probabilities are based upon past research, experience, or data that has determined the probability of an event.

For example, suppose we own an auto dealership and are interested in the probability that a customer who purchases a new car will select dealer financing.  Over the past two years, 210 out of 350 new car purchasers selected dealer financing.  210 / 350 = .60.  Therefore 60% of the customers over the last two years selected dealer financing.

To determine the probability of a new customer using dealer financing, we simply use formula 5.1 and apply it to our example.  Therefore the P(E) = .60.  This probability is based upon the past data.  That is why it is called a relative frequency probability.  It is based upon the frequency that an event occurred in the past.

We see relative frequency probabilities used in many ways.  The insurance industry uses them to determine the probability a customer will be in an auto accident.  In the education arena, high school graduation rates are simply percentages of high school students that graduate.  This percentage is used to predict future graduation rates and for comparisons among schools.

Subjective/Intuition - Single Events

Subjective/Intuition is the third type of method used to develop a probability.  Intuition is a subjective method, meaning that it is based upon a guess or an estimate based upon a person's past experience.  Unlike a relative frequency probability, there is usually no past research data or experience that can determine the probability of an event.

Often we say that a person using intuition to determine a probability is making an "educated guess".  Simply stated, the person is using his or her knowledge and/or experience to estimate a probability.

Examples of using intuition include guessing who will win a sporting event or predicting who will get an A in this course.  Another great example of the application of intuition is driving above the speed limit in a car.  Many of us on occasion drive above the speed limit.  We know that we might get caught and receive a ticket for speeding, but we will do it anyway because we believe that the probability of getting caught and receiving a ticket is very low.

PROBABILITY - IMPORTANT CONCEPTS

When we are working with probabilities there are some important rules that must be understood.

• Probability is measured on a scale of 0 to 1.  The probability of a particular event occurring P(E) will have a value from 0 to 1.  If the event has a probability value of 0, then the event will not occur.  If the event has a probability value of 1, then the event will always occur.

• The probabilities of all simple events related to a specific sample will add up to 1.  Simply put, an activity or sample will likely have a number of different possible outcomes.  All of the probabilities associated with these possible outcomes added together will equal 1.

• A probability associated with an event indicates that the event is likely to occur a certain number of times, but it may not actually occur the number of times indicated by the probability.  Simply put, lets suppose we flip a coin and know that the probability of getting a head is .50.  If we flip the coin ten times we expect to get an equal number of heads and tails (5 of each).  But if we actually flip the coin ten times, we might get something different (say 6 heads and four tails).  The probability continues to be .50, but our experiment indicated a different percentage outcome.  If we continue to flip the coin 1000 times we are likely to find our percentage becoming closer to .50.  Therefore probabilities are estimates - not definite outcomes.

PROBABILITIES FOR COMPOUND EVENTS

We have already examined the probabilities associated with a single event.  Now we shall turn our discussion to a compound event which relates to two or more events.

When we look at compound events we need to ask ourselves a question.  Specifically looking at a compound event involving two simple events:  Does the outcome of the second event change as a result of the outcome of the first event?  Another question to consider is:  Are we looking for the probability of both events occurring or the probability of either event occurring?  Your answers to these questions will direct you to the appropriate formula to solve the probability for a compound event.

COMPOUND EVENTS - COMPUTING P(A or B) - The Addition Rules

In order to calculate the probability of either event A or event B occurring, we must first determine whether our two events are mutually exclusive.

Events A and B are mutually exclusive if the occurrence of either A or B during a single event prevents the occurrence of the other.  For example, if a person is female that prevents them from being male at the same time.  The two are mutually exclusive.  Looking at our die roll, if we wanted to calculate the probability of getting a one or a two on a single die roll, we would first need to determine if the two events are mutually exclusive.  On a single die roll it is impossible to get BOTH a one and a two.  Therefore the events are mutually exclusive.  Getting a one eliminates the chance of getting a two on a single die roll.

To calculate the probability of either event A or event B occurring, we use the following formula when the two events are mutually exclusive.

