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 Fundamental Counting Principle
Exercises
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We want to calculate the number of three-digit numbers where all the digits are odd. There are 5 odd digits. 1, 3, 5, 7, 9 Making a three-digit number can be seen as a compound event, where the simple events are choosing the first digit, the second digit, and then the third digit. Each of the simple events has 5 possible outcomes. To find the number of possible outcomes of the compound event, we can use the Fundamental Counting Principle. ccccc First Digit&Second Digit&Third Digit 5&5&5 The number of possible three-digit numbers with all digits odd is equal to this product. 5*5*5=125
Next, we will calculate the number of three digit numbers with all digits even. There are 5 even digits. 0, 2, 4, 6, 8 Writing a three-digit number can be seen as a compound event, consisting of three simple events — one for choosing each digit. There are 5 possible digits for the second and third digit. However, the first digit of a three-digit number cannot be 0, so there are only 4 possible digits for the first digit of the number. cccc First Digit&Second Digit&Third Digit 4&5&5 By the Fundamental Counting Principle, the number of possible three-digit numbers with all digits even is equal to the product of the number of options for each digit. 4*5*5=100
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A three-digit number is randomly chosen.
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We randomly choose a three-digit number and want to find the probability that the second digit is even. Randomly choosing a three-digit number can be seen as a compound event consisting of three simple events — choosing the first, second, and third digit.
There are 10 options for the second and third digits, but only 9 for the first digit, since 0 cannot be the first digit of a three-digit number. By the Fundamental Counting Principle, the number of possible outcomes of randomly choosing a three-digit number is the product of the numbers of possible first, second, and third digits. ccccc First Digit&&Second Digit&&Third Digit 9&*&10&*&10 This means that there are 900 possible three-digit numbers, so there are 900 possible outcomes. 9*10*10=900 Next, let's find the number of positive outcomes. Every positive outcome must have an even second digit. There are 5 even digits, so there are 5 positive outcomes when choosing the second digit. ccccc First Digit&&Second Digit&&Third Digit &&5&& There are no restrictions on the first and third digits, so there are 9 and 10 positive outcomes when choosing the first and third digits, respectively. By the Fundamental Counting Principle, the number of positive outcomes is the product of the numbers of positive outcomes of choosing the first, second, and third digits. ccccc First Digit&&Second Digit&&Third Digit 9&*&5&*&10 There are 450 positive outcomes. 9*5*10=450 The probability of a randomly chosen three-digit number having an even second digit is the quotient of the number of positive outcomes to the total number of possible outcomes. Remember, there are 450 positive outcomes and 900 possible outcomes.
The probability that the second digit is even is 450900, or 12, which is equal to 50 %.
Next, we will find the probability that the third digit of a randomly chosen three-digit number is 1 or 7. First, let's find the number of positive outcomes. The first digit cannot be 0 and the third digit must be 1 or 7, so there are 9 outcomes for choosing the first digit, 10 outcomes for choosing the second, and 2 outcomes for choosing the third. ccccc First Digit&&Second Digit&&Third Digit 9&& 10&& 2 By the Fundamental Counting Principle, the number of positive outcomes is equal to the product of the number of positive outcomes of the simple events. This means that there are 180 positive outcomes. 9*10*2=180 As we found in Part A, there are 900 possible outcomes. The probability of a randomly chosen three-digit number having 1 or 7 as the third digit is the quotient of the number of positive outcomes to the total number of possible outcomes. Remember, there are 180 positive outcomes.
The probability that the third digit is 1 or 7 is 180900, or 15, which is equal to 20 %.
Now we will find the probability that all the digits of a randomly chosen three-digit number are different. As before, we think of choosing a three-digit number as a compound event consisting of three simple events for choosing each digit. As we found in Part A, there are 900 possible outcomes. ccccc First Digit&&Second Digit&&Third Digit 9&*&10&*&10 Let's find the number of positive outcomes. When choosing the first digit, there are 9 possible outcomes, as the first digit cannot be 0. ccccc First Digit&&Second Digit&&Third Digit 9&&&& There are also 9 possible outcomes of choosing the second digit. This is because there are 10 possible digits to choose from, but the first digit cannot be repeated. ccccc First Digit&&Second Digit&&Third Digit 9&& 9&& Similarly, there are 8 outcomes when choosing the third digit — there are 10 possible digits, but the first or second digits cannot be repeated. ccccc First Digit&&Second Digit&&Third Digit 9&& 9&& 8 By the Fundamental Counting Principle, the number of positive outcomes of choosing the number is equal to the product of the numbers of positive outcomes of choosing the first, second, and third digits. 9* 9 * 8 = 648 There are 648 positive outcomes. The probability that a random three-digit number has three different digits is the quotient of the number of positive outcomes to the number of possible outcomes. Remember, there are 900 possible outcomes.
