3. The Fundamental Counting Principle
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Chapter 12
3.
The Fundamental Counting Principle
The fundamental counting principle is a valuable tool for solving problems involving multiple choices or events. This lesson explains how to use it to calculate outcomes for both independent and dependent events. It also introduces permutations, which are essential for understanding the arrangements of items or events. Students can apply these concepts to real-world scenarios, such as planning schedules, designing systems, or analyzing probabilities. Mastery of these topics provides a strong foundation for tackling complex problems in statistics, probability, and decision-making processes.
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| | 10 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
The Fundamental Counting Principle
Slide of 10
When calculating probabilities, it is important to be able to find the number of possible outcomes of events. Listing all the possible outcomes of compound events can be very time-consuming, though. This is where the Fundamental Counting Principle comes in handy, making it possible to find the number of outcomes of a combination of events without listing them.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Charity Raffle
Tiffaniqua is helping organize a charity event. Her job is to prepare a raffle! She decides that everyone who buys a raffle ticket will first roll a die and then spin a spinner with 4 fields.
Their prize will be determined by the combination of the results of the coin toss and the spin.
a How many possible outcomes are there?
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b What is the probability of getting a six and the spinner landing on the purple field? Write the answer as a fraction.
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Discussion
Independent Events
Two events A and B are independent events if the occurrence of either of these events does not affect the occurrence of the other. It is also said that they are independent if and only if the probability that both events occur is equal to the product of the individual probabilities.
P(A∩B)=P(A)⋅P(B)
Why
For example, consider drawing two marbles from a bowl, one at a time.
Let G, B, and O be the events of drawing green, blue, and orange marbles, respectively. The probability of first picking a green marble can be calculated by dividing the favorable outcomes by the possible outcomes. There is 1 green marble and 3 marbles in total.
P(G)=31
Suppose that the first marble is replaced before the second draw. Therefore, after the replacement there is 1 orange marble, and 3 marbles in total.
P(O)=31
Note that there are 9 possible outcomes for drawing two marbles one at a time. Only 1 of these options corresponds to an event of drawing a green marble and then an orange marble.
GGGBGOBBBGBOOOOGOB
Therefore, the combined probability of picking a green marble first and an orange marble second is 91. Since the probability that both events occur is equal to the product of the individual probabilities, these events can be considered as independent events.
31⋅31=91⇓P(G)⋅P(O)=P(G∩O)
Discussion
Dependent Events
Two events A and B are considered dependent events if the occurrence of either of these events affects the occurrence of the other. If the events are dependent, the probability that both events occur is equal to the product of the probability of the first event occurring and the probability of the second event occurring after the first event.
P(A∩B)=P(A)⋅P(B∣A)
Why
For example, consider drawing two marbles from a bowl, one at a time.
Let G, B, and O be the events of drawing green, blue, and orange marbles, respectively. The probability of first picking a green marble can be calculated by dividing the favorable outcomes by the possible outcomes. There is 1 green marble and 3 marbles in total. 
This affects the probability of picking an orange marble on the second draw. Now there is still 1 orange marble, but instead of 3, there are 2 marbles in total.
P(G)=31
Suppose that after the green marble is picked, it is not replaced in the bowl. P(O∣G)=21
Using this information, the sample space of the described situation can be found.
GBGOBGBOOGOB
Out of 6, there is only 1 outcome that corresponds to first drawing a green marble and then an orange marble. Therefore, the probability of picking a green and then an orange marble is 61. 31⋅21=61⇓P(G)⋅P(O∣G)=P(G∩O)
Because the occurrence of the first event affects the occurrence of the second, these events can be concluded to be dependent.
Discussion
Fundamental Counting Principle
If an event A has n possible outcomes and an event B has m possible outcomes, then the total number of different outcomes for A and B combined is n×m. This principle is used to find the number of possible outcomes for a combination of independent events.
Proof
Informal JustificationConsider an arbitrary process that can be divided into two tasks. Now imagine that there are n different ways of completing the first task and m different ways of completing the second task. To complete the whole process, first, one of the n ways to start it should be chosen. Then, there will be m possible ways to finish it. 
