The Fundamental Counting Principle - Introduction to Probability and Statistics (Pre-Algebra)

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?

b What is the probability of getting a six and the spinner landing on the purple field? Write the answer as a fraction.

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 \cap B) = P (A) \cdot 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) = \frac{1}{3}

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) = \frac{1}{3}

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.

G G G B G O B B B G B O O O O G O B

Therefore, the combined probability of picking a green marble first and an orange marble second is

\frac{1}{9} .

Since the probability that both events occur is equal to the product of the individual probabilities, these events can be considered as independent events.

\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} ⇓ P (G) \cdot P (O) = P (G \cap 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 \cap B) = P (A) \cdot P (B ∣ A)

Why

For example, consider drawing two marbles from a bowl, one at a time.

Let

G,

B,

and

O

1

green marble and

3

marbles in total.

P (G) = \frac{1}{3}

Suppose that after the green marble is picked, it is not replaced in the bowl.

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 (O ∣ G) = \frac{1}{2}

Using this information, the sample space of the described situation can be found.

G B G O B G B O O G O B

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

\frac{1}{6} .

\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} ⇓ P (G) \cdot P (O ∣ G) = P (G \cap 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 \times m .

This principle is used to find the number of possible outcomes for a combination of independent events.

Proof

Informal Justification

Consider 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 \times 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 \times 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 Events

As 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 $2 \times 3 = 6 .$ 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?

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.

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.

A standard die has

6

sides, so there are

6

possible outcomes of rolling the die. There are

2

sides of a coin, so there are

2

possible outcomes of tossing the coin. To find the number of possible outcomes of rolling the die and then tossing the coin, consider 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 $n \cdot m .$

In our case, event

A

is rolling the die and event

B

is tossing the coin. The total number of different outcomes for rolling the die and tossing the coin is the product of the number of possible outcomes of rolling the die and the number of possible outcomes of tossing the coin.

6 \cdot 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.

The probability of winning a teddy bear is equal to the number of favorable outcomes divided by the number of possible outcomes. There is only

1

favorable outcome — rolling a

5

and getting tails. There are

12

possible outcomes, as shown in Part A.

P (Teddy bear) P (Teddy bear) = \frac{Favorable outcomes}{Possible outcomes} ⇓ = \frac{1}{1 2}

The probability of winning a teddy bear is

\frac{1}{1 2} .

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.

In this case, there are six possible permutations.

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 \cdot 2 \cdot 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!

Tiffaniqua behind the stand on which the five prizes are displayed

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?

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.

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

By the Fundamental Counting Principle, the total number of possible orders of the prizes is the product of the numbers of possible outcomes of each choice.

5 \cdot 4 \cdot 3 \cdot 2 \cdot 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

The Fundamental Counting Principle says that the number of possible orders of the second, fourth, and fifth prizes is the product of the numbers of options for each choice.

3 \cdot 2 \cdot 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) = \frac{6}{1 2 0} ⇕ P (Teddy bear first, CD third) = \frac{1}{2 0}

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

\frac{1}{2 0} .

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.

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.

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.

By the Fundamental Counting Principle, there are

400

possible outcomes of drawing a prize, putting it back in the box, and then drawing the prize again.

20 \cdot 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) = \frac{1}{4 0 0}

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.

This is because the prize that was drawn first is no longer in the box. By the Fundamental Counting Principle, there are

380

possible outcomes when drawing one prize and then another.

20 \cdot 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) = \frac{2}{3 8 0} ⇕ P (socks and gloves) = \frac{1}{1 9 0}

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 \cdot 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) = \frac{1}{2 4}

The probability of getting a six and the purple field when rolling a die and spinning the spinner is

\frac{1}{2 4} .

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Catch-Up and Review

Why

Why

Proof

Extra

Hint

Solution

Hint

Solution

Hint

Solution