12.8: Independent and Dependent Events
Introduction
The Finalists
The Talent Show has been a huge success and the judges have had a very difficult task. There are three prizes that will be given. After a lot of deliberation, the judges have narrowed it down to the following five finalists.
2 Sixth Graders
2 Seventh Graders
1 Eighth Grader
Given these standings, what is the probability that a seventh grader and an eighth grader will be selected for an award?
This is best solved by thinking about dependent and independent events. Pay close attention to this lesson and you will know how to figure out the probability by the end of it.
What You Will Learn
In this lesson, you will learn how to use the following skills.
- Recognize independent events as compound events in which the occurrence of one event does not affect the likelihood of another event.
- Recognize dependent events as compound events in which the occurrence of one event does effect the likelihood of the other.
- Find the probability of two independent events both occurring.
- Find the probability of two dependent events both occurring.
Teaching Time
I. Recognize Independent Events as Compound Events in which the Occurrence of One Event Does Not Affect the Likelihood of Another Event
Suppose you have two events:
Event \begin{align*}A\end{align*}: Spin red on spinner \begin{align*}A\end{align*}
Event \begin{align*}B\end{align*}: Spin purple on spinner \begin{align*}B\end{align*}
The probability of these events is easy enough to compute. In general:
\begin{align*}P (\text{event}) = \frac{\text{favorable outcomes}}{\text{total outcomes}}\end{align*}
So:
\begin{align*}P(\text{red}) & = \frac{1}{4}\\ P (\text{purple}) & = \frac{1}{3}\end{align*}
Now a question arises. Does event \begin{align*}A\end{align*} affect the probability of event \begin{align*}B\end{align*} in any way? That is, does the arrow landing on red in the first spinner affect the way the arrow lands in the second spinner? If not, then the two events are said to be independent events.
If the outcome of one event has no effect on the outcome of a second event, then the two events are independent events.
Events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} above are independent events. No matter how many times you spin spinner \begin{align*}A\end{align*}, the outcome of spinning spinner \begin{align*}A\end{align*} does not affect the outcome of spinning spinner \begin{align*}B\end{align*}.
Example
Jeremy flips a coin two times. Event \begin{align*}A\end{align*} is the first coin flip. Event \begin{align*}B\end{align*} is the second coin flip. Are the two coin flips independent events?
Ask yourself, Can the outcome of the first event in any way change the outcome of the second event? If not, then the two events are independent.
Suppose Jeremy’s first flip comes up heads. Does that in any way affect the outcome of the second flip? Is it now more likely to come up heads or tails?
In fact, the first flip does not affect the second flip. The probability of heads in the second flip is \begin{align*}\frac{1}{2}\end{align*}, no matter what the first flip was. Similarly, the probability of tails in the second flip is also \begin{align*}\frac{1}{2}\end{align*}, no matter what the first flip was. So the two events are independent.
II. Recognize Dependent Events as Compound Events in which the Occurrence of One Event Does Effect the Likelihood of the Other
Now that you know about independent events, you can learn about dependent events. If one event does depend on another event, then the events depend on each other. Let’s look at an example to understand this further.
Example
Mariko pulls a red sock from the laundry bag. Does this change the probability that the next sock Mariko pulls out of the bag will be red?
Here, the act of taking a sock out of the bag changes the situation. For the first sock, the probability of pulling out a red sock was:
\begin{align*}P(\text{red}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{3}{6} = \frac{1}{2}\end{align*}
For the second sock, there are now only 5 socks left in the bag and only 2 of them are red. So the probability of pulling out a red sock now for the second sock is:
\begin{align*}P(\text{red}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{2}{5}\end{align*}
Clearly, the first event affected the outcome of the second event in this situation. So the two events are NOT independent. In other words, they are dependent events.
If the outcome of one event has an effect on the outcome of a second event, then the two events are dependent events.
Write the difference between dependent events and independent events in your notebook.
12N. Lesson Exercises
Try a few of these on your own. Determine whether the events described are dependent or independent events.
- A box contains a penny, a nickel, a dime, and a quarter. What is the probability of pulling a quarter out of the box, putting it in your pocket, then pulling a penny out of the box?