This happens for each of the n different ways in which the process can be started. Therefore, there are n×m different ways of completing the process. This is a generic argument that can be applied in multiple scenarios. For example, the following diagram shows the different choices of notebooks that a store sells. 
In this example, the store sells 2 types of notebooks — one with a spiral binding and one without. Each notebook type comes in 3 different colors — blue, red, and green. According to the Fundamental Counting Principle, there are 6 different outcomes for what notebook a customer may buy.
2×3=6
It should be noted that this is an informal justification and should not be taken as a formal proof.
Extra
Counting the Outcomes of Dependent EventsAs mentioned above, this principle holds true only if the events are independent of each other. If the events are dependent, multiplying the number of possible outcomes for each event will not be correct. Considering the previous example, suppose now that the spiral-bound notebooks came only in red.
There are still 2 types of notebooks and a total of 3 colors for the non spiral bound notebooks. However, the possible number of different notebooks a customer may buy is not 2×3=6. Rather, it is 4. This happens because, in this case, the possible colors for a notebook depend on the type of notebook.
Example
Coming up With Ideas for the Raffle
While preparing the raffle, Tiffaniqua considered inviting everyone who buys a ticket to roll a die and toss a coin.
a How many different prizes would Tiffaniqua have to prepare?
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b One of the prizes is a teddy bear. To win the teddy bear, someone would have to roll a 5 and get tails in the coin toss. What is the probability of winning the teddy bear? Write the answer as a fraction.
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Hint
a There are 6 sides on a die and 2 sides of a coin.
b There is one outcome that results in winning the teddy bear. How many possible outcomes are there in total?
Solution
a Drawing a prize in Tiffaniqua's raffle can be considered a compound event consisting of two simple events. The first simple event is rolling the die, and the second one is tossing the coin.
|
Fundamental Counting Principle |
|
If an event A has n possible outcomes and an event B has m possible outcomes, then the total number of different outcomes for A and B combined is n⋅m. |
6⋅2=12
There are 12 possible outcomes, so Tiffaniqua would need to come up with 12 prizes for the raffle. The number of possible outcomes can also be found using a tree diagram. 
According to the tree diagram, there are 12 possible outcomes. This is the same number as the one found through the Fundamental Counting Principle. Tiffaniqua wanted to have more possible prizes, so she had to come up with an idea for the raffle with more possible outcomes!
b To win a teddy bear, the result of rolling the die must be a 5 and the result of tossing the coin must be tails.
P(Teddy bear)P(Teddy bear)=Possible outcomesFavorable outcomes⇓=121
The probability of winning a teddy bear is 121.
Discussion
Permutation
A permutation is an arrangement of objects in which the order is important. For example, consider constructing a number using only the digits 4, 5, and 6 without repetitions. Any of the three digits can be picked for the first position, leaving two choices for the second position, then only one choice for the third position.
456465546564645654
Although all these numbers are formed with the same three digits, the order in which the digits appear affects the number produced. Each different order of the digits creates a different number. The number of permutations can be calculated by using the Fundamental Counting Principle. Number of permutations=3⋅2⋅1⇕Number of permutations=6
Example
Lining up the Prizes
Tiffaniqua chose 5 prizes to display at her stall — a cinema ticket, a CD, a basketball, a teddy bear, and a pair of sunglasses. She is sure they will attract many people to her raffle!
Now all she has to do is decide in which order she should line the prizes up.
a In how many ways can Tiffaniqua order the prizes?
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b Tiffaniqua decides to choose the order of the prizes at random. What is the probability that the teddy bear will be the first prize and the CD will be third? Write the answer as a fraction in the simplest form.
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Hint
a When choosing the first prize to display, Tiffaniqua has 5 choices. When choosing the second prize, she has 4 choices. How many choices does she have when choosing the third, fourth, and fifth prize?
b The favorable outcomes are the outcomes where the teddy bear is the first prize and the CD is the third prize.
