Jasmine loves her funky ponytails. She has a box of 12 hair elastics and another box of 10 hair ribbons. In the box of hair elastics, there are 3 pink, 4 yellow, 2 green, and 3 purple elastics. In the ribbon box, she has 2 pink, 1 yellow, 1 green, and 6 purple ribbons. What are the chances that Jasmine will pull a purple elastic and a purple ribbon?

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

### Calculating Probabilities of Independent Events

What is the probability of two independent events both occurring? For example, what is the probability of spinner \begin{align*}A\end{align*} landing on red and spinner \begin{align*}B\end{align*} landing on blue?

We could create a tree diagram to show all of the possible options and figure out the probability, but that is very complicated. There is a simpler way.

Notice that this probability equals the product of the two independent probabilities.

\begin{align*}P(\text{red-blue}) & = P(\text{red}) \cdot P(\text{blue})\\ & = \frac{1}{4} \cdot \frac{1}{3}\\ & = \frac{1}{12}\end{align*}

The first spinner has four possible options, so the probability is \begin{align*}\frac{1}{4}\end{align*}. The second spinner has three possible options, so the probability is \begin{align*}\frac{1}{3}\end{align*}.

In fact, this method works for any independent events as summarized by the **Probability Rule**. The **Probability Rule** is the probability that two independent events, \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, will both occur and is written as the formula:

\begin{align*}P(A \ \text{and} \ B) = P (A) \cdot P (B)\end{align*}

What is the probability that if you spin the spinner two times, it will land on yellow on the first spin and red on the second spin?

First, define the value for each probability:

\begin{align*}P(\text{yellow and red}) & = P(\text{yellow}) \cdot P(\text{red})\\ P (\text{yellow}) & = \frac{3}{5}\\ P (\text{red}) & = \frac{2}{5}\end{align*}

Next, substitute the values into the Probability Rule formula:

\begin{align*}P(\text{yellow and red}) & = \frac{3}{5} \cdot \frac{2}{5}\\\end{align*}

Then, multiply for the probability of both events occurring:

\begin{align*}P(\text{yellow and red}) & = \frac{3}{5} \cdot \frac{2}{5}\\ & = \frac{6}{25}\end{align*}

The answer is the probability of both of these events occurring is \begin{align*}\frac{6}{25}\end{align*}.

### Examples

#### Example 1

Earlier, you were given a problem about Jasmine and her box of ribbons and box of elastics for her funky ponytails.

If the box of ribbons contains 2 pink, 1 yellow, 1 green, and 6 purple ribbons, and the box of elastics contains 3 pink, 4 yellow, 2 green, and 3 purple elastics, what is the probability she will pull a purple ribbon and purple elastic out of the respective boxes?

First, define the value for each probability:\begin{align*}P(\text{purple ribbon and purple elastic}) & = P (\text{purple ribbon}) \cdot P (\text{purple elastic})\\ P (\text{purple ribbon}) & = \frac{6}{10} = \frac{3}{5}\\ P (\text{purple elastic}) & = \frac{3}{12} = \frac{1}{4}\end{align*}Next, substitute the values into the Probability Rule formula:\begin{align*}P (\text{purple ribbon and purple elastic}) & = \frac{3}{5} \cdot \frac{1}{4}\\\end{align*}Then, multiply for the probability of both events occurring:\begin{align*}P (\text{purple ribbon and purple elastic}) & = \frac{3}{5} \cdot \frac{1}{4}\\ & = \frac{3}{20}\end{align*}The answer is the probability of both of these events occurring is \begin{align*}\frac{3}{20}\end{align*}.

#### Example 2

The probability of rain tomorrow is 40 percent. The probability that Jeff’s car will break down tomorrow is 3 percent. What is the probability that Jeff’s car will break down in the rain tomorrow?

First, define the value for each probability:

\begin{align*}P(\text{rain and break}) & = P (\text{rain}) \cdot P (\text{break})\\ P (\text{rain}) & = 40\% = \frac{40}{100} = \frac{2}{5}\\ P (\text{break}) & = 3\% = \frac{3}{100}\end{align*}

Next, substitute the values into the Probability Rule formula:

\begin{align*}P (\text{rain and break}) & = \frac{2}{5} \cdot \frac{3}{100}\\\end{align*}

Then, multiply for the probability of both events occurring:

\begin{align*}P (\text{rain and break}) & = \frac{2}{5} \cdot \frac{3}{100}\\ & = \frac{3}{250}\end{align*}

The answer is the probability of both of these events occurring is \begin{align*}\frac{3}{250}\end{align*}.

