sample space probability homework 2

Chapter Review

3.1 terminology.

In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.

3.2 Independent and Mutually Exclusive Events

Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent

In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.

3.3 Two Basic Rules of Probability

The multiplication rule and the addition rule are used for computing the probability of A and B , as well as the probability of A or B for two given events A , B defined on the sample space. In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common.

3.4 Contingency Tables

There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables, also known as two-way tables, help display data and are particularly useful when calculating probabilites that have multiple dependent variables.

3.5 Tree and Venn Diagrams

A tree diagram uses branches to show the different outcomes of experiments and makes complex probability questions easy to visualize.

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.

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Calcworkshop

Sample Space in Statistics Explained w/ 15 Examples!

// Last Updated: September 26, 2020 - Watch Video //

What is a sample space?

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

It’s a fundamental aspect of statistics and that’s what we’re going to discuss in today’s lesson.

So jump on in!

Law Of Large Numbers

Did you know that chance behavior is unpredictable in the short run but has a regular, predictable pattern in the long term?

For example, if you flip a coin ten times, the number of heads you get can vary. But if you flip a coin 100 times, the number of heads will be close to 50.

It’s called the Law of Large Numbers .

All this means is that the more repetitions that occur, the proportion of times that a specific event (i.e., head or tails) will occur approaches a single value. So, for our coin-flipping example, the percentage of heads we would expect to see if we flip the coin 100 times is approximately 0.5 of 50%.

Additionally, this means that probability , or the likelihood of something happening, is a number between 0 and 1 and describes the proportion of times that an outcome will occur over many repetitions (i.e., in the long run).

Sample Space And Events

And this leads us to sample spaces and events.

What is the sample space? And what is an event?

A sample space is the set of all possible outcomes of a statistical experiment, and it is sometimes referred to as a probability space. And outcomes are observations of the experiment, and they are sometimes referred to as sample points. An event is a subset of a sample space as discussed by Shafer and Zhang .

So, in our coin-flipping example, the probability space for flipping a coin one time is “Heads or Tails,” and we write this as S = {H,T}.

How To Find Sample Space?

The three most common ways to find a sample space are:

• To List All the Possible Outcomes.
• Create a Tree-Diagram.
• Use a Venn Diagram.

For example, let’s suppose we flip a coin and roll a die.

• How many outcomes are possible?
• What is the probability space?
• Identify the events.

When we flip a coin, there are only two possible outcomes {heads or tails}, and when we roll a die, there are six possible outcomes {1,2,3,4,5,6}.

That means we have two events:

• One event consists of “heads and tails.”
• The other event consists of {1,2,3,4,5,6}.

1. List Of All Possible Outcomes

Now, let’s see if we can find the sample space.

Example – Flipping A Coin And Rolling A Die

If we flip and roll, then we can get any of the following scenarios:

• Heads and 1
• Heads and 2
• Heads and 3
• Heads and 4
• Heads and 5
• Heads and 6
• Tails and 1
• Tails and 2
• Tails and 3
• Tails and 4
• Tails and 5
• Tails and 6

Or more simply stated in a sample space {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

What we’ve just done is to list all the possible outcomes of flipping a coin and rolling a die.

And if you notice, we were able to find 12 possible outcomes for this particular experiment.

2. Tree Diagram

As previously stated, this can also be represented in the form of a Tree diagram , as illustrated below. Each branch indicates a way of achieving the desired outcome and nicely demonstrates all the possible results clearly.

Sample Space Tree Diagram

3. Venn Diagram

Moreover, we could have illustrated the sample space by using a Venn Diagram as well. We would let one circle represent all the possible outcomes of flipping a coin {H, T}, and the other circle represent the six possible outcomes for rolling a die {1,2,3,4,5,6}. The overlap demonstrates the possibility for a single event to occur, such as H2 or T5.

Sample Space Venn Diagram

Fundamental Counting Principle

And this brings us to an important concept: The Counting Principle .

The counting principle, sometimes referred to as the counting rule or multiplication principle, is an easy way to figure out the number of possible outcomes (i.e., sample space). In other words, it is how we calculate the sample space.

All we have to do is multiply the events together to get the total number of outcomes. Using our example above, notice that flipping a coin has two possible results, and rolling a die has six possible outcomes. If we multiply them together, we get the total number of outcomes for the sample space: 2 x 6 = 12!

However, there are times when listing the probabilities is too difficult as the sample space is vast. When we encounter large sample spaces, with a large or infinite number of sample points, it is best to describe outcomes by statement or rule.

Finite Vs Infinite

But first, let’s notice the difference between finite sample spaces and infinite sample spaces .

