applying probability rules assignment

Module 6: Probability and Probability Distributions

Introduction to Probability Rules

What you’ll learn to do: reason from probability distributions, using probability rules, to answer probability questions..

In this section, we introduce probability rules and properties. These rules can make evaluating probabilities far simpler and can also help catch mistakes if results are nonsensical (for example, a 140% chance is impossible). We revisit conditional probabilities, which are a fundamental concept in understanding how to interpret results from hypothesis testing. Finally, we introduce the notion of independence, joint, and marginal probabilities, and present a useful rule that ties these concepts together.

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5.3: Probability Rules- “And” and “Or”

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• Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
• Coconino Community College

Learning Objectives

Students will be able to:

• Determine if two events are mutually exclusive and/or independent.
• Apply the "Or" rule to calculate the probability that either of two events occurs.
• Apply the "And" rule to calculate the probability that both of two events occurs.

Many probabilities in real life involve more than one event. If we draw a single card from a deck we might want to know the probability that it is either red or a jack. If we look at a group of students, we might want to know the probability that a single student has brown hair and blue eyes. When we combine two events we make a single event called a compound event . To create a compound event, we can use the word “and” or the word “or” to combine events. It is very important in probability to pay attention to the words “and” and “or” if they appear in a problem. The word “and” restricts the field of possible outcomes to only those outcomes that simultaneously describe all events. The word “or” broadens the field of possible outcomes to those that describe one or more events.

Example \(\PageIndex{1}\): Counting Students

Suppose a teacher wants to know the probability that a single student in her class of 30 students is taking either Art or English. She asks the class to raise their hands if they are taking Art and counts 13 hands. Then she asks the class to raise their hands if they are taking English and counts 21 hands. The teacher then calculates

\[P(\text{Art or English}) = \dfrac{13+21}{30} = \dfrac{33}{30} \nonumber\]

The teacher knows that this is wrong because probabilities must be between zero and one, inclusive. After thinking about it she remembers that nine students are taking both Art and English. These students raised their hands each time she counted, so the teacher counted them twice. When we calculate probabilities we have to be careful to count each outcome only once.

Mutually Exclusive Events

An experiment consists of drawing one card from a well shuffled deck of 52 cards. Consider the events E : the card is red, F : the card is a five, and G : the card is a spade. It is possible for a card to be both red and a five at the same time but it is not possible for a card to be both red and a spade at the same time. It would be easy to accidentally count a red five twice by mistake. It is not possible to count a red spade twice.

Definition: Mutually Exclusive

Two events are mutually exclusive if they have no outcomes in common.

Example \(\PageIndex{2}\): Mutually Exclusive with Dice

Two fair dice are tossed and different events are recorded. Let the events E , F and G be as follows:

• E = {the sum is five} = {(1, 4), (2, 3), (3, 2), (4, 1)}
• F = {both numbers are even} = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
• G = {both numbers are less than five} = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4,1), (4, 2), (4, 3), (4,4)}
• Are events E and F mutually exclusive?

Yes. E and F are mutually exclusive because they have no outcomes in common. It is not possible to add two even numbers to get a sum of five.

• Are events E and G mutually exclusive?

No. E and G are not mutually exclusive because they have some outcomes in common. The pairs (1, 4), (2, 3), (3, 2) and (4, 1) all have sums of 5 and both numbers are less than five.

• Are events F and G mutually exclusive?

No. F and G are not mutually exclusive because they have some outcomes in common. The pairs (2, 2), (2, 4), (4, 2) and (4, 4) all have two even numbers that are less than five.

Addition Rule for “Or” Probabilities

The addition rule for probabilities is used when the events are connected by the word “or”. Remember our teacher in Example \(\PageIndex{1}\) at the beginning of the section? She wanted to know the probability that her students were taking either art or English. Her problem was that she counted some students twice. She needed to add the number of students taking art to the number of students taking English and then subtract the number of students she counted twice. After dividing the result by the total number of students she will find the desired probability. The calculation is as follows:

\[ \begin{align*} P(\text{art or English}) &= \dfrac{\# \text{ taking art + } \# \text{ taking English - } \# \text{ taking both}}{\text{total number of students}} \\[4pt] &= \dfrac{13+21-9}{30} \\[4pt] &= \dfrac{25}{30} \approx {0.833} \end{align*}\]

The probability that a student is taking art or English is 0.833 or 83.3%.