The probability that all the digits of a random three-digit number are different is 1825, or 72 %.
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{12,58,79,20}
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We are given a set of 4 numbers. {12, 58, 79, 20} We want to find the number of ways in which the numbers in the set can be ordered. Every ordering is a permutation of the set. The number of permutations can be calculated using the Fundamental Counting Principle.
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.
Ordering the set can be seen as a compound event that consists of 4 simple events — choosing the first number, choosing the second number, choosing the third number, and choosing the fourth number. When choosing the first number, there are 4 options. ccccccc First Number && Second Number && Third Number && Fourth Number 4&&&& When choosing the second number, there are only 3 options, as one of the numbers in the set has already been chosen. Similarly, there are 2 options when choosing the third number and just 1 number when choosing the fourth number. ccccccc First Number && Second Number && Third Number && Fourth Number 4&& 3&& 2&& 1 By the Fundamental Counting Principle, the total number of outcomes is equal to the product of the numbers of outcomes of the simple events. ccccccc First Number && Second Number && Third Number && Fourth Number 4&*&3&*&2&*&1 The total number of outcomes is 24. There are 24 permutations of the given set, so there are 24 possible orders of the four numbers.
We want to find the probability that, when randomly ordering the given set, we end up with the set ordered in decreasing order. There is only 1 way for the set to be ordered in decreasing order. {79 58 20 12} As we found in Part A, there are 24 ways to order the given set. The probability that the set is in decreasing order is equal to the quotient of the number of positive outcomes to the number of possible outcomes. P(decreasing order)=1/24
Finally, we will find the probability that 12 is not the first number. To find the number of positive outcomes, we will once again consider ordering the given set as a compound event that consists of 4 simple events for choosing each number. When choosing the first number, there are 3 options, as 12 cannot be chosen. ccccccc First Number && Second Number && Third Number && Fourth Number 3&&&& When choosing the second number, there are also 3 options. This is because, while the first number cannot be chosen anymore, 12 becomes available. There are 2 options when choosing the third number and just 1 number when choosing the fourth number. ccccccc First Number && Second Number && Third Number &&s Fourth Number 3&& 3&& 2&& 1 Let's use the Fundamental Counting Principle! ccccccc First Number && Second Number && Third Number && Fourth Number 3&*&3&*&2&*&1 The total number of outcomes is 18. There are 18 positive outcomes. In Part A, we found that there are 24 total outcomes. P(12not first) = 18/24 ⇓ P(12not first) = 0.75 The probability that 12 is not the first number is 0.75.
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{1,2,3,4,5,6,7,8,9,10}
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We are given a set with 10 elements. Three numbers are drawn with replacement from the set. We want to find the probability that all three numbers are even. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Drawing three numbers with replacement is a compound event consisting of three simple events - drawing the first, second, and third numbers. The Fundamental Counting Principle tells us that the total number of possible outcomes is equal to the product of the number of outcomes of each simple event.
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.