A person plays the lottery, which means they will either win or lose. This is a finite space with two sample points. Or let’s say a survey is taken in a class ranking the course on a scale from 0 to 5. This is also a finite probability space, as there are six sample points. But what if you want to determine the set of all points inside a hot air balloon? Technically there is an infinite number of possible points inside the balloon; thus, making the sample space limitless. When this happens, we will use a rule or statement like x > 0 or “all positive real numbers.”

Complement Rule

And sometimes, we want to identify those elements that are not part of an event. This is called the complement rule .

So, for our coin-flipping example. If we flip a coin and want to count the number of heads, then the complement is the number of tails (i.e., NOT heads).

Here’s another example. Suppose we roll a fair die, so the sample space is S={1,2,3,4,5,6}, and we want to find all the even numbers. So, we define event A to be all the even numbers, which is A={2,4,6}. This means the complement of A would be all the elements that are NOT even, so we say A’={1,3,5}.

We denote the complement of an event as:

Complement In Probability

Union Vs Intersection

And sometimes we want to know the intersection or union of two events .

Venn Diagram Union And Intersection

Example – Rolling a Die

This is best illustrated with an example.

Using our previous example of rolling a fair die, which gives us a sample space of S={1,2,3,4,5,6}, let’s find the following two events: A={all the numbers less than 4} and B={all odd numbers}. This means that A={1,2,3} and B={1,3,5}.

So, what’s the intersection of A and B, and what is the union of A and B?

The intersection of A and B, are all the elements that they have in common, which would be {1,3}.

And the union of A and B are a complete list of elements that are in A or B or in both A and B. This means the union of A and B is {1,2,3,5}

Disjoint Events

But what happens if the two events have nothing in common?

This is called disjoint or mutually exclusive events . The two events have no outcomes in common and therefore, can never occur simultaneously.

For example, if we rolled a fair die with the sample space {1,2,3,4,5,6} and if event A represents the all even outcome (i.e., A={2,4,6}) and event B represents the odd outcomes (i.e., B={1,3,5}). What is the possibility that A and B happen at the same time?

Since A and B do not have any common numbers, we would say that these events are mutually exclusive (disjoint), and the likelihood that A and B can happen simultaneously is zero.

Together we will learn to identify sample spaces, events, and sample points, as well as determine the number of outcomes using the counting principle. We will learn how to create tree diagrams and Venn diagrams, identify intersections, unions, and disjoint events, and use the complement rule as they relate to sample spaces and probabilities.

Sample Space – Lesson & Examples (Video)

• Introduction to Video: Sample Space and Events
• 00:00:39 – Identifying sample points and sample spaces (Examples #1-3)
• Exclusive Content for Members Only
• 00:07:16 – Identifying statements and rules for large or infinite sample spaces (Examples #4-5)
• 00:14:07 – What is an event? How do we identify subspaces (Example #6)
• 00:18:27 – Identify the sample space and desired events (Examples #7-8)
• 00:25:33 – Finding complements of an event (Example #9)
• 00:28:31 – Identify the complement of each event (Examples #10-11)
• 00:31:19 – Overview of Intersection, Union and Disjoint Events and Venn Diagrams
• 00:38:06 – Identify the Intersection and Union or Shade the Venn Diagram (Examples #12-15)
• Practice Problems with Step-by-Step Solutions
• Chapter Tests with Video Solutions

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• Math Article

Sample Space

A sample space is a collection or a set of possible outcomes of a random experiment. The sample space is represented using the symbol, “S”. The subset of possible outcomes of an experiment is called events. A sample space may contain a number of outcomes that depends on the experiment. If it contains a finite number of outcomes, then it is known as discrete or finite sample spaces.

The samples spaces for a random experiment is written within curly braces “ { } “. There is a difference between the sample space and the events. For rolling a die, we will get the sample space, S as {1, 2, 3, 4, 5, 6 } whereas the event can be written as {1, 3, 5 } which represents the set of odd numbers and { 2, 4, 6 } which represents the set of even numbers. The outcomes of an experiment are random and the sample space becomes the universal set for some particular experiments. Some of the examples are as follows:

Tossing a Coin

When flipping a coin, two outcomes are possible, such as head and tail. Therefore the sample space for this experiment is given as

Sample Space, S = { H, T } = { Head, Tail }

Tossing Two Coins

When flipping two coins, the number of possible outcomes are four. Let, H 1 and T 1 be the head and tail of the first coin and H 2 and T 2 be the head and tail of the second coin respectively and the sample space can be written as

Sample Space, S = { (H 1 , H 2 ), (H 1 , T 2 ), (T 1 , H 2 ), (T 1 , T 2 ) }

In general, if you have “n” coins, then the possible number of outcomes will be 2 n .