When we calculate the probability for compound events connected by the word “or” we need to be careful not to count the same thing twice. If we want the probability of drawing a red card or a five we cannot count the red fives twice. If we want the probability a person is blonde-haired or blue-eyed we cannot count the blue-eyed blondes twice. The addition rule for probabilities adds the number of blonde-haired people to the number of blue-eyed people then subtracts the number of people we counted twice.

If A and B are any events then

\[P(A\, \text{or}\, B) = P(A) + P(B) – P(A \,\text{and}\, B).\]

If A and B are mutually exclusive events then \(P(A \,\text{and}\, B) = 0\), so then

\[P(A \, \text{or}\, B) = P(A) + P(B).\]

Example \(\PageIndex{3}\): Additional Rule for Drawing Cards

A single card is drawn from a well shuffled deck of 52 cards. Find the probability that the card is a club or a face card.

There are 13 cards that are clubs, 12 face cards (J, Q, K in each suit) and 3 face cards that are clubs.

\[ \begin{align*} P(\text{club or face card}) &= P(\text{club}) + P(\text{face card}) - P(\text{club and face card}) \\[4pt] &= \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} \\[4pt] &= \dfrac{22}{52} = \dfrac{11}{26} \approx {0.423} \end{align*}\]

The probability that the card is a club or a face card is approximately 0.423 or 42.3%.

A simple way to check this answer is to take the 52 card deck and count the number of physical cards that are either clubs or face cards. If you were to set aside all of the clubs and face cards in the deck, you would end up with the following:

{2 Clubs, 3 Clubs, 4 Clubs, 5 Clubs, 6 Clubs, 7 Clubs, 8 Clubs, 9 Clubs, 10 Clubs, J Clubs, Q Clubs, K Clubs, A Clubs, J Hearts, Q Hearts, K Hearts, J Spades, Q Spades, K Spades, J Diamonds, Q Diamonds, K Diamonds}

That is 22 cards out of the 52 card deck, which gives us a probably of: \[ \begin{align*} \dfrac{22}{52} = \dfrac{11}{26} \approx {0.423} \end{align*}\]

This confirms our earlier answer using the formal Addition Rule.

Example \(\PageIndex{4}\): Addition Rule for Tossing a Coin and Rolling a Die

An experiment consists of tossing a coin then rolling a die. Find the probability that the coin lands heads up or the number is five.

Let H represent heads up and T represent tails up. The sample space for this experiment is S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

• There are six ways the coin can land heads up, {H1, H2, H3, H4, H5, H6}.
• There are two ways the die can land on five, {H5, T5}.
• There is one way for the coin to land heads up and the die to land on five, {H5}.

\[ \begin{align*} P(\text{heads or five}) &= P(\text{heads}) + P(\text{five}) - P(\text{both heads and five}) \\[4pt] &= \dfrac{6}{12} + \dfrac{2}{12} - \dfrac{1}{12} \\[4pt] &= \dfrac{7}{12} = \approx {0.583} \end{align*}\]

The probability that the coin lands heads up or the number is five is approximately 0.583 or 58.3%.

Example \(\PageIndex{5}\): Addition Rule for Satisfaction of Car Buyers

Two hundred fifty people who recently purchased a car were questioned and the results are summarized in the following table.

Find the probability that a person bought a new car or was not satisfied.

\[\begin{align*} P(\text{new car or not satisfied}) &= P(\text{new car}) + P(\text{not satisfied}) - P(\text{new car and not satisfied}) \\[4pt] &= \dfrac{120}{250} + \dfrac{75}{250} - \dfrac{28}{250} = \dfrac{167}{250} \approx 0.668 \end{align*}\]

The probability that a person bought a new car or was not satisfied is approximately 0.668 or 66.8%.

Independent Events

Sometimes we need to calculate probabilities for compound events that are connected by the word “and.” Tossing a coin multiple times or rolling dice are independent events. Each time you toss a fair coin the probability of getting heads is ½. It does not matter what happened the last time you tossed the coin. It’s similar for dice. If you rolled double sixes last time that does not change the probability that you will roll double sixes this time. Drawing two cards without replacement is not an independent event. When you draw the first card and set it aside, the probability for the second card is now out of 51 cards not 52 cards.

Definition: Independent Events

Two events are independent events if the occurrence of one event has no effect on the probability of the occurrence of the other event.