Each of the simple events has 10 possible outcomes because the set has 10 elements and each number is drawn from the full set. ccccccc First Number && Second Number && Third Number 10&*& 10&*& 10 This means that there are 1000 possible outcomes. Next, let's find the number of positive outcomes. There are 5 even numbers in the set. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} This means that there are 5 positive outcomes for choosing each number. Using the Fundamental Counting Principle again, the number of positive outcomes when drawing 3 numbers is the product of the numbers of positive outcomes for choosing each number. ccccccc First Number && Second Number && Third Number 5&*& 5&*& 5 There are 125 positive outcomes. The probability that all drawn numbers are even is the quotient of the number of positive outcomes and the number of possible outcomes. P(all numbers even) = 125/1000 ⇕ P(all numbers even) = 0.125
Next, we will find the probability that the first number drawn is divisible by 3 and the third number is greater than 6. In the given set, there are 3 numbers divisible by 3. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} This means that there are 3 positive outcomes when choosing the first number. There are no restrictions on the second number, so there are 10 positive outcomes of choosing the second number. Finally, there are 4 numbers greater than 6 in the set, so there are 4 positive outcomes when choosing the third number. ccccccc First Number && Second Number && Third Number 3&*& 10&*& 4 By the Fundamental Counting Principle, there are 3*10* 4, or 120, positive outcomes. We know from Part A that there are 1000 possible outcomes. The probability that the first number is divisible by 3 and the third number is greater than 6 is the quotient of the number of positive outcomes and the number of possible outcomes. P(1st divisible by3, 3rd greater than6) = 120/1000 ⇕ P(1st divisible by3, 3rd greater than6) = 0.12
Finally, we will find the probability that the first number drawn is odd, the second number is divisible by 3, and the third number is not greater than 4. In the given set, there are 5 odd numbers, 3 numbers that are divisible by 3, and 4 numbers not greater than 4. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} This means that there are 5 positive outcomes when choosing the first number, 3 positive outcomes of choosing the second number, and 4 positive outcomes when choosing the third number. Let's use the Fundamental Counting Principle. ccccccc First Number && Second Number && Third Number 5&*& 3&*& 4 There are 60 positive outcomes. We know from Part A that there are 1000 possible outcomes. The probability that the first number drawn is odd, the second number is divisible by 3, and the third number is not greater than 4 is the quotient of the number of positive outcomes and the number of possible outcomes. 60/1000 = 0.06
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{1,2,3,4,5,6,7,8,9,10}
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We are given a set with 10 elements. Three numbers are drawn without replacement from the set. We want to find the probability that all three numbers are even. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Drawing three numbers without replacement is a compound event consisting of three simple events - drawing the first, second, and third numbers. The Fundamental Counting Principle tells us that the total number of possible outcomes is equal to the product of the number of outcomes of each simple event.
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.
Drawing the first number has 10 possible outcomes because the set has 10 elements. Drawing the second number has 9 possible outcomes, as the first number drawn is removed from the set. Similarly, drawing the third number has 8 possible outcomes. ccccccc First Number && Second Number && Third Number 10&*& 9&*& 8 This means that there are 720 possible outcomes. Next, let's find the number of positive outcomes. There are 5 even numbers in the set. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} This means that there are 5 positive outcomes for choosing the first number. After the first even number is drawn, only 4 even numbers remain, so there are 4 positive outcomes for drawing the second number. Similarly, there are 3 positive outcomes for choosing the third number. ccccccc First Number && Second Number && Third Number 5&*& 4&*& 3 By the Fundamental Counting Principle, the number of positive outcomes is the product of the numbers of positive outcomes for choosing each number. This means that there are 60 positive outcomes. The probability that all drawn numbers are even is the quotient of the number of positive outcomes and the number of possible outcomes. P(all numbers even) = 60/720 ⇕ P(all numbers even) = 1/12
Next, we will find the probability that the first two numbers drawn are both greater than 6. Let's find the number of positive outcomes! In the set, there are 4 numbers greater than 6. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} This means that there are 4 positive outcomes when choosing the first number. When choosing the second number, only 3 numbers greater than 6 remain, so there are 3 positive outcomes. There are no extra restrictions on the last number, so there are 8 positive outcomes of choosing the last number. ccccccc First Number && Second Number && Third Number 4&*& 3&*& 8 By the Fundamental Counting Principle, there are 4*3* 8, or 96, positive outcomes. We know from Part A that there are 720 possible outcomes. The probability that the first two numbers drawn are greater than 6 is the quotient of the number of positive outcomes and the number of possible outcomes. P(first two numbers greater than6) = 96/720 ⇕ P(first two numbers greater than6) = 2/15
Finally, we will find the probability that all three numbers drawn are divisible by 4. There are 2 numbers divisible by 4 in the given set. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} There are 2 positive outcomes when choosing the first number. After the first number is drawn, 1 number divisible by 4 remains in the set, so there is only 1 positive outcome for choosing the second number. There are no numbers divisible by 4 left when drawing the last number, so there are 0 positive outcomes for choosing the third number. ccccccc First Number && Second Number && Third Number 2&*& 1&*& 0 By the Fundamental Counting Principle, there are 0 positive outcomes! We know from Part A that there are 720 possible outcomes. The probability all the numbers drawn are divisible by 4 is the quotient of the number of positive outcomes to the number of possible outcomes. There are 0 positive outcomes, so the probability is 0. P(all numbers divisible by4) = 0
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The Fundamental Counting Principle
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