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# 1.2: Independent Events and Sample Spaces

Difficulty Level: Basic Created by: CK-12
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What is the probability that it will rain today? What is the probability that the school cafeteria will serve pizza for lunch? Are these probabilities dependent or independent of each other?

### Watch This

First watch this video to learn about independent events.

Then watch this video to see some examples.

Watch this video for more help.

### Guidance

What’s in a Word?

The words dependent and independent are used by students and teachers on a daily basis. In fact, they are probably used quite frequently. You may tell your parent or guardian that you are independent enough to go to the movies on your own with your friends. You could say that when you bake a cake or make a cup of hot chocolate, the taste of these are dependent on what ingredients you use. In the English language, the term dependent means to be unable to do without, whereas independent means to be free from any outside influence.

What about in mathematics? What do the terms dependent and independent actually mean? This concept will explore the mathematics of independence and dependence.

Independent Events

In mathematics, the term independent means to have one event not dependent on the other. It is similar to the English definition. Suppose you are trying to convince your parent/guardian to let you go to the movies on your own. Your parent/guardian is thinking that if you go, you will not have time to finish your homework. For this reason, you have to convince him/her that you are independent enough to go to the movies and finish your homework. Therefore, you are trying to convince your parent/guardian that the 2 events, going to the movies and finishing your homework, are independent events. This is similar to the mathematical definition. Say you were asked to pick a particular card from a deck of cards and roll a 6 on a die. It does not matter if you choose the card first and roll a 6 second, or vice versa. The probability of rolling the 6 would remain the same, as would the probability of choosing the card.

#### Example A

In ABC\begin{align*}ABC\end{align*} High School, 30 percent of the students have a part-time job, and 25 percent of the students from the high school are on the honor roll. Event A\begin{align*}A\end{align*} represents randomly choosing a student holding a part-time job. Event B\begin{align*}B\end{align*} represents randomly choosing a student on the honor roll. What is the probability of both events occurring?

Event A\begin{align*}A\end{align*} is randomly choosing a student holding a part-time job, and event B\begin{align*}B\end{align*} is randomly choosing a student on the honor roll. These 2 events are independent of each other. In other words, whether you hold a part-time job is not dependent on your being on the honor roll, or vice versa. The outcome of one event is not dependent on the outcome of the second event. To calculate the probability, you would look at the overlapping part of the Venn diagram. The region representing A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} is the probability of both events occurring. Let’s look at the probability calculation, which is done with the Multiplication Rule:

P(A)P(B)P(A and B)P(A and B)P(A and B)=30% or 0.30=25% or 0.25=P(A)×P(B)=0.30×0.25=0.075\begin{align*}P(A) &= 30\% \ \text{or} \ 0.30\\ P(B) &= 25\% \ \text{or} \ 0.25\\ P(A \ \text{and} \ B) &= P(A) \times P(B)\\ P(A \ \text{and} \ B) &= 0.30 \times 0.25\\ P(A \ \text{and} \ B) &= 0.075\end{align*}

In other words, 7.5% of the students of ABC\begin{align*}ABC\end{align*} high school are both on the honor roll and have a part-time job.

#### Example B

2 coins are tossed one after the other. Event A\begin{align*}A\end{align*} consists of the outcomes when tossing heads on the first toss. Event B\begin{align*}B\end{align*} consists of the outcomes when tossing heads on the second toss. What is the probability of both events occurring?

Event A\begin{align*}A\end{align*} consists of the outcomes when getting heads on the first toss, and event B\begin{align*}B\end{align*} consists of the outcomes when getting heads on the second toss. What would be the probability of tossing the coins and getting a head on both the first coin and the second coin? We know that the probability of getting a head on a coin toss is 12\begin{align*}\frac{1}{2}\end{align*}, or 50%. In other words, we have a 50% chance of getting a head on a toss of a fair coin and a 50% chance of getting a tail.

P(A)P(B)P(A and B)P(A and B)P(A and B)=50% or 0.50=50% or 0.50=P(A)×P(B)=0.50×0.50=0.25\begin{align*}P(A) &= 50\% \ \text{or} \ 0.50\\ P(B) &= 50\% \ \text{or} \ 0.50\\ \\ P(A \ \text{and} \ B) &= P(A) \times P(B)\\ P(A \ \text{and} \ B) &= 0.50 \times 0.50\\ P(A \ \text{and} \ B) &= 0.25\end{align*}

Therefore, there is a 25% chance of getting 2 heads when tossing 2 fair coins.

#### Example C

The following table represents data collected from a grade 12 class in DEF High School.

Plans after High School
Gender University Community College Total
Males 28 56 84
Females 43 37 80
Total 71 93 164

Suppose 1 student was chosen at random from the grade 12 class.

(a) What is the probability that the student is female?

(b) What is the probability that the student is going to university?