Example \(\PageIndex{6}\): Determining When Events are Independent

Are these events independent?

a) A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.

b) The two events (1) “It will rain tomorrow in Houston” and (2) “It will rain tomorrow in Galveston” (a city near Houston).

c) You draw a card from a deck, then draw a second card without replacing the first.

a) The probability that a head comes up on the second toss is \(\frac{1}{2}\) regardless of whether or not a head came up on the first toss, so these events are independent .

b) These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.

c) The probability of the second card being red depends on whether the first card is red or not, so these events are not independent .

Multiplication Rule for “And” Probabilities: Independent Events

If events A and B are independent events, then \( P(\text{A and B}) = P(A) \cdot P(B)\).

Example \(\PageIndex{7}\): Independent Events for Tossing Coins

Suppose a fair coin is tossed four times. What is the probability that all four tosses land heads up?

The tosses of the coins are independent events. Knowing a head was tossed on the first trial does not change the probability of tossing a head on the second trial.

\(P(\text{four heads in a row}) = P(\text{1st heads and 2nd heads and 3rd heads and 4th heads})\)

\( = P(\text{1st heads}) \cdot P(\text{2nd heads}) \cdot P(\text{3rd heads}) \cdot P(\text{4th heads})\)

\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}\)

\( = \dfrac{1}{16}\)

The probability that all four tosses land heads up is \(\dfrac{1}{16}\).

Example \(\PageIndex{8}\): Independent Events for Drawing Marbles

A bag contains five red and four white marbles. A marble is drawn from the bag, its color recorded and the marble is returned to the bag. A second marble is then drawn. What is the probability that the first marble is red and the second marble is white?

Since the first marble is put back in the bag before the second marble is drawn these are independent events.

\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{9} = \dfrac{20}{81}\end{align*}\]

The probability that the first marble is red and the second marble is white is \(\dfrac{20}{81}\).

Example \(\PageIndex{9}\): Independent Events for Faulty Alarm Clocks

Abby has an important meeting in the morning. She sets three battery-powered alarm clocks just to be safe. If each alarm clock has a 0.03 probability of malfunctioning, what is the probability that all three alarm clocks fail at the same time?

Since the clocks are battery powered we can assume that one failing will have no effect on the operation of the other two clocks. The functioning of the clocks is independent.

\[\begin{align*} P(\text{all three fail}) &= P(\text{first fails}) \cdot P(\text{second fails})\cdot P(\text{third fails}) \\[4pt] &= (0.03)(0.03)(0.03) \\[4pt] &= 2.7 \times 10^{-5} \end{align*}\]

The probability that all three clocks will fail is approximately 0.000027 or 0.0027%. It is very unlikely that all three alarm clocks will fail.

At Least Once Rule for Independent Events

Many times we need to calculate the probability that an event will happen at least once in many trials. The calculation can get quite complicated if there are more than a couple of trials. Using the complement to calculate the probability can simplify the problem considerably. The following example will help you understand the formula.

Example \(\PageIndex{10}\): At Least Once Rule

The probability that a child forgets her homework on a given day is 0.15. What is the probability that she will forget her homework at least once in the next five days?

Assume that whether she forgets or not one day has no effect on whether she forgets or not the second day.

If P (forgets) = 0.15, then P (not forgets) = 0.85.

\[\begin{align*} P(\text{forgets at least once in 5 tries}) &= P(\text{forgets 1, 2, 3, 4 or 5 times in 5 tries}) \\[4pt] & = 1 - P(\text{forgets 0 times in 5 tries}) \\[4pt] &= 1 - P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \\[4pt] &= 1 - (0.85)(0.85)(0.85)(0.85)(0.85) \\[4pt] & = 1 - (0.85)^{5} = 0.556 \end{align*}\]

The probability that the child will forget her homework at least one day in the next five days is 0.556 or 55.6%

The idea in Example \(\PageIndex{9}\) can be generalized to get the At Least Once Rule.

Definition: At Least Once Rule

If an experiment is repeated n times, the n trials are independent and the probability of event A occurring one time is P(A) then the probability that A occurs at least one time is: \(P(\text{A occurs at least once in n trials}) = 1 - P(\overline{A})^{n}\)

Example \(\PageIndex{11}\): At Least Once Rule for Bird Watching

The probability of seeing a falcon near the lake during a day of bird watching is 0.21. What is the probability that a birdwatcher will see a falcon at least once in eight trips to the lake?

