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# Probability of Independent Events

## Two outcomes both occurring independently.

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Probability of Independent Events

At the end of the school year, Mr. Fike collected 80 surveys to find out which sports had been the most popular during the school year. On the survey he asked students to list their three favorite sports. He found that football was chosen as a favorite sport by 60 out of 80 students. Basketball was chosen as a favorite by 40 students. He was surprised to find out that 20 students chose tennis as a favorite while only 15 chose baseball. Mr. Fike wondered what the likelihood was of a student choosing football, basketball, and tennis as their favorite sports. How can Mr. Fike figure out the probability of a student selecting all three?

In this concept, you will learn how to calculate the probability of independent events.

### Calculating the Probability of Independent Events

An independent event is an event that does not depend on another event to determine its outcome. When there are two independent events, one outcome does not impact the outcome of the second event. Think about choosing a card from a deck of cards. There are 52 cards in a deck. You can choose a card and there is a certain probability that it will be a red card. Then you can put the card back and choose again. The first outcome of choosing a card has nothing to do with the second outcome because they are independent events.

You can calculate the probability of an independent event occurring.

Let’s say there are four tiles face down on a table.

On the side you cannot see are different images. Three of them have a sun and one of them has a four-leaf clover.
What is the probability of selecting a sun on the first flip?
First, because this is an independent event, nothing else impacts the outcome except the flip. If there are three suns then you have a 3 out of 4 chance of flipping a sun.

P=34\begin{align*}P=\frac{3}{4}\end{align*}

You can also have two independent events. The events have nothing to do with each other, and yet you can calculate the probability of them occurring.

Take a look at the following example.

There are 9 marbles in a bag. There are three blue, three green and three orange. What is the probability of selecting one blue marble, then putting that one back and selecting one green marble?

There are two independent events in this example. The first is choosing one blue marble. The next is choosing one green marble.

Also notice that the first marble is put back into the bag before the next one is taken out. This is a key for independent events.

What is the probability of choosing a blue one first?

There are 9 marbles and 3 are blue. That means 3 out of 9 chances.

39=13\begin{align*}\frac{3}{9}=\frac{1}{3}\end{align*}

What is the probability of choosing a green one next?

There are 9 marbles and 3 are green.

39=13\begin{align*}\frac{3}{9}=\frac{1}{3}\end{align*}

To find the likelihood of these two independent events occurring, multiply the two probabilities.

P(B and G)=1313=19\begin{align*}P(B \ \text{and} \ G)= \frac{1}{3} \cdot \frac{1}{3}=\frac{1}{9}\end{align*}

Take another look at the example, this time with three independent events occurring.

There are 9 marbles in a bag. There are three blue, three green and three orange. What is the probability of selecting one blue marble, then putting that one back and selecting one green marble, then putting that one back and selecting one orange marble?

Since you already know the probability of the first two independent events, find the probability of choosing an orange one next.

There are 9 marbles and 3 are orange.

39=13\begin{align*}\frac{3}{9}=\frac{1}{3}\end{align*}

Next, figure the likelihood of three independent events occurring by multiplying each probability.

P(B, then G, then O)=131313=127\begin{align*}P(B, \ \text{then} \ G, \ \text{then} \ O)= \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}= \frac{1}{27}\end{align*}There is a very small likelihood that this event will occur.

### Examples

#### Example 1

Earlier, you were given a problem about Mr. Fike and his sports survey.

Mr. Fike surveyed a representative sample of students from grades 6, 7, and 8 to find out what their three favorite sports were at school. He collected 80 surveys and found that football was chosen as a favorite sport by 60 students. Basketball was chosen as a favorite by 40 students. And tennis was the next top favorite with 20 votes.

Mr. Fike wants to know the likelihood of a student choosing football, basketball, and tennis as their favorite sports.

First, figure out the probability of the first event, 60 out of 80 students chose football.

6080=34\begin{align*}\frac{60}{80}=\frac{3}{4}\end{align*}

Next, figure out the probability of the second event, 40 out of 80 students chose basketball.

4080=12\begin{align*}\frac{40}{80}=\frac{1}{2}\end{align*}

Then, figure out the probability of the third event, 20 out of 80 students chose tennis.

2080=14\begin{align*}\frac{20}{80}=\frac{1}{4}\end{align*}

Finally, multiply the probabilities of the three independent events.

341214=332\begin{align*}\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{4}=\frac{3}{32}\end{align*}

There is a 3 out of 32 chance that a student would choose all three.