Formula Box 4.2.  Addition Rule for Two Mutually Exclusive Events P(A or B)

P(A or B) = P(A) + P(B)

Notice that we are now adding the probabilities of each event together instead of multiplying.  Using our example of the die roll, if we are trying to find the probability of getting a one or a two on a single die roll, we must first determine the probability of event A and of event B.  Since the die has six sides, the probability of any one side occurring is 1 / 6 or .1667.  Therefore the P(A or B) = .1667 + .1667 = .3334.  The probability value of .3334 indicates that we have a .3334 probability that we will get a one or a two on a single roll of a die.

How about another example?  Lets take a look at a deck of cards.  What is the probability that if you draw a single card, it will be an Ace or a Queen?

In order to answer the question, we need to know some basic facts about a deck of cards.  A standard deck of cards contains 52 cards, divided into four suites:  clubs, diamonds, hearts, and spades.  There are 13 cards in each suite.  Each suite contains the following cards:  ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, 2.  Therefore in any deck of cards, there are four aces, kings, queens, jacks, 10s, etc.

In this example we are trying to find the probability of getting an ace or a queen.  This is represented as P(A or B).  Using the formula in formula box 4.2, we know that the P(A or B) is equal to P(A) + P(B).  So to solve this problem we simply need to determine the probability of getting an ace and the probability of getting a queen, and add them together.  To find the probability of getting an ace, we simply use formula 4.1 which is:

P(E) = f / n

In this case our event (E) is the probability of getting an ace.  The value f represents the number of favorable or possible outcomes.  Since there are four aces in a deck of cards, f = 4.  The value n is the total number of possible outcomes.  Since there are 52 cards in a deck, n = 52.  Therefore the P(E) = 4 / 52 or .0769.

We are still not finished with our problem since we must also find the probability of getting a queen.  Since there are four queens in a deck, the process is identical to solving for the aces.  Here the P(E) also equals 4 / 52 or .0769.

Now we can solve our problem.  What is the probability of getting an ace or a queen on a single draw?  P(A or B) = 4 / 52 + 4 / 52 = 8 / 52 = .1538.

COMPOUND EVENTS - COMPUTING P(A or B) NOT MUTUALLY EXCLUSIVE - The Addition Rules

What happens if event A and event B are NOT mutually exclusive?  First of all if they are not mutually exclusive it means that the two events can occur at the same time.  How can this happen?  Lets go back to our deck of cards.  What if you were asked what is the probability of drawing a jack or a diamond from a deck of cards?  To determine if two events are mutually exclusive we must ask the question:  can event A and event B both occur at the same time?

As we examine our deck of cards we find that it is possible for a card to be both a jack and a diamond:  the jack of diamonds.  Therefore the two events are not mutually exclusive.  When we have two events that are not mutually exclusive and we are looking for the probability that either event A or event B will occur, the following formula is used.

Formula Box 4.3.  Addition Rule for Two Events That are NOT Mutually Exclusive  P(A or B).

P(A or B) = P(A) + P(B) - P(A and B)

Did you notice what is different about this formula compared to the formula found in 4.3?  We take the probability of event A and add to it the probability of event B, and then we subtract the probability of both A and B occurring.  The difference is we must consider the overlap or the part of each event that can occur at the same time.

Lets apply the formula to our last example.  The probability of A which is the probability of drawing a jack is equal to 4 / 52.  The probability of drawing a diamond is 13 / 52.  Now we need to find the probability of drawing a jack and a diamond.  Since there is only one jack of diamonds in our deck, the probability equals 1 / 52.  Therefore we solve the problem using the following equation:

P(jack or diamond) = P(jack) + P(diamond) - P( jack and diamond)

P(jack or diamond) = 4 / 52 + 13 / 52 - 1 / 52 = 16 / 52 = .3077

The reason why we need to subtract the probability of being a jack and a diamond is to eliminate any double counting.  In our example, the jack of diamonds is counted as part of the jack event and it is also counted as part of the diamond event.  Therefore it is necessary to eliminate counting it twice.

Lets do another problem.  Take a look at the data in table 4.1.  The data was obtained from a survey of registered voters in a school district in regards to a proposed tax millage increase to finance a new high school.

Table 4.1.  Survey of Registered Voters - Millage Increase.