Let A be the event that he sees a falcon so P(A) = 0.21. Then, \(P(\overline{A}) = 1 - 0.21 = 0.79\).

\(P(\text{at least once in eight tries}) = 1 - P(\overline{A})^{8}\)

\( = 1 - (0.79)^{8}\)

\( = 1 - (0.152) = 0.848\)

The probability of seeing a falcon at least once in eight trips to the lake is approximately 0.848 or 84.8%.

Example \(\PageIndex{12}\): At Least Once Rule for Guessing on Multiple Choice Tests

A multiple choice test consists of six questions. Each question has four choices for answers, only one of which is correct. A student guesses on all six questions. What is the probability that he gets at least one answer correct?

Let A be the event that the answer to a question is correct. Since each question has four choices and only one correct choice, \(P(\text{correct}) = \dfrac{1}{4}\).

That means \(P(\text{not correct}) =1 - \dfrac{1}{4} = \dfrac{3}{4}\).

\[ \begin{align*} P(\text{at least one correct in six trials}) &= 1 - P(\text{not correct})^{6} \\[4pt] &= 1 - \left(\dfrac{3}{4}\right)^{6} \\[4pt] &= 1 - (0.178) = 0.822 \end{align*}\]

The probability that he gets at least one answer correct is 0.822 or 82.2%.

Probabilities from Two-Way Tables

Two-way tables can be used to define events and find their probabilities using two different approaches: intuitively or using the probability rules. We can calculate “and” and "or" probabilities by combining the data in relevant cells.

Example \(\PageIndex{13}\): Probabilities from a Two-Way Table

Continuation of Example \(\PageIndex{5}\):

A person is chosen at random. Find the probability that the person:

• bought a new car

\[\begin{align*} P(\text{new car}) &= \dfrac{\text{number of new car}}{\text{number of people}} \\[4pt] &= \dfrac{120}{250} = 0.480 = 48.0 \% \end{align*} \]

• was satisfied

\[\begin{align*} P(\text{satisfied}) &= \dfrac{\text{number of satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{175}{250} = 0.700 = 70.0 \% \end{align*} \]

• bought a new car and was satisfied

\[\begin{align*} P(\text{new car and satisfied}) &= \dfrac{\text{number of new car and satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{92}{250} = 0.368 = 36.8 \% \end{align*} \]

• bought a new car or was satisfied

\[\begin{align*} P(\text{new car or satisfied}) &= \dfrac{\text{number of new car + number of satisfied - number of new car and satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{120 + 175 - 92}{250} = \dfrac{203}{250} = 0.812 = 81.2 \% \end{align*} \]

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2.4 - how to assign probability to events.

We know that probability is a number between 0 and 1. How does an event get assigned a particular probability value? Well, there are three ways of doing so:

• the personal opinion approach
• the relative frequency approach
• the classical approach

On this page, we'll take a look at each approach.

The Personal Opinion Approach Section  

This approach is the simplest in practice, but therefore it also the least reliable. You might think of it as the "whatever it is to you" approach. Here are some examples:

• "I think there is an 80% chance of rain today."
• "I think there is a 50% chance that the world's oil reserves will be depleted by the year 2100."
• "I think there is a 1% chance that the men's basketball team will end up in the Final Four sometime this decade."

Example 2-4 Section  

At which end of the probability scale would you put the probability that:

• one day you will die?
• you can swim around the world in 30 hours?
• you will win the lottery someday?
• a randomly selected student will get an A in this course?
• you will get an A in this course?

The Relative Frequency Approach Section  

The relative frequency approach involves taking the follow three steps in order to determine P ( A ), the probability of an event A :

• Perform an experiment a large number of times, n , say.
• Count the number of times the event A of interest occurs, call the number N ( A ), say.
• Then, the probability of event A equals:

\(P(A)=\dfrac{N(A)}{n}\)

The relative frequency approach is useful when the classical approach that is described next can't be used.

Example 2-5 Section  

When you toss a fair coin with one side designated as a "head" and the other side designated as a "tail", what is the probability of getting a head?