Example: If you toss 3 coins, “n” is taken as 3.

Therefore, the possible number of outcomes will be 2 3 = 8 outcomes

Sample space for tossing three coins is written as

Sample space S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

A Die is Thrown

When a single die is thrown, it has 6 outcomes since it has 6 faces. Therefore, the sample is given as

S = { 1, 2, 3, 4, 5, 6}

Two Dice are Thrown

When two dice are thrown together, we will get 36 pairs of possible outcomes. Each face of the first die can fall with all the six faces of the second die. As there are 6 x 6 possible pairs, it becomes 36 outcomes. The 36 outcome pairs are written as:

If three dice are thrown, it should have the possible outcomes of 216 where n in the experiment is taken as 3, so it becomes 6 3 = 216.

Sample Problem

Write the sample space for the given interval [3,9].

As the integers given are in the closed interval, we can take the value from 3 to 9.

Therefore, the sample space for the given interval is:

Sample space = { 3, 4, 5, 6, 7, 8, 9 }

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Probability Worksheets

Navigate through this assortment of printable probability worksheets that includes exercises on basic probability based on more likely, less likely, equally likely, certain and impossible events, pdf worksheets based on identifying suitable events, simple spinner problems, for students in grade 4, grade 5, and grade 6. With the required introduction, the beginners get to further their knowledge with skills like probability on single coin, two coins, days in a week, months in a year, fair die, pair of dice, deck of cards, numbers and more. Mutually exclusive and inclusive events, probability on odds and other challenging probability worksheets are useful for grade 7, grade 8, and high school. Grab some of these probability worksheets for free!

Probability on Coins

Simple probability worksheets based on tossing single coin or two coins. Identify the proper sample space before finding probability.

Probability in a single coin toss

Probability in pair of coin - 1

Probability in pair of coin - 2

Probability on Days and Months

Fun filled worksheet pdfs based on days in a week and months in a year. Sample space is easy to find but care is required in identifying like events.

Days of a week

Months of a year - 1

Months of a year - 2

Probability on Fair Die

Fair die is numbered from 1 to 6. Understand the multiples, divisors and factors and apply it on these probability worksheets.

Simple numbers

Multiples and divisors

Mutually exclusive and inclusive

Probability on Pair of Dice

Sample space is little large which contains 36 elements. Write all of them in papers before start answering on probability questions for grade 7 and grade 8.

Based on numbers

Based on sum and difference

Based on multiples and divisors

Based on factors

Probability on Numbers

Students should learn the concepts of multiples, divisors and factors before start practicing these printable worksheets.

Probability on numbers - 1

Probability on numbers - 2

Probability on numbers - 3

Probability on numbers - 4

Probability on numbers - 5

Probability on Deck of Cards

Deck of cards contain 52 cards, 26 are black, 26 are red, four different flowers, each flower contain 13 cards such as A, 1, 2, ..., 10, J, Q, K.

Deck of cards worksheet - 1

Deck of cards worksheet - 2

Deck of cards worksheet - 3

Probability on Spinners

Interactive worksheets for 4th grade and 5th grade kids to understand the probability using spinners. Colorful spinners are included for more fun.

Spinner worksheets on numbers

Spinner worksheets on colors

Probability on Odds

Probability on odds worksheets can be broadly classifieds as favorable to the events or against the events.

Odds worksheet - 1

Odds worksheet - 2

Odds worksheet - 3

Probability on Independent and Dependent

Here comes our challenging probability worksheets set for 8th grade and high school students based on dependent and independent events with various real-life applications.

Based on deck of cards

Based on marbles

Based on cards

Probability on Different Events

Basic probability worksheets for beginners in 6th grade and 7th grade to understand the different type of events such as more likely, less likely, equally likely and so on.

Balls in container

Identify suitable events

Mutually inclusive and exclusive events

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8.1: Sample Spaces and Probability

• Last updated
• Save as PDF
• Page ID 37906

• Rupinder Sekhon and Roberta Bloom
• De Anza College

Learning Objectives

In this section, you will learn to:

• Write sample spaces.
• Calculate probabilities by examining simple events in sample spaces.

If two coins are tossed, what is the probability that both coins will fall heads? The problem seems simple enough, but it is not uncommon to hear the incorrect answer 1/3. A student may incorrectly reason that if two coins are tossed there are three possibilities, one head, two heads, or no heads. Therefore, the probability of two heads is one out of three. The answer is wrong because if we toss two coins there are four possibilities and not three. For clarity, assume that one coin is a penny and the other a nickel. Then we have the following four possibilities.