- In a laundry bag with 3 red socks and 3 blue socks, Mariko pulls out a blue sock, sees it’s the wrong sock and returns it to the bag. Now Mariko pulls out a second sock. What is the probability that it will be red?
Discuss your answers with a partner. Explain in your own words why each event is dependent or independent.
III. Find the Probability of Two Independent Events Both Occurring
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*} and \begin{align*}B\end{align*}. The result of spinning spinner \begin{align*}A\end{align*} does not affect the result of spinning spinner \begin{align*}B\end{align*}.
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*} 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*}
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*}. The second spinner has three possible options, so the probability is \begin{align*}\frac{1}{3}\end{align*}. The Probability Rule takes care of the rest.
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*} and \begin{align*}B\end{align*}, will both occur is:
\begin{align*}P(A \ \text{and} \ B) = P (A) \cdot P (B)\end{align*}
Write the Probability Rule in your notebook.
Example
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*}.
Example
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?
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.
Note that the probability rule can work for any number of probabilities.
II. Find the Probability of Two Dependent Events Occurring
You just learned how to find the probability of two independent events. Well, we can use a similar method to find the probability of two dependent events.
If we go back to our laundry bag full of socks and ask: What is the probability of picking two socks from the bag and have them both turn out to be red?
The probability of the first sock being red is:
\begin{align*}P (\text{red 1st sock}) = \frac{3}{6} = \frac{1}{2}\end{align*}
What about the second sock?
Having removed the first sock from the bag, we have changed the number of red socks and the total number of socks in the bag. So now instead of there being 3 red socks out of 6 total socks, there are only 2 red socks left out of 5 total socks:
\begin{align*}P (\text{red 2nd sock}) = \frac{2}{5}\end{align*}
This gives a probability of both events occurring as:
\begin{align*}P (\text{red 1st sock and red 2nd sock}) & = P (\text{red 1st sock}) \cdot P (\text{red 2nd sock})\\ & = \frac{1}{2} \cdot \frac{2}{5}\\ & = \frac{2}{10}\\ & = \frac{1}{5}\end{align*}
The same general method works for calculating any two (or more) dependent events.
Example
A stack of 8 cards has 5 Jacks and 3 Queens. What is the probability of picking 2 Jacks from the stack at random?
Use the Probability Rule to find the probability of the two dependent events.
The probability of the 1st Jack is:
\begin{align*}P (\text{1st Jack}) = \frac{5}{8}\end{align*}
Once the 1st Jack is taken, the probability of 2nd Jack is only 4 of 7 because there are only 4 Jacks left out of 7 total cards:
\begin{align*}P (\text{2nd Jack}) = \frac{4}{7}\end{align*}
So:
\begin{align*}P (\text{1st Jack and 2nd Jack}) & = P (\text{1st Jack}) \cdot P (\text{2nd Jack})\\ & = \frac{5}{8} \cdot \frac{4}{7}\\ & = \frac{5}{14}\end{align*}
The answer is \begin{align*}\frac{5}{14}\end{align*}.
We can use the Probability Rule to find the probability of both independent and dependent events!
Real–Life Example Completed
The Finalists
Here is the original problem once again. Reread it and then begin to work on figuring out the probability question.
The Talent Show has been a huge success and the judges have had a very difficult task. There are three prizes that will be given. After a lot of deliberation, the judges have narrowed it down to the following five finalists.
2 Sixth Graders
2 Seventh Graders
1 Eighth Grader
Given these standings, what is the probability that a seventh grader and an eighth grader will be selected for an award?
Before we even begin to figure out the probability, we first need to decide if these are independent events or dependent events.
Think about it this way, if we select a seventh grader for an award, then the number of possible finalists for the next award changes from 5 to 4. One event depends on the other event. Therefore, these are dependent events.
To figure out the probability of dependent events, we can multiply. We want to figure out the probability of a seventh and an eighth grader being selected for an award.
\begin{align*}P(7^{th} \text{and} \ 8^{th}) = \frac{2}{5} \cdot \frac{1}{4}\end{align*}
We have 2 out of 5 for seventh graders. Then once a seventh grader is selected, we go to four finalists. One is an eighth grader, so we have a 1 out of 4 chance of having an eighth grader for an award.
