12.17: Probability of Independent Events
Have you ever wondered if two things can happen at once?
Jana has two decks of cards. Each deck has ten cards in it. There are three face cards in the first deck and four in the second. What are the chances that Jana will draw a face card from both decks?
This is an independent event. In this Concept, you will learn how to figure out probabilities like this one.
Guidance
You now know the difference between independent and dependent events. Think about independent events. If one event does not impact the result of a second event, then the two events are independent. For example, there are two different spinners \begin{align*}A\end{align*}
But now we ask a new question. What is the probability of two completely independent events both occurring? For example, what is the probability of spinner \begin{align*}A\end{align*}
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{redblue}) & = P(\text{red}) \cdot P(\text{blue})\\
& = \frac{1}{4} \cdot \frac{1}{3}\\
& = \frac{1}{12}\end{align*}
Where did these fractions come from?
They came from the probability of the sample space of each spinner. The first spinner has four possible options, so the probability is \begin{align*}\frac{1}{4}\end{align*}
In fact, this method works for any independent events as summarized in this rule.
Probability Rule: The probability that two independent events, \begin{align*}A\end{align*}
\begin{align*}P(A \ \text{and} \ B) = P (A) \cdot P (B)\end{align*}
Write the Probability Rule in your notebook.
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?
To find the solution, use the rule.
\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*}
So:
\begin{align*}P(\text{yellow and red}) & = \frac{3}{5} \cdot \frac{2}{5}\\
& = \frac{6}{25}\end{align*}
The probability of both of these events occurring is \begin{align*}\frac{6}{25}\end{align*}
Now it's time for you to try a few on your own.
Example A
What is the probability of spinner A landing on red and spinner B landing on red?
Solution:\begin{align*}\frac{1}{12}\end{align*}
Example B
What is the probability of spinner A landing on blue or yellow and spinner B landing on blue?
Solution:\begin{align*}\frac{1}{6}\end{align*}
Example C
What is the probability of spinner A landing on yellow and spinner B landing on red or green?
Solution:\begin{align*}\frac{1}{6}\end{align*}
Here is the original problem once again.
Jana has two decks of cards. Each deck has ten cards in it. There are three face cards in the first deck and four in the second. What are the chances that Jana will draw a face card from both decks?
To figure this out, let's write the probability of picking a face card from the first deck.
\begin{align*}\frac{3}{10}\end{align*}
Now let's write the probability of picking a face card from the second deck.
\begin{align*}\frac{4}{10} = \frac{2}{5}\end{align*}
Now we can multiply and simplify.
\begin{align*}\frac{3}{10} \times \frac{2}{5}\end{align*}
\begin{align*}\frac{3}{25}\end{align*}
This is our answer.
Vocabulary
Here are the vocabulary words in this Concept.
 Independent Events
 events where one event does not impact the result of another.
 Probability Rule

\begin{align*}P(A) \cdot P(B) = \text{Probability of} \ A \ \text{and} \ B\end{align*}
P(A)⋅P(B)=Probability of A and B
Guided Practice
Here is one for you to try on your own.
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?
Answer
To find the solution, use the rule.
\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*}
So:
\begin{align*}P (\text{rain and break}) & = \frac{2}{5} \cdot \frac{3}{100}\\
& = \frac{3}{250}\end{align*}
You can see that the probability is very small.
Video Review
Here is a video for review.
 This is a Khan Academy video on the probability of independent events.
Practice
Directions: Solve each problem.
1. Mia spins the spinner two times. What is the probability that the arrow will land on 2 both times?
2. 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?
3. Mia spins the spinner two times. What is the probability that the arrow will land on an even number on the first spin and and an odd number on the second spin?
4. 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?
5. 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.
6. 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.
7. 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.
8. Dirk has two 52card 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?
9. Dirk has two 52card 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?
10. Dirk has two 52card 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?
11. Karina flips a coin 3 times. What is the probability that she will flip heads 3 times in a row?
12. Karina flips a coin 4 times. What is the probability that she will flip heads 4 times in a row?
13. Karina flips a coin 4 times. What is the probability that she will NOT flip heads 4 times in a row?
14. Karina flips a coin 5 times. What is the probability that she will flip heads twice in a row?
15. Karina flips a coin 5 times. What is the probability that she will flip tails four times in a row?
Independent Events
Two events are independent if the occurrence of one event does not impact the probability of the other event.Probability Rule
The probability of two independent events A and B both occurring is P( A and B) = P(A) P(B).Image Attributions
Here you'll learn to find the probability of two independent events both occurring.
Concept Nodes:
Independent Events
Two events are independent if the occurrence of one event does not impact the probability of the other event.Probability Rule
The probability of two independent events A and B both occurring is P( A and B) = P(A) P(B).