Solution
a Choosing the order of the prizes can be considered as five separate choices: choosing the first prize, choosing the second prize, and so on. When choosing the first prize, there are 5 possible outcomes, as there are five available prizes.
| First Prize | 5 choices |
|---|---|
| Second Prize | |
| Third Prize | |
| Fourth Prize | |
| Fifth Prize |
When choosing the second prize, there are only 4 possible outcomes. This is because one of the five prizes has already been chosen for the first prize.
| First Prize | 5 choices |
|---|---|
| Second Prize | 4 choices |
| Third Prize | |
| Fourth Prize | |
| Fifth Prize |
For the same reason, there are 3 choices for the third prize, 2 choices for the fourth prize, and only 1 choice for the fifth prize.
| First Prize | 5 choices |
|---|---|
| Second Prize | 4 choices |
| Third Prize | 3 choices |
| Fourth Prize | 2 choices |
| Fifth Prize | 1 choice |
5⋅4⋅3⋅2⋅1=120
There are 120 possible ways for Tiffaniqua to order the five prizes. In other words, there are 120 permutations of the 5 prizes.
b The probability that the teddy bear will be the first prize and the CD will be the third prize is equal to the quotient of the number of favorable outcomes and the total number of possible outcomes. In any favorable outcome, the teddy bear must be the first prize and the CD must be the third prize.
| First Prize | Teddy Bear |
|---|---|
| Second Prize | |
| Third Prize | CD |
| Fourth Prize | |
| Fifth Prize |
The second prize could be the sunglasses, the basketball, or the cinema ticket. There are 3 options for the second prize.
| First Prize | Teddy Bear |
|---|---|
| Second Prize | 3 options |
| Third Prize | CD |
| Fourth Prize | |
| Fifth Prize |
The fourth prize cannot be the teddy bear, the CD, or the second prize. This means that there are 2 options left for the fourth prize. Similarly, there is only 1 option left for the fifth prize after the fourth prize is chosen.
| First Prize | Teddy Bear |
|---|---|
| Second Prize | 3 options |
| Third Prize | CD |
| Fourth Prize | 2 options |
| Fifth Prize | 1 option |
3⋅2⋅1=6
There are 6 possible outcomes of choosing the second, fourth, and fifth prizes when the first prize is the teddy bear and the third prize is the CD. This means that when choosing the order of the prizes, there are 6 outcomes where the first prize is the teddy bear and the third prize is the CD. In Part A, it was shown that there are 120 possible outcomes.
P(Teddy bear first, CD third)=1206⇕P(Teddy bear first, CD third)=201
When choosing the order of the prizes at random, the probability that the teddy bear is the first prize and the CD is the third prize is 201.
Example
A Random Prize
Right before the event, a new sponsor decided to give Tiffaniqua a bunch of new prizes for her raffle! It was too late to change the way the raffle works to accommodate the new prizes. Instead, Tiffaniqua decided to have a side raffle!
When someone wants to take part in the side raffle, they draw a random prize from a box of 20 prizes. The big box contains prizes like a pair of gloves and a pair of socks.
a Tiffaniqua was worried that someone might not like the prize they draw. She decided that, for an extra donation, it should be possible to put the random prize back in the box and draw a prize again. What is the probability that a person taking part draws a pair of gloves and then a pair of socks in the second attempt? Express the answer as a fraction.
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b Zain buys two tickets for the side raffle, so they need to draw two prizes from the box! The box is not refilled between Zain drawing the first prize and the second prize. They draw one prize and then the other. What is the probability that they draw a pair of socks and a pair of gloves? Express the answer as a fraction in the simplest form.
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Hint
a In both attempts, there are 20 possible prizes to draw.
b When drawing the first prize, there are 20 prizes in the box. When drawing the second prize, there are only 19 prizes left in the box. Zain can draw either the socks or the gloves first.
Solution
a Each person taking part in the side raffle draws a prize from 20 available prizes. If they do not like their prize, they can make an extra donation, put their prize back in the box, and draw another prize. When drawing a prize again, there are also 20 possibilities.