HH HT TH TT

The possibility HT, for example, indicates a head on the penny and a tail on the nickel, while TH represents a tail on the penny and a head on the nickel. It is for this reason, we emphasize the need for understanding sample spaces.

Sample Spaces

An act of flipping coins, rolling dice, drawing cards, or surveying people are referred to as a probability experiment . A sample space of an experiment is the set of all possible outcomes.

Example \(\PageIndex{1}\)

If a die is rolled, write a sample space.

A die has six faces each having an equally likely chance of appearing. Therefore, the set of all possible outcomes \(S\) is

{ 1, 2, 3, 4, 5, 6 }.

Example \(\PageIndex{2}\)

A family has three children. Write a sample space.

The sample space consists of eight possibilities.

{ BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG }

The possibility BGB, for example, indicates that the first born is a boy, the second born a girl, and the third a boy.

We illustrate these possibilities with a tree diagram.

Example \(\PageIndex{3}\)

Two dice are rolled. Write the sample space.

We assume one of the dice is red, and the other green. We have the following 36 possibilities.

The entry (2, 5), for example, indicates that the red die shows a 2, and the green a 5.

Probability

Now that we understand the concept of a sample space, we will define probability.

Definition: Probability

For a sample space \(S\), and an outcome \(A\) of \(S\), the following two properties are satisfied.

• If \(A\) is an outcome of a sample space, then the probability of \(A\), denoted by \(P(A)\), is between 0 and 1, inclusive. \[0 ≤ P(A) ≤ 1 \nonumber \]
• The sum of the probabilities of all the outcomes in \(S\) equals 1.

The probability \(P(A)\) of an event \(A\) describes the chance or likelihood of that event occurring.

• If \(P(A) = 0\), event A is certain not to occur. If \(P(A) = 1\), event \(A\) is certain to occur.
• If \(P(A) = 0.5\), then event A is equally likely to occur or not occur.
• If we toss a fair coin that is equally likely to land on heads or tails, then P(Head) = 0.50.
• If the weather forecast says there is a 70% chance of rain today, then P(Rain) = 0.70, indicating is it more likely to rain than to not rain.

Example \(\PageIndex{4}\)

If two dice, one red and one green, are rolled, find the probability that the red die shows a 3 and the green shows a six.

Since two dice are rolled, there are 36 possibilities. The probability of each outcome, listed in Example \(\PageIndex{3}\), is equally likely.

Since (3, 6) is one such outcome, the probability of obtaining (3, 6) is 1/36.

The example we just considered consisted of only one outcome of the sample space. We are often interested in finding probabilities of several outcomes represented by an event.

An event is a subset of a sample space. If an event consists of only one outcome, it is called a simple event.

Example \(\PageIndex{5}\)

If two dice are rolled, find the probability that the sum of the faces of the dice is 7.

Let E represent the event that the sum of the faces of two dice is 7.

The possible cases for the sum to be equal to 7 are: (1, 6), (2,5), (3, 4), (4, 3), (5, 2), and (6, 1), so event E is

E = {(1, 6), (2,5), (3, 4), (4, 3), (5, 2), (6, 1)}

The probability of the event E is

P(E) = 6/36 or 1/6.

Example \(\PageIndex{6}\)

A jar contains 3 red, 4 white, and 3 blue marbles. If a marble is chosen at random, what is the probability that the marble is a red marble or a blue marble?

We assume the marbles are \(r_1\), \(r_2\), \(r_3\), \(w_1\), \(w_2\), \(w_3\), \(w_4\), \(b_1\), \(b_2\), \(b_3\). Let the event \(\mathrm{C}\) represent that the marble is red or blue.

The sample space \(\mathrm{S}=\left\{\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}, \mathrm{w}_{1}, \mathrm{w}_{2}, \mathrm{w}_{3}, \mathrm{w}_{4}, \mathrm{b}_{1}, \mathrm{b}_{2}, \mathrm{b}_{3}\right\} \).

And the event \(\mathrm{C}=\left\{\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}, \mathrm{b}_{1}, \mathrm{b}_{2}, \mathrm{b}_{3}\right\}\)

Therefore, the probability of \(\mathrm{C}\),

\[\mathrm{P}(\mathrm{C})=6 / 10 \text { or } 3 / 5 \nonumber \]

Example \(\PageIndex{7}\)

A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn without replacement , what is the probability that the sum of the numbers is 5?

Note: The two marbles in this example are drawn consecutively without replacement . That means that after a marble is drawn it is not replaced in the jar, and therefore is no longer available to select on the second draw.

