Probability
Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one above.[footnote]The figure is for illustrative purposes only and does not model any particular storm.[/footnote] They combine a collection of weather data to predict the most likely path of a hurricane. Each colored line represents one possible path. The group of squiggly lines can begin to resemble strands of spaghetti, hence the name. In this section, we will investigate methods for making these types of predictions.
An example of a "spaghetti model," which can be used to predict possible paths of a tropical storm.Construct Probability Models
Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment, or an activity with an observable result. The numbers on the cube are possible results, or outcomes, of this experiment. The set of all possible outcomes of an experiment is called the sample space of the experiment. The sample space for this experiment is [latex]\left\{1,2,3,4,5,6\right\}[/latex]. An event is any subset of a sample space. The likelihood of an event is known as probability. The probability of an event [latex]p[/latex] is a number that always satisfies [latex]0\le p\le 1[/latex], where 0 indicates an impossible event and 1 indicates a certain event. A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities. For instance, if there is a 1% chance of winning a raffle and a 99% chance of losing the raffle, a probability model would look much like the table below.| Outcome | Probability |
|---|---|
| Winning the raffle | 1% |
| Losing the raffle | 99% |
How To: Given a probability event where each event is equally likely, construct a probability model.
- Identify every outcome.
- Determine the total number of possible outcomes.
- Compare each outcome to the total number of possible outcomes.
Example: Constructing a Probability Model
Construct a probability model for rolling a single, fair die, with the event being the number shown on the die.Answer: Begin by making a list of all possible outcomes for the experiment. The possible outcomes are the numbers that can be rolled: 1, 2, 3, 4, 5, and 6. There are six possible outcomes that make up the sample space. Assign probabilities to each outcome in the sample space by determining a ratio of the outcome to the number of possible outcomes. There is one of each of the six numbers on the cube, and there is no reason to think that any particular face is more likely to show up than any other one, so the probability of rolling any number is [latex]\frac{1}{6}[/latex].
| Outcome | Roll of 1 | Roll of 2 | Roll of 3 | Roll of 4 | Roll of 5 | Roll of 6 |
| Probability | [latex]\frac{1}{6}[/latex] | [latex]\frac{1}{6}[/latex] | [latex]\frac{1}{6}[/latex] | [latex]\frac{1}{6}[/latex] | [latex]\frac{1}{6}[/latex] | [latex]\frac{1}{6}[/latex] |
Q & A
Do probabilities always have to be expressed as fractions?
No. Probabilities can be expressed as fractions, decimals, or percents. Probability must always be a number between 0 and 1, inclusive of 0 and 1.Try It
Construct a probability model for tossing a fair coin.Answer:
| Outcome | Probability |
|---|---|
| Roll of 1 | |
| Roll of 2 | |
| Roll of 3 | |
| Roll of 4 | |
| Roll of 5 | |
| Roll of 6 |
Computing Probabilities of Equally Likely Outcomes
Let [latex]S[/latex] be a sample space for an experiment. When investigating probability, an event is any subset of [latex]S[/latex]. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in [latex]S[/latex]. Suppose a number cube is rolled, and we are interested in finding the probability of the event "rolling a number less than or equal to 4." There are 4 possible outcomes in the event and 6 possible outcomes in [latex]S[/latex], so the probability of the event is [latex]\frac{4}{6}=\frac{2}{3}[/latex].A General Note: Computing the Probability of an Event with Equally Likely Outcomes
The probability of an event [latex]E[/latex] in an experiment with sample space [latex]S[/latex] with equally likely outcomes is given by[latex]P\left(E\right)=\frac{\text{number of elements in }E}{\text{number of elements in }S}=\frac{n\left(E\right)}{n\left(S\right)}[/latex]
[latex]E[/latex] is a subset of [latex]S[/latex], so it is always true that [latex]0\le P\left(E\right)\le 1[/latex].Example: Computing the Probability of an Event with Equally Likely Outcomes
A number cube is rolled. Find the probability of rolling an odd number.Answer: The event "rolling an odd number" contains three outcomes. There are 6 equally likely outcomes in the sample space. Divide to find the probability of the event.