20⋅20=400
There is only one favorable outcome — drawing a pair of gloves and then drawing a pair of socks. The probability of drawing a pair of gloves and then a pair of socks is the quotient of the number of favorable outcomes and the number of all possible outcomes.
P(pair of gloves, then pair of socks)=4001
b Zain bought two tickets for the side raffle. They draw one prize and then another. We want to know the probability that they draw a pair of socks and a pair of gloves. When drawing the first prize, there are 20 prizes available. When drawing the second prize, there are only 19 prizes in the box.
20⋅19=380
There are 2 favorable outcomes — drawing the pair of socks first and then the pair of gloves, or drawing the pair of gloves first and then the pair of socks. The probability that Zain wins the pair of socks and the pair of gloves is the quotient of the number of favorable outcomes and the number of all possible outcomes.
P(socks and gloves)=3802⇕P(socks and gloves)=1901
Closure
Combining Events
In Tiffaniqua's raffle, the prize is determined by the combination of the results of a die roll and a spin of a spinner.
The Fundamental Counting Principle can be used to find the number of all possible outcomes. There are 6 possible outcomes of rolling a die and 4 possible outcomes of spinning the spinner.
6⋅4=24
There are 24 possible outcomes. The probability of getting a six and the purple field is the quotient of the number of positive outcomes to the number of possible outcomes. There is only 1 positive outcome — getting a six on the die roll and the purple field on the spinner.
P(6, purple)=241
The probability of getting a six and the purple field when rolling a die and spinning the spinner is 241.
The winner of a lottery is chosen by drawing 6 out of 30 numbered balls. The order in which the balls are drawn does not matter.
Heichi chooses to play the following numbers in the lottery.
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Heichi bought a lottery ticket and chose the following numbers. We want to find the probability that he wins the jackpot.
There are 30 possible outcomes of drawing the first ball, 29 possible outcomes of choosing the second ball, 28 possible outcomes of choosing the third ball, 27 possible outcomes of choosing the fourth ball, 26 possible outcomes of choosing the fifth ball, and 25 possible outcomes of drawing the sixth ball. cccccc First [-0.4em]ball & Second [-0.4em]ball & Third [-0.4em]ball & Fourth [-0.4em]ball & Fifth [-0.4em]ball & Sixth [-0.4em]ball 30& 29& 28& 27& 26& 25 By the Fundamental Counting Principle, the number of possible results of the whole draw is equal to the product of the numbers of possible outcomes of drawing each ball. 30* 29* 28* 27* 26* 25=427 518 000 The total number of possible outcomes is 427 518 000. For Heichi to win the lottery, every number drawn must be one of the 6 numbers he chose. This means that there are 6 positive outcomes of choosing the first ball, 5 possible outcomes of choosing the second ball, and so on. cccccc First [-0.4em]ball & Second [-0.4em]ball & Third [-0.4em]ball & Fourth [-0.4em]ball & Fifth [-0.4em]ball & Sixth [-0.4em]ball 6& 5& 4& 3& 2& 1 By the Fundamental Counting Principle, there are 6* 5* 4* 3* 2* 1, or 720, positive outcomes. The probability that Heichi gets all the numbers right is the quotient of the number of positive outcomes to the number of all positive outcomes. 720/427 518 000 = 1/593 775 The probability that Heichi guesses all six numbers correctly is 1593 775.