Since two marbles are drawn without replacement, the sample space consists of the following six possibilities.

\[\mathrm{S}=\{(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)\} \nonumber \]

Note that (1,1), (2,2) and (3,3) are not listed in the sample space. These outcomes are not possible when drawing without replacement, because once the first marble is drawn but not replaced into the jar, that marble is not available in the jar to be selected again on the second draw.

Let the event \(\mathrm{E}\) represent that the sum of the numbers is five. Then

\[\mathrm{E}=\{(2,3),(3,2)\} \nonumber \]

Therefore, the probability of \(\mathrm{E}\) is

\[\mathrm{P}(\mathrm{E})=2 / 6 \text { or } 1 / 3 \nonumber. \nonumber \]

Example \(\PageIndex{8}\)

A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn without replacement , what is the probability that the sum of the numbers is at least 4?

The sample space, as in Example \(\PageIndex{7}\), consists of the following six possibilities.

Let the event \(\mathrm{F}\) represent that the sum of the numbers is at least four. Then

\[\mathrm{F}=\{(1,3),(3,1),(2,3),(3,2)\} \nonumber \]

Therefore, the probability of \(\mathrm{F}\) is

\[\mathrm{P}(\mathrm{F})=4 / 6 \text { or } 2 / 3 \nonumber \]

Example \(\PageIndex{9}\)

A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn with replacement , what is the probability that the sum of the numbers is 5?

Note: The two marbles in this example are drawn consecutively with replacement . That means that after a marble is drawn it IS replaced in the jar, and therefore is available to select again on the second draw.

When two marbles are drawn with replacement, the sample space consists of the following nine possibilities.

\[\mathrm{S}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\} \nonumber \]

Note that (1,1), (2,2) and (3,3) are listed in the sample space. These outcomes are possible when drawing with replacement, because once the first marble is drawn and replaced, that marble is not available in the jar to be drawn again.

Let the event E represent that the sum of the numbers is four. Then

\[ \mathrm{E} = {(2, 3), (3, 2) } \nonumber \]

Therefore, the probability of \(\mathrm{F}\) is \(\mathrm{P}(\mathrm{E}) = 2/9\)

Note that in Example \(\PageIndex{9}\) when we selected marbles with replacement, the probability has changed from Example \(\PageIndex{7}\) where we selected marbles without replacement.

Example \(\PageIndex{10}\)

A jar contains three marbles numbered 1, 2, and 3. If two marbles are drawn with replacement , what is the probability that the sum of the numbers is at least 4?

The sample space when drawing with replacement consists of the following nine possibilities.

\[ \mathrm{S} = {(1,1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3,3)} \nonumber \]

\[ \mathrm{F} = {(1, 3), (3, 1), (2, 3), (3, 2), (2,2), (3,3)} \nonumber \]

\[\mathrm{P}(\mathrm{F}) = 6/9 \text{ or } 2/3 \nonumber. \nonumber \]

Note that in Example \(\PageIndex{10}\) when we selected marbles with replacement, the probability is the same as in Example \(\PageIndex{8}\) where we selected marbles without replacement.

Thus sampling with or without replacement MAY change the probabilities, but may not, depending on the situation in the particular problem under consideration. We’ll re-examine the concepts of sampling with and without replacement in Section 8.3.

Example \(\PageIndex{11}\)

One 6 sided die is rolled once. Find the probability that the result is greater than 4.

The sample space consists of the following six possibilities in set \(\mathrm{S}\): \(\mathrm{S}={1,2,3,4,5,6}\)

Let \(\mathrm{E}\) be the event that the number rolled is greater than four: \(\mathrm{E}={5,6} \)

Therefore, the probability of \(\mathrm{E}\) is: \(\mathrm{P}(\mathrm{E}) = 2/6 \text{ or } 1/3\).

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Into Math Grade 7 Module 15 Lesson 2 Answer Key Find Theoretical Probability of Compound Events

We included  HMH Into Math Grade 7 Answer Key PDF   Module 15 Lesson 2 Find Theoretical Probability of Compound Events to make students experts in learning maths.

HMH Into Math Grade 7 Module 15 Lesson 2 Answer Key Find Theoretical Probability of Compound Events

I Can find the theoretical probability of a compound event.

Step It Out

Explanation: Lucas made a table to represent the sample space of sums. Each entry in the table is the sum of the numbers in the corresponding row and column. Completed the remainder of the table as shown above.

B. The table in Part A shows:

• There are _________ possible outcomes with a sum of 3.
• There are __________ equally likely outcomes in the sample space.