I think you all might instinctively reply \(\dfrac{1}{2}\). Of course, right? Well, there are three people who once felt compelled to determine the probability of getting a head using the relative frequency approach:

As you can see, the relative frequency approach yields a pretty good approximation to the 0.50 probability that we would all expect of a fair coin. Perhaps this example also illustrates the large number of times an experiment has to be conducted in order to get reliable results when using the relative frequency approach.

By the way, Count Buffon (1707-1788) was a French naturalist and mathematician who often pondered interesting probability problems. His most famous question

Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips?

came to be known as Buffon's needle problem. Karl Pearson (1857-1936) effectively established the field of mathematical statistics. And, once you hear John Kerrich's story, you might understand why he, of all people, carried out such a mind-numbing experiment. He was an English mathematician who was lecturing at the University of Copenhagen when World War II broke out. He was arrested by the Germans and spent the war interned in a prison camp in Denmark. To help pass the time he performed a number of probability experiments, such as this coin-tossing one.

Example 2-6 Section  

Some trees in a forest were showing signs of disease. A random sample of 200 trees of various sizes was examined yielding the following results:

What is the probability that one tree selected at random is large?

There are 68 large trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is large is 68/200 = 0.34.

What is the probability that one tree selected at random is diseased?

There are 37 diseased trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is diseased is 37/200 = 0.185.

What is the probability that one tree selected at random is both small and diseased?

There are 8 small, diseased trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is small and diseased is 8/200 = 0.04.

What is the probability that one tree selected at random is either small or disease-free?

There are 121 trees (35 + 46 + 24 + 8 + 8) out of 200 total trees that are either small or disease-free, so the relative frequency approach would tell us that the probability that a tree selected at random is either small or disease-free is 121/200 = 0.605.

What is the probability that one tree selected at random from the population of medium trees is doubtful of disease?

There are 92 medium trees in the sample. Of those 92 medium trees, 32 have been identified as being doubtful of disease. Therefore, the relative frequency approach would tell us that the probability that a medium tree selected at random is doubtful of disease is 32/92 = 0.348.

The Classical Approach Section  

The classical approach is the method that we will investigate quite extensively in the next lesson. As long as the outcomes in the sample space are equally likely (!!!), the probability of event \(A\) is:

\(P(A)=\dfrac{N(A)}{N(\mathbf{S})}\)

where \(N(A)\) is the number of elements in the event \(A\), and \(N(\mathbf{S})\) is the number of elements in the sample space \(\mathbf{S}\). Let's take a look at an example.

Example 2-7 Section  

Suppose you draw one card at random from a standard deck of 52 cards. Recall that a standard deck of cards contains 13 face values (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King) in 4 different suits (Clubs, Diamonds, Hearts, and Spades) for a total of 52 cards. Assume the cards were manufactured to ensure that each outcome is equally likely with a probability of 1/52. Let \(A\) be the event that the card drawn is a 2, 3, or 7. Let \(B\) be the event that the card is a 2 of hearts (H), 3 of diamonds (D), 8 of spades (S) or king of clubs (C). That is:

• \(A= \{x: x \text{ is a }2, 3,\text{ or }7\}\)
• \(B = \{x: x\text{ is 2H, 3D, 8S, or KC}\}\)
• What is the probability that a 2, 3, or 7 is drawn?
• What is the probability that the card is a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
• What is the probability that the card is either a 2, 3, or 7 or a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
• What is \(P(A\cap B)\)?

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3.7: Counting Rules

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• Page ID 36659

• Rachel Webb
• Portland State University

Learning Objectives

• Use the Fundamental Counting Rule
• Compute factorials, permutations and combinations
• Understand when to use each method
• Find probabilities using counting rules

There are times when the sample space is very large and is not feasible to write out. In that case, it helps to have mathematical tools for counting the size of the sample space. These tools are known as counting techniques or counting rules.

Fundamental Counting Rule

Definition: fundamental counting rule.

Fundamental Counting Rule: If event 1 can be done m 1 ways, event 2 can be done m 2 ways, and so forth to event n being done m n ways, then the number of ways to do event 1, followed by event 2,…, followed by event n together would be to multiply the number of ways for each event: m 1 · m 2 ··· m n .

Example \(\PageIndex{1}\)

A menu offers a choice of 3 salads, 8 main dishes, and 5 desserts. How many different meals consisting of one salad, one main dish, and one dessert are possible?