You can look at this as a percent too to get an idea of how small the likelihood is.

332=9.38%\begin{align*}\frac{3}{32}=9.38 \%\end{align*}The answer is there is less than a 1 in 10 chance of a student choosing all three sports as a favorite.

In the following examples, calculate the likelihood of the independent event using a percent.

#### Example 2

Find the probability of the independents events occurring in the following scenario.

Out of 100 students at Riverview Middle School, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf?

50100=12\begin{align*}\frac{50}{100}=\frac{1}{2}\end{align*}

Next, figure out the probability of the second event, 25 out of 100 students enjoy golf.

25100=14\begin{align*}\frac{25}{100}=\frac{1}{4}\end{align*}

Then, multiply the probabilities of the two independent events.

1214=18\begin{align*}\frac{1}{2} \cdot \frac{1}{4}=\frac{1}{8}\end{align*}

The answer is there is a one out of eight chance that this will happen.

Out of 100 students at Riverview Middle, 50 enjoy swimming. Out of the same 100 students, 25 enjoy golf. What is the probability that a student would enjoy both swimming and golf?

Let’s add to this problem that 40 out of 100 enjoy basketball.

What is the likelihood that a student would enjoy all three?

First, find the probability of the third event, 40 out of 100 enjoy basketball.

40100=25\begin{align*}\frac{40}{100}=\frac{2}{5}\end{align*}

Remember the probability of the first and second events.50100=1225100=14\begin{align*}\begin{array}{rcl} \frac{50}{100}=\frac{1}{2}\\ \frac{25}{100}=\frac{1}{4} \end{array}\end{align*}

Next, multiply the third event’s probability by the first and second independent events.

121425=240=120\begin{align*}\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{5}=\frac{2}{40}=\frac{1}{20}\end{align*}

The answer is there is a one out of 20 chance that a student would like all three.

You can look at this as a percent too to get an idea of how small the likelihood is.

120=5%\begin{align*}\frac{1}{20}=5 \%\end{align*}

#### Example 3

If 6 out of 24 students enjoy opera in a class, what is the probability that a student selected at random from the class will enjoy the opera?

First, express the probability as a fraction, 6 out of 24, and then simplify.

624=14\begin{align*}\frac{6}{24}=\frac{1}{4}\end{align*}

Next, divide 1 by 4 to get a percent.

25%\begin{align*}25 \%\end{align*}

The answer is 25%\begin{align*}25 \%\end{align*}.

#### Example 4

If 4 out of 12 students like Math class the best, what is the probability that a student selected at random from the class will enjoy Math the best?

First, express the probability as a fraction, 4 out of 12, and then simplify.

412=13\begin{align*}\frac{4}{12}=\frac{1}{3}\end{align*}

Next, divide 1 by 3 to get a percent.

33.33%\begin{align*}33.33 \%\end{align*}

The answer is 33.33%\begin{align*}33.33 \%\end{align*}.

#### Example 5

If 5 out of 10 like Drama best, what is this probability?

First, express the probability as a fraction, 5 out of 10, and then simplify.

510=12\begin{align*}\frac{5}{10}=\frac{1}{2}\end{align*}

Next, divide 1 by 2 to get a percent.

50%\begin{align*}50 \%\end{align*}

The answer is 50%\begin{align*}50 \%\end{align*}.

### Review

A bag has sixteen marbles in it. There are four of each color. There are four red, four blue, four green and four yellow.

1. What is the probability of pulling a green marble out of the bag?
2. What is the probability of pulling a yellow or a green marble out of the bag?
3. What is the probability of pulling a red marble out of the bag?
4. What is the probability of pulling a red, yellow or green marble out of the bag?
5. What is the probability of not pulling a red marble out of the bag?
6. What is the probability of pulling a green marble, then a red marble out of the bag?
7. What is the probability of pulling a yellow or blue marble out of the bag?
8. What is the probability of pulling a red, then a blue then a yellow out of the bag?
9. If I take out a red marble and a blue marble, what is the probability of pulling a green marble out of the bag?
10. What is the probability that you will pull a red marble out of the bag?

For questions 10-20, look at the probability fractions that you wrote for 1-10. Convert each one to a decimal and then a percent. You may round when necessary.

To see the Review answers, open this PDF file and look for section 12.20.

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### Vocabulary Language: English

Probability

Probability is the chance that something will happen. It can be written as a fraction, decimal or percent.