Answer: There are 2 possible outcomes with a sum of 3, There are 5 equally likely outcomes in the sample space,

Explanation: Using the data of table in Part A there are 2 possible outcomes with a sum of 3 one is 1 + 2 = 3 and other is 2 + 1 = 3. There are 5 equally likely outcomes in the same sample space they are 1. 3, 3, 2. 4, 4 , 4, 3. 5, 5, 5, 5,  4. 6, 6, 6, 5. 7, 7 respectively.

Explanation: The theoretical probability of spinning a sum of 3 is p(sum of 3) = number of equally likely outcomes that add to 3/ total number of equally likely outcomes in the sample space = 2/16 = 1/8.

D. Determine another possible outcome in the sample space that has the same probability as that of spinning a sum of 3. P (sum of 3) = P (___________) Answer: P (sum of 3) = 1/8,

Explanation: Another possible outcome in the sample space that has the same probability as that of spinning a sum of 3 is number of equally likely outcomes that add to 3/ total number of equally likely outcomes in the sample space = 2/16 = 1/8.

Explanation: Completed the tree diagram to represent the sample space as shown above.

B. The tree diagram shows that there are _________ equally likely outcomes in the sample space. Answer: No, equally likely outcomes in the sample space,

Explanation: The tree diagram shows that there are no equally likely outcomes in the sample space.

Explanation: The probability P(white, blue, tan) = number of equally likely outcomes in the event(white, blue, tan)/ total number of equally likely outcomes in the sample space = 1/3.

Explanation: The probability of selecting white shorts and a blue shirt when selecting clothes at random is number of equally likely outcomes in the event(white, blue)/ total number of equally likely outcomes in the sample space = 1/3.

Explanation: Given Ali draws cards from a hat that contains three cards labeled Winning Combination A, B, and C. He has an equal chance of drawing each card. He draws one card at a time, records the result, and replaces it before the next draw. Ali wins a prize for drawing the A-card 3 times in a row. Completed the organized lists to represent the sample space as shown above.

B. The organized lists show that there are ____________ equally likely outcomes in the sample space. Answer: The organized lists show that there are no equally likely outcomes in the sample space.

Explanation: The probability P(A, A, A) is number of equally likely outcomes in the event(A,A,A)/ Tootal number of equally likely outcomes in the sample space = 1/27 or 0.0370 or 3.7% or 4%.

Turn and Talk How do you think you could use the organized lists to find the probability of drawing 2 or more A-cards in any order? Answer: By checking the organized list and by counting the probability of drawing 2 or more A-cards in any order,

Explanation: Simply by checking the organized list and by counting the probability of drawing 2 or more A-cards in any order we get 1. A, A, A, 2. A, A, B, 3. A, A, C, 4. B, A, A, 5. C, A, A.

Check Understanding

Question 1. What are some ways that you can represent the sample space of a compound event? Answer: The three most common ways to find a sample space are: To list all the possible outcomes, Create a Tree- Diagram, Use a Venn diagram,

Explanation: If an event has more than one possible outcome, the event is called a compound event. A sample space is the set of all possible outcomes for an experiment. When determining a sample space, we must be careful to include all possibilities. Therefore the three most common ways to find a sample space are: To list all the possible outcomes, Create a Tree- Diagram, Use a Venn diagram.

Question 2. Mila spins two spinners. One spinner has two equal sections labeled 1 and 2. The other spinner has three equal sections labeled 1, 2, and 3. What is the probability that the sum of the two spins is 4? Answer: The probability that the sum of the two spins is 4 is 2,

Explanation: Given Mila spins two spinners. One spinner has two equal sections labeled 1 and 2. The other spinner has three equal sections labeled 1, 2, and 3. The probability that the sum of the two spins is 4 is one spinner    – 1, 2, 1 other spinner – 1, 2, 3                       (+)  4, 4 is 2.

On Your Own

Explanation: Completed the table to represent the sample space for the sum of these two number prisms as shown above.

B. What is the probability that Wyatt rolls a sum of 3? Answer: The probability that Wyatt rolls a sum of 3 is 2,

Explanation: The probability that Wyatt rolls a sum of 3 is 1 + 2 = 3 and 2 + 1 = 3 so it is 2.

C. What event has the same probability as rolling a sum of 3? Explain. Answer: Event 1 + 2 = 3,

Explanation: Event 1 + 2 = 3 has the same probability as rolling a sum of 3.

D. Name an event that is less likely than a sum of 3. Explain. Answer: Event 1 + 1 = 2,

Explanation: Event 1 + 1 = 2 has less likely than a sum of 3.

