8.4.1: Probability
- Page ID
- 63948
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Define event, outcome, trial, simple event, and sample space, and calculate the probability that an event will occur.
- Calculate the probability of events for more complex outcomes.
- Solve applications involving probabilities.
Introduction
Probability provides a measure of how likely it is that something will occur. It is a number between and including the numbers and 1. It can be written as a fraction, a decimal, or a percent.
Picking numbers randomly means that there is no specific order in which they are chosen. Many games use dice or spinners to generate numbers randomly. If you understand how to calculate probabilities, you can make thoughtful decisions about how to play these games by knowing the likelihood of various outcomes.
Definitions
First you need to know some terms related to probability. When working with probability, a random action or series of actions is called a trial. An outcome is the result of a trial, and an event is a particular collection of outcomes. Events are usually described using a common characteristic of the outcomes.
Let's apply this language to see how the terms work in practice. Some games require rolling a die with six sides, numbered from 1 to 6. (Dice is the plural of die.) The chart below illustrates the use of trial, outcome, and event for such a game:
| Trial | Outcomes | Examples of Events |
| Rolling a die | There are 6 possible outcomes: {1, 2, 3, 4, 5, 6} |
Rolling an even number: {2, 4, 6} Rolling a 3: {3} Rolling a 1 or a 3: {1, 3} Rolling a 1 and a 3: {} (Only one number can be rolled, so this outcome is impossible. The event has no outcomes in it.) |
Notice that a collection of outcomes is put in braces and separated by commas.
A simple event is an event with only one outcome. Rolling a 1 would be a simple event, because there is only one outcome that works: 1! Rolling more than a 5 would also be a simple event, because the event includes only 6 as a valid outcome. A compound event is an event with more than one outcome. For example, in rolling one six-sided die, rolling an even number could occur with one of three outcomes: 2, 4, and 6.
When you roll a six-sided die many times, you should not expect any outcome to happen more often than another (assuming that it is a fair die). The outcomes in a situation like this are said to be equally likely. It’s very important to recognize when outcomes are equally likely when calculating probability. Since each outcome in the die-rolling trial is equally likely, you would expect to get each outcome \(\ \frac{1}{6}\) of the rolls. That is, you'd expect \(\ \frac{1}{6}\) of the rolls to be 1, \(\ \frac{1}{6}\) of the rolls to be 2, \(\ \frac{1}{6}\) of the rolls to be 3, and so on.
A spinner is divided into four equal parts, each colored with a different color as shown below. When this spinner is spun, the arrow points to one of the colors. Are the outcomes equally likely?
- Yes, they are equally likely.
- No, they are not equally likely.
- Answer
- All the outcomes are equally likely. Each color provides a different outcome, and each color takes up \(\ \frac{1}{4}\) of the circle. You would expect the arrow to point to each color \(\ \frac{1}{4}\) of the time.
Probability of Events
The probability of an event is how often it is expected to occur. It is the ratio of the size of the event space to the size of the sample space.
First, you need to determine the size of the sample space. The size of the sample space is the total number of possible outcomes. For example, when you roll 1 die, the sample space is 1, 2, 3, 4, 5, or 6. So the size of the sample space is 6.
Then you need to determine the size of the event space. The event space is the number of outcomes in the event you are interested in. The event space for rolling a number less than three is 1 or 2. So the size of the event space is 2.
For equally likely outcomes, the probability of an event E can be written P(E).
\(\ P(E)=\frac{\text { size of the event space }}{\text { size of the sample space }}=\frac{\text { number of outcomes in the event }}{\text { total number of possible outcomes }}\)
A game requires rolling a six-sided die numbered from 1 to 6. What is the probability of rolling an even number?
Solution
|
\(\ \text { Sample space }=\{1,2,3,4,5,6\}\) \(\ \text { Event space }=\{2,4,6\}\) |
First, find the sample space and the event space. The sample space is all the possible outcomes, and the event space is the outcomes in the event. In this case, the event is “rolling an even number.” |
| \(\ P(\text { even number })=P(\{2,4,6\})=\frac{3}{6}=\frac{1}{2}\) | Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. |
\(\ P(\text { even number })=\frac{1}{2}\)
It is a common practice with probabilities, as with fractions in general, to simplify a probability into lowest terms since that makes it easier for most people to get a sense of how great it is. Unless there is reason not to do so, express all final probabilities in lowest terms.