There are three events: choosing a salad, a main dish, and a dessert. There are 3 choices for salad, 8 choices for the main dish, and 5 choices for dessert. The ways to choose a salad, main dish, and dessert are: \(\underset{\overline{\text { salad }}} {3} \times \underset{\overline{\text { main dish }}}{8} \times \underset{\overline{\text{ dessert }}}{5}\) = 120 different meals.

Example \(\PageIndex{2}\)

How many 4-digit debit card personal identification numbers (PIN) can be made?

There are four events in this example. The events are picking the first number, then the second number, then the third number, and then the fourth number. The first event can be done 10 ways since the choices are the numbers 1 through 9 and 0. We can use the same numbers over again (repeats are allowed) for the second number, so it can also be done 10 ways. The same with the third and fourth numbers, which also have 10 choices.

There are \(\underset{\overline{\text { first number }}}{10} \times \underset{\overline{\text { second number }}}{10} \times \underset{\overline{\text { third number }}}{10} \times \underset{\overline{\text { fourth number }}}{10} \) = 10,000 possible PINs.

Example \(\PageIndex{3}\)

How many ways can the three letters a , b , and c be arranged with no letters repeating?

There are three events in this case. The events are to pick the first letter, then the second letter, and then the third letter. The first event can be done 3 ways since there are 3 letters. The second event can be done 2 ways, since the first event took one of the letters (repeats are not allowed). The third event can be done 1 way, since the first and second events took 2 of the letters.

There are \(\underset{\overline{\text { first letter }}}{3} \times \underset{\overline{\text { second letter }}}{2} \times \underset{\overline{\text { third letter }}}{1} \) = 6 ways to arrange the letters.

You can also use the tree diagram in Figure \(\PageIndex{1}\) to visualize the arrangements. There are 6 different arrangements of the letters, which can be found by multiplying 3·2·1 = 6.

Figure \(\PageIndex{1}\)

Factorials, Permutations and Combinations

If we have 10 different letters for, say, a password, the tree diagram would be very time-consuming to make because of the length of options and tasks, so we have some shortcut formulas that help count these arrangements.

Many counting problems involve multiplying a list of decreasing numbers, which is called a factorial . The factorial is represented mathematically by the starting number followed by an exclamation point, in this case 3! = 3·2·1 = 6. There is a special symbol for this and a special button on your calculator for the factorial.

Formula: Factorial Rule

Factorial Rule: The number of different ways to arrange n distinct objects is

\(n! = n \cdot (n – 1) \cdot (n – 2) \ldots \cdot 3 \cdot 2\cdot 1\), where repetitions are not allowed.

0 factorial is defined to be 0! = 1 and 1 factorial is defined to be 1! = 1.

TI-84 Plus: Enter the number of which you would like to find the factorial. Press [MATH]. Use cursor keys to move to the PRB (or PROB) menu. Press 4 (4:!). Press [ENTER] to calculate.

Excel : In an empty cell type in =FACT( n ), where n is the number of which you would like to find the factorial. Thus 4! would be =FACT(4).

Example \(\PageIndex{4}\)

How many ways can you arrange 5 people standing in line?

No repeats are allowed since you cannot reuse a person twice in the line. Order is important since the first person is first in line and will be selected first. This meets the requirements for the factorial rule. There are 5! = 5·4·3·2·1 = 120 ways to arrange 5 people standing in line.

Sometimes we do not want to select the entire group but only select r objects from n total objects. The number of ways to do this depends on if the order you choose the r objects matters or if it does not matter. As an example, if you are trying to call a person on the phone, you have to have the digits of their number in the correct order. In this case, the order of the numbers matters. If you were picking random numbers for the lottery, it does not matter which number you pick first since they always arrange the numbers from the smallest to largest once the numbers are drawn. As long as you have the same numbers that the lottery officials pick, you win. In this case, the order does not matter.

A permutation is an arrangement of items with a specific order. You use permutations to count items when the order matters.

Formula: Permutation Rule

Permutation Rule: The number of different ways of picking r objects from n distinct total objects when repeats are not allowed and order matters is \({ }_{n} P_{r} = P(n, r) = \dfrac{n !}{(n-r) !}\).

When the order does not matter, you use combinations. A combination is an arrangement of items when order is not important. When you do a counting problem, the first thing you should ask yourself is “Are repeats allowed?”, then ask yourself “Does order matter?”