E. Name an event that is more likely than a sum of 3. Explain. Answer: Event 1 + 3, Event 2 + 2, Event 2 + 3, Event 3 + 1, Event 3 + 2, Event 3 + 3,

Explanation: Event 1 + 3 = 4, Event 2 + 2 = 4, Event 2 + 3 = 5, Event 3 + 1 = 4, Event 3 + 2 = 5, Event 3 +3 = 6 are more likely than a sum of 3.

Model with Mathematics Find each probability. Give your answer as a simplified fraction, a decimal to the nearest hundredth, and a percent to the nearest whole percent.

F. P(sum of 4) Answer: 3,

Explanation: The P(sum of 4) is Event 1 + 3 = 4, Event 2 + 2 = 4 and Event 3 + 1 = 4.

G. P(sum of 6) Answer: Event 3 + 3,

Explanation: The P(sum of 6) is Event 3 + 3 = 6.

H. P(sum < 10) Answer: Event 1 + 1, Event  1 + 2, Event 1 + 3, Event 2 + 1, Event 2 + 2, Event 2 + 3, Event 3 + 1, Event 3 + 2, Event 3 + 3,

Explanation: P(sum < 10) all these events have probability of sum less than 10 Event 1 + 1 = 2, Event  1 + 2 = 3, Event 1 + 3 = 4, Event 2 + 1 = 3, Event 2 + 2 = 4, Event 2 + 3 = 5, Event 3 + 1 = 4, Event 3 + 2 = 5, Event 3 + 3 = 6.

I. P(sum > 10) Answer: Zero,

Explanation: There is no event where the probability of sum is greater than 10.

Explanation: Completed the tree diagram to show the sample space for how the coins can land. H represent heads and T represents tails above.

B. What is the probability that exactly two coins land heads up? Write the probability as a fraction, decimal, and percent. Answer: 2/8, 0.25, 25%,

Explanation: The probability that exactly two coins land heads up is 1/8 and 1/8 so the probability as a fraction is 1/8 + 1/8 = 2/8 = 1/4 as decimal is 0.25 and percent is 0.25 X 100 = 25%.

C. What is the probability that exactly two coins land tails up? Compare this probability to the probability that exactly two coins land heads up. Answer: 3/8 and 2/8 probability of tails up is more than heads up,

Explanation: The probability that exactly two coins land tails up is 1/8 + 2/8 = 3/8, now comparing this probability to the probability that exactly two coins land heads up is 3/8 > 2/8.

D. What is the probability that exactly three coins land tails up? Write the probability as a fraction, decimal, and percent. Answer: 1/8, 0.125, 12.5,

Explanation: The probability that exactly three coins land tails up is 1/8, the probability as a fraction is 1/8 decimal, is 0.125 and percent is 12.5.

E. Name an event with the same probability as that of three tails? Explain. Answer: Event (T, T, T),

Explanation: An event with the same probability as that of three tails is event (T, T, T).

F. Compare the theoretical probability of getting at least one head and the theoretical probability of getting at least two tails. Answer: The theoretical probability of getting at least one head is less than the theoretical probability of getting at least two tails,

Explanation: The theoretical probability of getting at least one head is 1/8 and the theoretical probability of getting at least two tails is 3/8 now comparing the theoretical probability of getting at least one head is less than the theoretical probability of getting at least two tails is 1/8 < 3/8.

Explanation: Made an organized list to show the sample space for spinning the spinner and rolling the number cube above.

B. What is the probability that Daniel gets a sum of 5? Answer: 3/21 or 1/7,

Explanation: The probability that Daniel gets a sum of 5 is 1 + 4, 2 + 3, 3 + 2 so it is 3/21 or 1/7.

C. What is the probability that Daniel gets a sum of 3? Answer: 2/21,

Explanation: The probability that Daniel gets a sum of 3 is 1 + 2, 2 + 1 is 2/21.

D. Compare the probability of getting a sum of 5 and getting a sum of 3. Answer: The probability of getting a sum of 5 is more likely than getting a sum of 3,

Explanation: The probability of getting a sum of 5 and getting a sum of 3 is the probability that Daniel gets a sum of 5 is 1 + 4, 2 + 3, 3 + 2 so it is 3/21 and the probability that Daniel gets a sum of 3 is 1 + 2, 2 + 1 is 2/21 so comparing 3/21 is more likely than 2/21 means sum of 5 is more likely than getting a sum of 3.

E. Open-Ended Give an example of an event that is more likely than getting a sum of 8. Explain your reasoning. Answer: Event 2 + 6 or Event 3 + 5,

Explanation: An event that is more likely than getting a sum of 8 is event 2 + 6 or event 3 + 5.