A spinner is divided into equal parts, each colored with a different color as shown below. Find the probability of spinning blue or green on this spinner:
- \(\ \frac{1}{6}\)
- \(\ \frac{1}{3}\)
- 2
- 6
- Answer
-
- Incorrect. There are 6 equally likely outcomes, so the probability of each outcome is \(\ \frac{1}{6}\). However, the event has two outcomes in it, {blue, green}. The probability is \(\ \frac{\text { size of event space }}{\text { size of sample space }}=\frac{2}{6}\). The correct answer is \(\ \frac{1}{3}\).
- Correct. There are 6 equally likely outcomes, and the event has two outcomes in it, {blue, green}. The probability is \(\ \frac{\text { size of event space }}{\text { size of sample space }}=\frac{2}{6}=\frac{1}{3}\).
- Incorrect. The highest number that a probability value can be is 1! There are 2 outcomes in the event space, but the probability of the event is the ratio of outcomes in the event space to the total number of equally likely outcomes. There are 6 equally likely outcomes, so the probability is \(\ \frac{\text { size of event space }}{\text { size of sample space }}=\frac{2}{6}=\frac{1}{3}\).
- Incorrect. The highest number that a probability value can be is 1! There are 6 equally likely outcomes in the sample space, but the probability of the event is the ratio of outcomes in the event space to the total number of equally likely outcomes. The probability is \(\ \frac{\text { size of event space }}{\text { size of sample space }}=\frac{2}{6}=\frac{1}{3}\).
Counting Methods to find Sample Spaces
The most difficult thing for calculating a probability can be finding the size of the sample space, especially if there are two or more trials. There are several counting methods that can help.
The first one to look at is making a chart. In the example below, Tori is flipping two coins. So you need to determine the sample space carefully. A chart such as the one shown in the example that follows is a good approach.
Tori is flipping a pair of coins and noting how many flips of “heads” she gets. What is the probability that she flips 2 heads? What is the probability that she flips only 1 head?
Solution
|
Outcomes:
sample space: {HH, HT, TH, TT} event space for 2 heads: {HH} event space for 1 head: {HT, TH} |
Create a chart to record the results of flipping the first coin, followed by the result of flipping the second coin. | |||||||||||||||
|
\(\ P(2 \text { heads })=P(\{\mathrm{HH}\})=\frac{1}{4}\) \(\ P(1 \text { head })=P(\{\mathrm{HT}, \mathrm{TH}\})=\frac{2}{4}=\frac{1}{2}\) |
Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. |
\(\ P(2 \text { heads })=\frac{1}{4}\)
\(\ P(1 \text { head })=\frac{1}{2}\)
In the example below, the sample space for Tori is simple as only one die is being rolled. However, since James is rolling two die, a chart helps to organize the information.
Tori rolled a six-sided die and wanted to get a result of either 1 or 4. James rolled two six-sided dice, one blue and one red, and wanted to get a result of both a 1 and a 3, at the same time. Which event has a greater probability?
Solution
|
Tori's sample space: {1, 2, 3, 4, 5, 6} Tori's event space: {1, 4} |
First, find the sample space and the event space for the two trials. For Tori's trial, this is straightforward. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| \(\ \text { Tori: } P(1 \text { or } 4)=\frac{2}{6}=\frac{1}{3}\) | Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
It's not so obvious for James’ trial, since he is rolling two dice. He can create and use a chart to find the possibilities. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
James' sample space has 36 outcomes. James' event space has 2 outcomes. |
There are 36 outcomes. Of these, there are 2 that have both 1 and 3. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| \(\ \text { James: } P(1 \text { and } 3)=\frac{2}{36}=\frac{1}{18}\) | Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Tori's event has a greater probability.
You can also use a tree diagram to determine the sample space. A tree diagram has a branch for every possible outcome for each event.
Suppose a closet has three pairs of pants (black, white, and green), four shirts (green, white, purple, and yellow), and two pairs of shoes (black and white). How many different outfits can be made? There are 3 choices for pants, 4 choices for shirts, and 2 choices for shoes. For our tree diagram, let's use B for black, W for white, G for green, P for purple, and Y for yellow.
You can see from the tree diagram that there are 24 possible outfits (some perhaps not great choices) in the sample space.
Now you could fairly easily solve some probability problems. For example, what is the probability that if you close your eyes and choose randomly you would choose pants and shoes with the same color? In the chart there are 8 outfits where the pants and the shoes match.
\(\ P(\text { same color })=\frac{8}{24}=\frac{1}{3}\)
As you've seen, when a trial involves more than one random element, such as flipping more than one coin or rolling more than one die, you don't always need to identify every outcome in the sample space to calculate a probability. You only need the number of outcomes.