Next, we want to find the probability that Heichi has a chance to win the jackpot until the last ball is drawn. This means that the first 5 balls drawn must all be numbers that Heichi chose. From Part A, we know that there are 427 518 000 possible results of the draw. Let's find the number of positive outcomes. cccccc First [-0.4em]ball & Second [-0.4em]ball & Third [-0.4em]ball & Fourth [-0.4em]ball & Fifth [-0.4em]ball & Sixth [-0.4em]ball The first number must be one of the 6 that Heichi chose, so there are 6 positive outcomes of drawing the first ball. The second number must be one of Heichi's numbers and cannot be the number that was drawn first, so there are 5 positive outcomes. cccccc First [-0.4em]ball & Second [-0.4em]ball & Third [-0.4em]ball & Fourth [-0.4em]ball & Fifth [-0.4em]ball & Sixth [-0.4em]ball 6& 5&&&& Similarly, there are 4, 3, and 2 positive outcomes for drawing the third, fourth, and fifth ball, respectively. cccccc First [-0.4em]ball & Second [-0.4em]ball & Third [-0.4em]ball & Fourth [-0.4em]ball & Fifth [-0.4em]ball & Sixth [-0.4em]ball 6& 5& 4& 3& 2& The last ball must be one of the numers that Heichi did not choose — otherwise, he would win the jackpot! There are 24 numbers that Heichi didn't choose, so there are 24 positive outcomes of drawing the last ball. cccccc First [-0.4em]ball & Second [-0.4em]ball & Third [-0.4em]ball & Fourth [-0.4em]ball & Fifth [-0.4em]ball & Sixth [-0.4em]ball 6& 5& 4& 3& 2& 24 By the Fundamental Counting Principle, the total number of positive outcomes is the product of the numbers of positive outcomes of each event. 6* 5* 4* 3* 2* 24 = 17 280 The probability is the quotient of the number of positive outcomes to the number of possible outcomes. 17 280/427 518 000 = 8/197 925
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Ignacio's class has 4 maths lessons every week. During every lesson, one of the students in Ignacio's class is randomly chosen to present their solution to the homework given during the previous lesson. There are 25 students in Ignacio's class.
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In each math lesson, one of the 25 students in Ignacio's class is chosen to present their solution to the homework. We want to know the probability that Ignacio is chosen in each of 4 lessons in a week. Choosing students to present their solutions over a week can be seen as a compound event consisting of 4 simple events.
Fundamental Counting Principle |- If an event A has n possible outcomes and an event B has m possible outcomes, then the total number of different outcomes for A and B combined is n * m.
In each of the simple events, there are 25 possible outcomes, one for each student that can be chosen. By the Fundamental Counting Principle, the total number of outcomes of the combination of the 4 choices in a week is the following product. 25*25*25*25 = 390 625 There are 390 625 possible outcomes. There is only 1 outcome in which Ignacio is chosen every time, so there is 1 positive outcome. The probability that Ignacio is chosen every time is the quotient of the number of positive outcomes and the number of possible outcomes. P(Ignacio every time) = 1/390 625
Next, we will find the probability that one of 5 students in Ignacio's homework group is chosen to present their solution in every lesson in a week. Like in Part A, there are 390 625 possible outcomes. There are 5 positive outcomes for each choice. Let's find the number of positive outcomes using the Fundamental Counting Principle. 5* 5* 5* 5 = 625 There are 625 positive outcomes. The probability is the quotient of the number of positive outcomes and the number of possible outcomes. P(memeber of the group every time ) = 625/390 625 ⇕ P(member of the group every time) = 0.0016
Finally, we will find the probability that one of the students in Ignacio's group is chosen to present their solution in every lesson in a week, but no one is chosen more than once. For the first lesson, there are 5 positive outcomes, as any of the 5 students in the group can be chosen. cccc First [-0.4em]Lesson & Second [-0.4em]Lesson & Third [-0.4em]Lesson & Fourth [-0.4em]Lesson 5 For the second lesson, there are 4 positive outcomes, as the student who was chosen in the first lesson cannot be chosen again. Similarly, there are 3 positive outcomes for the third lesson and 2 positive outcomes for the fourth lesson. cccc First [-0.4em]Lesson & Second [-0.4em]Lesson & Third [-0.4em]Lesson & Fourth [-0.4em]Lesson 5& 4& 3& 2 By the Fundamental Counting Principle, the number of positive outcomes is equal to the product of the numbers of positive outcomes of each choice. 5* 4* 3 * 2 = 120 There are 120 positive outcomes. In Part A, we found that there are 390 625 possible outcomes. Let's calculate the probability! 120/390 625 = 24/78 125 The probability that in each lesson in a week, one of the members of Ignacio's group is chosen, but no one is chosen more than once is 2478 125.
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The Fundamental Counting Principle
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