[latex]P\left(E\right)=\frac{3}{6}=\frac{1}{2}[/latex]
Try It
A number cube is rolled. Find the probability of rolling a number greater than 2.Answer: [latex]\frac{2}{3}[/latex]
Computing Probability Using Counting Theory
Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems. Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are [latex]C\left(5,2\right)[/latex] ways to select 2 phones that are not defective. There are 8 phones, so there are [latex]C\left(8,2\right)[/latex] ways to select 2 phones. The probability of selecting 2 phones that are not defective is:[latex]\begin{array}{l}\frac{\text{ways to select 2 phones that are not defective}}{\text{ways to select 2 phones}}=\frac{C\left(5,2\right)}{C\left(8,2\right)}\hfill \\ \text{ }=\frac{10}{28}\hfill \\ \text{ }=\frac{5}{14}\hfill \end{array}[/latex]
Example: Computing Probability Using Counting Theory
A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears.- Find the probability that only bears are chosen.
- Find the probability that 2 bears and 3 dogs are chosen.
- Find the probability that at least 2 dogs are chosen.
Answer:
- We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are [latex]C\left(6,5\right)[/latex] ways to choose 5 bears. There are 14 toys, so there are [latex]C\left(14,5\right)[/latex] ways to choose any 5 toys.
[latex]\frac{C\left(6\text{,}5\right)}{C\left(14\text{,}5\right)}=\frac{6}{2\text{,}002}=\frac{3}{1\text{,}001}[/latex]
- We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are [latex]C\left(6,2\right)[/latex] ways to choose 2 bears. There are 5 dogs, so there are [latex]C\left(5,3\right)[/latex] ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are [latex]C\left(6,2\right)\cdot C\left(5,3\right)[/latex] ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
[latex]\frac{C\left(6\text{,}2\right)C\left(5\text{,}3\right)}{C\left(14\text{,}5\right)}=\frac{15\cdot 10}{2\text{,}002}=\frac{75}{1\text{,}001}[/latex]
- It is often easiest to solve "at least" problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen.When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are [latex]C\left(9,5\right)[/latex] ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are [latex]C\left(14,5\right)[/latex] ways to choose the 5 toys from all of the toys.
[latex]\frac{C\left(9\text{,}5\right)}{C\left(14\text{,}5\right)}=\frac{63}{1\text{,}001}[/latex]If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are [latex]C\left(5,1\right)\cdot C\left(9,4\right)[/latex] ways to choose 1 dog and 1 other toy.[latex]\frac{C\left(5\text{,}1\right)C\left(9\text{,}4\right)}{C\left(14\text{,}5\right)}=\frac{5\cdot 126}{2\text{,}002}=\frac{315}{1\text{,}001}[/latex]Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen.[latex]\frac{63}{1\text{,}001}+\frac{315}{1\text{,}001}=\frac{378}{1\text{,}001}[/latex]We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.[latex]1-\frac{378}{1\text{,}001}=\frac{623}{1\text{,}001}[/latex]
Try It
A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs.- Find the probability that all 3 gumballs selected are purple.
- Find the probability that no yellow gumballs are selected.
- Find the probability that at least 1 yellow gumball is selected.
Answer: [latex]\begin{array}{lll}\text{a}\text{. }\frac{1}{91};\hfill & \text{b}\text{. }\frac{\text{5}}{\text{91}};\hfill & \text{c}\text{. }\frac{86}{91}\hfill \end{array}[/latex]
Key Equations
| probability of an event with equally likely outcomes | [latex]P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}[/latex] |
| probability of the union of two events | [latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)-P\left(E\cap F\right)[/latex] |
| probability of the union of mutually exclusive events | [latex]P\left(E\cup F\right)=P\left(E\right)+P\left(F\right)[/latex] |
| probability of the complement of an event | [latex]P\left(E\text{'}\right)=1-P\left(E\right)[/latex] |
Key Concepts
- Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain.
- The probabilities in a probability model must sum to 1.
- When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment.
- To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously.
- To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events.
- The probability of the complement of an event is the difference between 1 and the probability that the event occurs.
- In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces.
Glossary
complement of an event the set of outcomes in the sample space that are not in the event [latex]E[/latex] event any subset of a sample space experiment an activity with an observable result mutually exclusive events events that have no outcomes in common outcomes the possible results of an experiment probability a number from 0 to 1 indicating the likelihood of an event probability model a mathematical description of an experiment listing all possible outcomes and their associated probabilities sample space the set of all possible outcomes of an experiment union of two events the event that occurs if either or both events occurLicenses & Attributions
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