The Fundamental Counting Principle is a way to find the number of outcomes without listing and counting every one of them.
If one event has \(\ p\) possible outcomes, and another event has \(\ m\) possible outcomes, then there are a total of \(\ p \cdot m\) possible outcomes for the two events.
Examples
- Rolling two six-sided dice: Each die has 6 equally likely outcomes, so the sample space is \(\ 6 \cdot 6\) or 36 equally likely outcomes.
- Flipping three coins: Each coin has 2 equally likely outcomes, so the sample space is \(\ 2 \cdot 2 \cdot 2\) or 8 equally likely outcomes.
- Rolling a six-sided die and flipping a coin: The sample space is \(\ 6 \cdot 2\) or 12 equally likely outcomes.
So you could use the Fundamental Counting Principle to find out how many outfits there are in the previous example. There are 3 choices for pants, 4 choices for shirts, and 2 choices for shoes. Using the Fundamental Counting Principle, you have \(\ 4 \cdot 3 \cdot 2=24\) different outfits.
Barry volunteers at a charity walk to make lunches for all the other volunteers. In each bag he puts:
- one of two sandwiches (peanut butter and jelly, or turkey and cheese),
- one of three chips (regular potato chips, baked potato chips, or corn chips),
- one piece of fruit (an apple or an orange).
He forgot to mark what was in the bags. Assuming that each choice is equally likely, what is the probability that the bag Therese gets holds a peanut butter and jelly sandwich and an apple?
Solution
|
Size of sample space: Multiply the (number of sandwich choices) by the (number of chip choices) by the (number of fruit choices) This gives you: \(\ 2 \cdot 3 \cdot 2=12\) |
First, use the Fundamental Counting Principle to find the size of the sample space. |
|
Size of event space: Multiply the (number of sandwich choices in the event) by the (number of chip choices in the event) by the (number of fruit choices in the event) This gives you: \(\ 1 \cdot 3 \cdot 1=3\) |
For the event space, follow the same principle. In this case, there is only one sandwich and one piece of fruit of interest, but any of the three types of chips are acceptable. |
| \(\ P(\mathrm{~PB} \& \mathrm{~J} \text { and apple })=\frac{\text { size of event space }}{\text { size of sample space }}=\frac{3}{12}=\frac{1}{4}\) | Use the ratio to find the probability. |
Carrie flips four coins and counts the number of tails. There are four ways to get exactly one tail: HHHT, HHTH, HTHH, and THHH. What is the probability that Carrie gets exactly one tail?
- \(\ \frac{1}{16}\)
- \(\ \frac{1}{8}\)
- \(\ \frac{1}{4}\)
- \(\ \frac{1}{2}\)
- Answer
-
- Incorrect. Since there are two outcomes for each coin, there are 16 possible outcomes since \(\ 2 \cdot 2 \cdot 2 \cdot 2=16\). However, there are four outcomes in the event, so the probability is \(\ \frac{4}{16}\), or \(\ \frac{1}{4}\).
- Incorrect. Since there are two outcomes for each coin, there are 16 possible outcomes since \(\ 2 \cdot 2 \cdot 2 \cdot 2=16\). There are four outcomes in the event, so the probability is \(\ \frac{4}{16}\), or \(\ \frac{1}{4}\).
- Correct. Since there are two outcomes for each coin, there are 16 possible outcomes since \(\ 2 \cdot 2 \cdot 2 \cdot 2=16\). There are four outcomes in the event, so the probability is \(\ \frac{4}{16}\), or \(\ \frac{1}{4}\).
- Incorrect. There are two outcomes for each coin, but there are 4 coins. That means there are 16 possible outcomes since \(\ 2 \cdot 2 \cdot 2 \cdot 2=16\). There are four outcomes in the event, so the probability is \(\ \frac{4}{16}\), or \(\ \frac{1}{4}\).
Summary
Probability helps you understand random, unpredictable situations where multiple outcomes are possible. It is a measure of the likelihood of an event, and it depends on the ratio of event and possible outcomes, if all those outcomes are equally likely.
\(\ P(E)=\frac{\text { size of the event space }}{\text { size of the sample space }}=\frac{\text { number of outcomes in the event }}{\text { total number of possible outcomes }}\)
The Fundamental Counting Principle is a shortcut to finding the size of the sample space when there are many trials and outcomes:
If one event has \(\ p\) possible outcomes, and another event has \(\ m\) possible outcomes, then there are a total of \(\ p \cdot m\) possible outcomes for the two events.