What is the probability that a voter in this survey is married with children or voting for the school millage increase?

The first step you must take is to determine if the two events, married with children and voting for the school millage, are mutually exclusive.  Is it possible to be married with children and voting for the school millage?  The answer is yes.  Therefore the two events are NOT mutually exclusive and you use formula 4.3 to answer the question.

Using formula 4.3, we first need to find the probability of being married with children.  Looking at the data, there is a total of 2125 people who are married with children out of a total of 5882 people who were surveyed.  Therefore our P(A), which is being married with children, is equal to 2125 / 5882 = .3613.

To find the P(B), which is the probability of being in favor of the school millage, we take a look at the total number of people in favor of the millage which is 2666.  We then find the total number of people who were surveyed which is 5882.  Therefore our P(B), which is being in favor of the millage increase, is equal to 2666 / 5882 = .4533.

Now we must solve for the third part of the formula.  To find P(A and B) we must look at the table and find the number of people who are married with children AND in favor of the school millage increase.  That number is 1125 out of a total of 5882.  Therefore our P(A and B) = 1125 / 5882 = .1913.

Finally we simply put all of our pieces together.   The probability that a voter in this survey is married with children or for the school millage increase is:

P(married with children or for the school millage increase) = P(married with children) + P(for the school millage) - P(being married with children and for the school millage)

.3613 + .4533 - .1913 = .6233

Once again the third part of our formula eliminates the double counting that occurs since the two events are not mutually exclusive.

COMPOUND INDEPENDENT EVENTS - COMPUTING P(A and B) - The Multiplication Rule

Now lets examine the situation where we want to determine the probability that both event A and event B will occur.  Looking at our previous example, in this situation we are looking at the probability of getting a head on a single flip of a coin, AND the probability of getting a one on a single die roll.  These two events continue to be independent:  the flip of a coin does not affect the outcome of a die roll.

There is a specific formula that is used to determine the probability of two independent events occurring.  This formula is called the special multiplication rule for independent events.

Formula Box 4.4.  Special Multiplication Rule for Independent Events P(A and B)

P(A and B) = P(A) X P(B)

So in order for us to determine the probability of event A and event B occurring, we need to multiply the probability of event A by the probability of event B.  Using our coin toss and die example, we are trying to find the probability of getting a head and the probability of rolling a one.  The P(A and B) = .50 X .1667 = .0834.

How did we get this number?  Remember that the probability of getting a head on a single flip of the coin was .50.  Also the probability of getting a one on a single roll of a die was .1667.  Therefore the probability of getting a head AND a one equals .50 X .1667 which equals .0834.  Did you notice that the probability is quite small?  Remember we are looking for the probability of two events occurring:  getting a head on a coin toss AND getting a one on a roll of a die.  Although there are only two possible outcomes from a coin toss, there are six possible outcomes associated with the die roll.  Therefore we are looking at 12 possible different combinations that could occur ( number of possible outcomes from event A multiplied by the number of possible outcomes from event B).

What happens if we look at three or more events?  Fortunately for you this is beyond the scope of this course.  However if the events are independent, you simply multiply the probability of each event together.

COMPOUND EVENTS - NOT INDEPENDENT - COMPUTING P(A and B) - The Multiplication Rule

If we need to compute the probability that two events will occur, and the two events are NOT independent, then the general rule of multiplication for non independent events applies.  The two events are said to have a joint probability.  The formula is given below.

Formula Box 4.5.  General Multiplication Rule for Non Independent Events P(A and B).

P(A and B) = P(A) X P(B | A)

This formula is interpreted as the probability of joint event A and B equals the probability of event A multiplied by the probability of event B given that event A has occurred.

Lets look at an example of how this formula could be used.  What if we have a deck of cards and we want to know the probability of drawing an ace followed by drawing another ace.

In order to answer the question, we need to know some basic facts about a deck of cards.  A standard deck of cards contains 52 cards, divided into four suites:  clubs, diamonds, hearts, and spades.  There are 13 cards in each suite.  Each suite contains the following cards:  ace, king, queen, jack, 10, 9, 8, 7, 6, 5, 4, 3, 2.  Therefore in any deck of cards, there are four aces, kings, queens, jacks, 10s, etc.