F. Open-Ended Give an example of an event that is as equally likely as getting a sum of 6. Explain your reasoning. Answer: Event 1 + 5 or Event 2 + 4 or Event 3 + 3,

Explanation: Event 1 + 5 or Event 2 + 4 or Event 3 + 3 are examples of an event that is as equally likely as getting a sum of 6.

Question 6. Reason How is finding the theoretical probability of a compound event similar to finding the theoretical probability of a simple event? Answer: Compound Probability describes the chances of more than one separate event occurring, for example flipping heads on a coin and pulling a 7 from a standard deck of cards, Simple probability expresses the probability of one event occurring and is often visually expressed using coins, dice, marbles or spinner. Number of favorable outcomes over the number of total outcomes.

Explanation: Compound event consists of two or more simple events. To find the probability of a compound event, we write a ratio of the number of ways the compound event can happen to the total number of equally likely possible outcomes. The probability of simple events is finding the probability of a single event occurring. When finding the probability of an event occurring, we will use the formula: number of favorable outcomes over the number of total outcomes. After calculating this probability.

Lesson 15.2 More Practice/Homework

Explanation: Made a table to represent the sample space for drawing a particular sum from the numbers in the two boxes as shown above.

B. What is the probability of drawing a sum of 7? Answer: 2/15,

Explanation: The probability of drawing a sum of 7 is 2/15.

C. Name an event that is more likely than drawing a sum of 7. Answer: Event 2 + 5 or Event 3 + 4,

Explanation: An event that is more likely than drawing a sum of 7 is Event 2 + 5 or Event 3 + 4.

Explanation: Made a tree diagram to find the Sandwich Side sample space as sown above.

B. What is the probability Lucy picks a ham sandwich and grapes? Answer: 1/2,

Explanation: The probability Lucy picks a ham sandwich and grapes is out of two sides it is one so it is 1/2.

C. What is the probability Lucy chooses milk with a turkey sandwich and apples? Answer: 1/8,

Explanation: The probability Lucy chooses milk with a turkey sandwich and apples is 1/8.

D. Reason Name an event that is equally as likely as choosing ham and apples. Answer: Event (ham, apples, milk) or Event (ham, apples, water),

Explanation: An event that is equally as likely as choosing ham and apples is Event (ham, apples, milk) or Event (ham, apples, water).

Explanation: Different possible outcomes are there for the experiment described are step 1 : tossing a number cube it is 1/6, step 2 : tossing a fair coin it is 1/2, step 3 : spinning the spinner it is 1/4 .

Explanation: Given Emma is going to flip a coin, resulting in H for heads or T for tails, and spin the spinner shown. the list correctly represents the sample space is (C) H-1, H-2, H-3, H-4, H-5, H-6, T-1, T-2, T-3, T-4, T-5, T-6.

Explanation: Given Luna rolls two number cubes labeled 1-6. Matching the probability to the correct event as P(sum of 5) is 4/36 = 1/9, P(sum less than 4) is 1/12, P(sum of 7) is 1/6, P(sum greater than 8) is 5/18 above.

Question 6. Carson can choose one vehicle (car or truck), one color (blue, red, or silver), and one type of transmission (standard or automatic). How many possible outcomes are there? Answer: 1/12,

Explanation: Given Carson can choose one vehicle (car or truck), one color (blue, red, or silver), and one type of transmission (standard or automatic). Number of possible outcomes are there are 1. car, blue, standard, 2. car, blue, automatic, 3. car, red, standard, 4. car, red, automatic, 5. car, silver, standard, 6. car, silver, automatic, 7. truck, blue, standard, 8. truck, blue, automatic, 9. truck, red, standard, 10. truck, red, automatic, 11. truck, silver, standard, 12. truck, silver, automatic.

Spiral Review

Explanation: Given Aaliyah spins the spinner shown. The theoretical probability that the spinner will land on an even number is out of 9 we have even numbers are 2,4,6,8 so it is 4/9.

Explanation: Given a relationship we check it is proportional or not as 1. 120/ 3 = 40, 2. 200/5 = 40, 3. 320/8 = 40, 4. 400/10 = 40 so it is clear that for 1 minute words typed are 40 therefore ii is proportional and the unit rate is 40.

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A) Make a table to display the sample space of the compound events representing the players making moves during the game. Make the columns correspond to the possible rolls of the dice and the rows correspond to the possible spinner numbers. For each cell entry, write the sum of the row value and the column value. B) Find the probability of getting a sum of 4, 5, or 6. C) Find the probability of the sum being a 7, 8, or 9 and at least one of the two

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Answer by Noah · Jun 7, 2023