Next, we multiply.
\begin{align*}P(7^{th} \text{and} \ 8^{th}) = \frac{2}{5} \cdot \frac{1}{4} = \frac{2}{20}\end{align*}
What is this as a percent?
\begin{align*}\frac{2}{20} = \frac{10}{100} = 10\%\end{align*}
There is only a 15% chance of both a seventh grader and an eighth grader as award winners!
Vocabulary
Here are the vocabulary words found in this lesson.
- Independent Events
- events where one event does not impact the result of another.
- Dependent Events
- events where one event does impact the result of another.
- Probability Rule
- \begin{align*}P(A) \cdot P(B) = \text{Probability of} \ A \ \text{and} \ B\end{align*}
Technology Integration
Khan Academy, Independent Events 1
Khan Academy, Independent Events 2
Khan Academy, Independent Events 3
Time to Practice
Directions: Write whether each pair of events is dependent or independent.
1. A: Mike rolls a number cube. B: Leah spins a red-blue-green spinner.
2. A: In a game of Go Fish, the probability of one player drawing a Queen from the deck. B: On the next player’s turn, the probability of drawing a Queen.
3. A: The probability that a randomly ordered pizza will be large. B: The probability that the same randomly ordered pizza will be deep-dish.
4. A: The probability that a randomly ordered 2-topping pizza will have pepperoni. B: The probability that the same randomly ordered 2-topping pizza will have mushrooms.
5. A: The probability of flipping a coin tails 5 times in a row. B: The probability of the sixth flip turning out to be heads.
6. A: In a 4-team league, the probability of the Rockets finishing in first place. B: In a 4-team league, the probability of the Sharks finishing in first place.
7. A: On a roll of a number cube, the probability of rolling 6. B: On a second roll of a number cube, the probability of rolling 6.
8. A: In a spelling bee, the probability of the first contestant being given the word khaki from a list of 10 words. B: In a spelling bee, the probability of the second contestant getting the word khaki from the same list of words.
9. A: The probability that it will snow on Tuesday. B: The probability that Tuesday will fall on an odd day of the month.
10. A: The probability that it will be below 32 degrees on Tuesday. B: The probability that it will snow on Tuesday.
11. A: The probability that it will snow on Tuesday. B: The probability that school will be cancelled on Tuesday.
12. A: The probability that the first Wednesday in June will fall on an even day of the month. B: The probability that the first Thursday in June will fall on an even day of the month.
13. A: The probability that the first Wednesday in June will fall on an even day of the month. B: The probability that the first Thursday in June will be sunny.
14. A: The probability that a coin will land on heads. B: The probability that a number cube will land on 5.
15. A: The probability that the first spin of a red-blue-green spinner will land on green. B: The probability that the second spin of a red-blue-green spinner will land on green.
Directions: Solve each problem.
16. Mia spins the spinner two times. What is the probability that the arrow will land on 2 both times?
17. 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?
18. 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?
19. 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?
20. 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.
21. 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.
22. 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.
23. 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?
24. 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?
25. 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?
26. Karina flips a coin 3 times. What is the probability that she will flip heads 3 times in a row?
27. Karina flips a coin 4 times. What is the probability that she will flip heads 4 times in a row?
28. Karina flips a coin 4 times. What is the probability that she will NOT flip heads 4 times in a row?
Directions: Solve each problem.
29. A clothes dryer contains 5 black socks and 1 white sock. What is the probability of taking two socks, one after another, out of the dryer and having them both be black?
30. A clothes dryer contains 4 black socks and 2 white socks. What is the probability of taking two socks out of the dryer and having them both be black?
31. A clothes dryer contains 3 black socks and 3 white socks. What is the probability of taking two socks out of the dryer and having them both be black?
32. A clothes dryer contains 3 black socks and 3 white socks. What is the probability of taking two socks out of the dryer and having the first one be black and the second one be white?
33. Bob bought two theater box tickets. The computer randomly assigns the tickets in one of 5 seats: end seat A, middle seat B, middle seat C, middle seat D, or end seat E. What is the probability that the first ticket is A and the second ticket is seat B?