Formula: Combination Rule

Combination Rule: The number of ways to select r objects from n distinct total objects when repeats are not allowed and order does not matter is \({ }_{n} C_{r} = C(n, r) = \dfrac{n !}{r !(n-r) !}\).

TI-84 Plus: Enter the number of total objects ( n ). Press [MATH]. Use cursor keys to move to the PRB (or PROB) menu. Press 2 for permutation (2: nPr), or 3 for combination (3: nCr). Enter the number of objects to be selected ( r ). Press [ENTER] to calculate.

Excel: In a blank cell, type the formula =COMBIN( n, r ) or =PERMUT( n, r ) where n is the total number of objects and r is the number of objects that you are selecting. For example, =COMBIN(8, 3).

The following flow chart in Figure \(\PageIndex{2}\) may help with deciding which counting rule to use.

Start at the top: ask yourself if the same item can be repeated. For instance, a person on a committee cannot be counted as two distinct people; however, a number on a car license plate may be used twice. If repeats are not allowed, then ask, does the order in which the item is chosen matter? If it does not, then use a combination. If it does, then ask if you are ordering the entire group. If so, use a factorial. If only ordering some from the total, use a permutation.

Figure \(\PageIndex{2}\)

Example \(\PageIndex{5}\)

Circle K International, a college community service club, has 15 members this year. How many ways can a board of officers consisting of a president, vice-president, secretary and treasurer be elected?

In this case, repeats are not allowed since the same member cannot hold more than one position. The order matters because if you elect person 1 for president and person 2 for vice-president, there would be different members in those positions than if you elect person 2 for president, person 1 for vice-president. Thus, this is a permutation problem with n = 15 and r = 4. There are \({ }_{15} P_{4}=\dfrac{15 !}{(15-4) !}=\dfrac{15 !}{11 !}=32,760 \) ways to elect the 4 officers.

In general, if you were selecting items that involve rank, a position title, 1 st , 2 nd , or 3 rd place or prize, etc., then the order in which the items are arranged is important and you would use permutation.

Example \(\PageIndex{6}\)

Circle K International, a college community service club, has 15 members this year. They need to select 2 members to be representatives for the school's Inter-Club Council. How many ways can the 2 members be chosen?

In this case, repeats are not allowed, since there must be 2 different members as representatives. The order in which the representatives are selected does not matter since they have the same position. Thus, this is a combination problem with n = 15 and r = 2. There are \({ }_{15} C_{2} =\dfrac{15 !}{2 !(15-2) !}=\dfrac{15 !}{2 ! \cdot 13 !} = 105\) ways to select 2 representatives.

The counting rules can be used in the probability rules to calculate the number of ways that an event can occur.

Example \(\PageIndex{7}\)

An iPhone has a 6-digit numerical password to unlock the phone. What is the probability of guessing the password on the first try?

Example \(\PageIndex{8}\)

What is the probability of winning the jackpot in the Powerball Lottery? To play, you must pick 5 numbers between 1 and 69, and pick one Power Number between 1 and 26. You must match all 6 numbers to win the jackpot. ( https://www.calottery.com/draw-games/powerball )

Example \(\PageIndex{9}\)

What is the probability of getting a full house if 5 cards are randomly dealt from a standard deck of cards?

A full house consists of 3-of-a-kind and a pair. For example, 3 queens and a pair of 2s. There are \({}_{13} C_{1}\) ways to choose a card between ace, 2, 3, … king for the 3-of-a-kind. Once a card is chosen, there are 4 cards with that rank and there are \({}_{4} C_{3}\) ways to choose a 3-of-a-kind from that rank. Since the card chosen for the 3-of-a-kind cannot be chosen for the pair, there are \({}_{12} C_{1}\) ways to choose a card for the pair. Once a card is chosen for the pair, there are \({}_{4} C_{2}\) ways to choose a pair from that rank. All together there are \({}_{52} C_{5}\) ways to randomly deal 5 cards from a deck. Using the Fundamental Counting Rule, the probability of getting a full house is \(P(\text{full house}) = \dfrac{ {}_{13} C_{1} \times {}_{4} C_{3} \times {}_{12} C_{1} \times {}_{4} C_{2} }{ {}_{52} C_{5} } = \dfrac{13 \times 4 \times 12 \times 6}{2,598,960} = \dfrac{3744}{2,598,960} = 0.00144\).