**Use the spinners below to answer the following questions.**

#### Example 3

What is the probability of spinner A landing on red and spinner B landing on red?

First, define the value for each probability:\begin{align*}P(\text{red and red}) & = P (\text{red}) \cdot P (\text{red})\\ P (\text{red}) & = \frac{1}{4}\\ P (\text{red}) & = \frac{1}{3}\end{align*}Next, substitute the values into the Probability Rule formula:\begin{align*}P (\text{red and red}) & = \frac{1}{4} \cdot \frac{1}{3}\\\end{align*}Then, multiply for the probability of both events occurring:\begin{align*}P (\text{red and red}) & = \frac{1}{4} \cdot \frac{1}{3}\\ & = \frac{1}{12}\end{align*}The answer is the probability of both of these events occurring is \begin{align*}\frac{1}{12}\end{align*}.

#### Example 4

What is the probability of spinner A landing on blue or yellow and spinner B landing on blue?

First, define the value for each probability:\begin{align*}P(\text{blue or yellow and blue}) & = P (\text{blue or yellow}) \cdot P (\text{blue})\\
P (\text{blue or yellow}) & = \frac{2}{4} = \frac{1}{2}\\
P (\text{blue}) & = \frac{1}{3}\end{align*}Next, substitute the values into the Probability Rule formula:\begin{align*}P (\text{blue or yellow and blue}) & = \frac{1}{2} \cdot \frac{1}{3}\\\end{align*}Then, multiply for the probability of both events occurring:\begin{align*}P (\text{blue or yellow and blue}) & = \frac{1}{2} \cdot \frac{1}{3}\\
& = \frac{1}{6}\end{align*}The answer is the probability of both of these events occurring is ** \begin{align*}\frac{1}{6}\end{align*}.**

#### Example 5

What is the probability of spinner A landing on yellow and spinner B landing on red or green?

First, define the value for each probability:\begin{align*}P(\text {yellow and red or green}) & = P (\text{yellow}) \cdot P (\text{red or green})\\
P (\text{yellow}) & = \frac{1}{4}\\
P (\text{red or green}) & =\frac{2}{3}\end{align*}Next, substitute the values into the Probability Rule formula:\begin{align*}P (\text{yellow and red or green}) & = \frac{1}{4} \cdot \frac{2}{3}\\\end{align*}Then, multiply for the probability of both events occurring:\begin{align*}P (\text{yellow and red or green}) & = \frac{1}{4} \cdot \frac{2}{3}\\
& = \frac{2}{12}= \frac{1}{6}\end{align*}The answer is the probability of both of these events occurring is ** \begin{align*}\frac{1}{6}\end{align*}.**

### Review

Solve each problem.

- Mia spins the spinner two times. What is the probability that the arrow will land on 2 both times?
- Mia spins the spinner two times. What is the probability that the arrow will land on 2 on the first spin and 3 on the second spin?
- Mia spins the spinner two times. What is the probability that the arrow will land on an even number on the first spin and an odd number on the second spin?
- Mia spins the spinner two times. What is the probability that the arrow will land on an odd number on the first spin and a number less than 4 on the second spin?
- A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that both socks will be black? Write your answer as a decimal.
- A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that both socks will be white? Write your answer as a decimal.
- A laundry bag has 8 black socks and 2 white socks. If you pull out a sock, then put it back in the bag and pull out a second sock, what is the probability that the first sock will be black and the second sock will be white? Write your answer as a decimal.
- Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick an Ace out of each deck?
- Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick a face card (Jack, Queen, King) out of each deck?
- Dirk has two 52-card decks. Each deck has 4 Aces, 4 Kings, 4 Queens, and so on. What is the probability that Dirk will pick a card lower than a Jack out of each deck?
- Karina flips a coin 3 times. What is the probability that she will flip heads 3 times in a row?
- Karina flips a coin 4 times. What is the probability that she will flip heads 4 times in a row?
- Karina flips a coin 4 times. What is the probability that she will NOT flip heads 4 times in a row?
- Karina flips a coin 5 times. What is the probability that she will flip heads once?
- Karina flips a coin 5 times. What is the probability that she will flip tails once?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 12.17.