Since there are four aces in a deck of cards and a total of 52 cards in a deck, the probability of drawing an ace on the first draw is 4 / 52 = .0769.  This is the P(A).  As we move to the second draw we assume that the first card is not returned to the deck.  Now we have 51 cards in the deck.  Therefore our probability of getting an ace on the second draw has been affected by the drawing of the first card and the fact that it was an ace.  The total number of cards has declined to 51, and the total number of aces has declined to 3.  Therefore our probability of getting an ace on the second draw is 3 / 51 = .0588.  This is the P(B | A).  Notice that this probability is conditional, indicating that the probability of B is influenced by what happens to A.

To answer the question we use formula 4.5.  P(drawing two aces) = P(ace) X P(ace given that an ace was already drawn)

P(A and B) = .0769 X .0588 = .0045       Quite a small probability!

Conditional Probability - Two Events

For situations where we have two events that are not independent of each other, and we are trying to find the probability of one event given that the other event has occurred, a special formula exists to help us.  Unlike the situation discussed above that involved a single deck of cards, in this situation we would have two distinct events where is is possible to be included in both items.  Let's look again at Table 4.1.

Table 4.1.  Survey of Registered Voters - Millage Increase.

Now let's ask the following question:  What is the probability that a person is against the school millage given that they are single?  The key term that gives you a clue that you have a conditional probability is the word "given".  In these instances you would use the following formula:

Formula 4.6.  Formula for Conditional Probability.

P(B|A) = P(A and B) / P(A)

Now we will apply this formula to answer the question.  The probability of A and B in this problem is the number of people against the millage who are single, divided by the total number of people.  There we have 965/5882.  For the probability of A, we use the probability of being single 1426/5882.  Now we divide 965/5882 by 1426/5882.  Converting to decimals we take .1640 and divide it by .2424 and we get .6765 our answer.  The probability that a person is against the school millage given that they are single is .6765.

How about another example looking at Table 4.1.  What is the probability of being Married w/ Children given that they are undecided?  The probability of A and B here is found by taking the number of people who are undecided and married with children (139) and dividing it by the total number of people (5882).  For the probability of A, we use the probability of being undecided, which is 237 / 5882.  So we need to divide 139 / 5882 by 237 / 5882.  Converting to decimals we take .0236 and divide it by .0403 and we get .5856 our answer.  The probability that a person is Married w/ Children given that they are undecided is .5856.

Is there another way to find the probabilities when working with a table like this?  Yes!  The clue is the word "given".  The word given means that you are changing your total possible outcomes.  For the first question, given indicated that you are only concerned with single people.  Your total sample space or outcomes becomes 1426.  Now you find the number of people against the millage in the single category, which is 965.  Take 965 and divide it by 1426 and you get .6767.  Essentially the same answer (it may be a little different due to rounding)!  This is an application of the basic formula P(E) = f / n.  For the second question, our total possible outcomes change to 237 and the number of Married w/ Children in this category are 139.  Take 139 and divide it by 237 and you get .5865.  Once again essentially the same answer that is only a little different due to rounding.

Joke Time!

There was a statistics instructor who, when driving his car, would always accelerate hard before coming to an intersection, whip straight through it, then slow down again one he'd got past it.  One day, he had a passenger with him who was understandably unnerved by his driving style, and asked him why he went fast through intersections.  The statistics instructor replied:  "Well, statistically speaking, you are far more likely to have an accident at an intersection, so I just make sure that I spend less time there."

Source:  G.C. Ramseyer

Assignment

The homework problems for unit 4 can be found by going to the Course Documents link in Blackboard, and clicking on the link for Unit 4.  Look for the Unit 4 - Assignment link.  Once you have completed your homework assignment, you will need to post your answers on Blackboard®.  Once you have signed into Blackboard, simply go to the Unit 4 - Assignment 4 - Post Answers Here link and post your answers.  Immediate feedback is provided once you have completed the posting of all of your answers and clicked on submit.  Make sure you print the entire submitted homework assignment to assist you with quizzes and tests.

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