Now suppose 2 people both randomly chose 1 student from the grade 12 class. Assume that it's possible for them to choose the same student.

(c) What is the probability that the first person chooses a student who is female and the second person chooses a student who is going to university?

Probabilities: P(female)P(female)P(going to university)P(female)×P(going to university)=80164164 \ \text{total students}=2041=71164=2041×71164=14206724=3551681=0.211\begin{align*}\text{Probabilities:} \ P(\text{female}) &= \frac{80}{164} \swarrow \fbox{164 \ \text{total students}}\\ P(\text{female}) &= \frac{20}{41}\\ P(\text{going to university}) &= \frac{71}{164}\\ \\ P(\text{female}) \times P(\text{going to university})&= \frac{20}{41} \times \frac{71}{164}\\ &= \frac{1420}{6724}\\ &= \frac{355}{1681}\\ &= 0.211\end{align*}

Therefore, there is a 21.1% probability that the first person chooses a student who is female and the second person chooses a student who is going to university.

### Vocabulary

2 or more events whose outcomes do not affect each other are called independent events. The Multiplication Rule states that for 2 events (A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}), the probability of A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} is given by: P(A and B)=P(A)×P(B)\begin{align*}P(A \ \text{and} \ B) = P(A) \times P(B)\end{align*}. When everyone or everything in a population has an equal chance of being selected, the selection can be said to occur at random.

### Guided Practice

2 cards are chosen from a deck of cards. The first card is replaced before choosing the second card. What is the probability that they both will be sevens?

Let A=1st\begin{align*}A = 1^{\text{st}}\end{align*} seven chosen.

Let B=2nd\begin{align*}B = 2^{\text{nd}}\end{align*} seven chosen.

A little note about a deck of cards

A deck of cards consists of 52 cards.

Each deck has 4 parts (suits) with 13 cards in them.

Each suit has 3 face cards.

The total number of sevens in the deck4 suits1 seven per suit =4×1=4.\begin{align*}& 4 \ \text{suits} \qquad 1 \ \text{seven} \ \text{per suit}\\ & \ \searrow \qquad \swarrow\\ \text{The total number of sevens in the deck} &= 4 \times 1=4.\end{align*}

Since the card was replaced, these events are independent:

P(A)P(B)P(A and B)P(AB)P(AB)=452  Note: The total number of cards is=452  52 after choosing the first card,  because the first card is replaced.=452×452 or P(AB)=452×452=162704=1169\begin{align*}P(A) &= \frac{4}{52}\\ \\ & \qquad \qquad \ \ \text{Note: The total number of cards is}\\ P(B) &= \frac{4}{52} \ \swarrow \ \text{52 after choosing the first card,}\\ & \qquad \qquad \ \ \text{because the first card is replaced.}\\ \\ P(A \ \text{and} \ B) &= \frac{4}{52} \times \frac{4}{52} \ \text{or} \ P(A \cap B)=\frac{4}{52} \times \frac{4}{52}\\ \\ P(A \cap B) &= \frac{16}{2704}\\ \\ P(A \cap B) &= \frac{1}{169}\end{align*}

### Practice

1. Determine which of the following are examples of independent events.
1. Rolling a 5 on one die and rolling a 5 on a second die.
2. Choosing a cookie from the cookie jar and choosing a jack from a deck of cards.
3. Winning a hockey game and scoring a goal.
2. Determine which of the following are examples of independent events.
1. Choosing an 8 from a deck of cards, replacing it, and choosing a face card.
2. Going to the beach and bringing an umbrella.
3. Getting gasoline for your car and getting diesel fuel for your car.
3. A coin and a die are tossed. Calculate the probability of getting tails and a 5.
4. In Tania's homeroom class, 9% of the students were born in March and 40% of the students have a blood type of O+. What is the probability of a student chosen at random from Tania's homeroom class being born in March and having a blood type of O+?
5. If a baseball player gets a hit in 31% of his at-bats, what it the probability that the baseball player will get a hit in 5 at-bats in a row?
6. What is the probability of tossing 2 coins one after the other and getting 1 head and 1 tail?
7. 2 cards are chosen from a deck of cards. The first card is replaced before choosing the second card. What is the probability that they both will be clubs?
8. 2 cards are chosen from a deck of cards. The first card is replaced before choosing the second card. What is the probability that they both will be face cards?
9. If the probability of receiving at least 1 piece of mail on any particular day is 22%, what is the probability of not receiving any mail for 3 days in a row?
10. Johnathan is rolling 2 dice and needs to roll an 11 to win the game he is playing. What is the probability that Johnathan wins the game?

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English Spanish

Independent Events

Two events are independent if the occurrence of one event does not impact the probability of the other event.

Multiplication Rule

States that for 2 events (A and B), the probability of A and B is given by: P(A and B) = P(A) x P(B).

Random

When everyone or everything in a population has an equal chance of being selected, the selection can be said to occur at random.

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