When this spinner is spun, the arrow points to one of the colors. Are the outcomes equally likely?
A) Yes, they are equally likely.
B) No, they are not equally likely.
Events and Outcomes (Counting)
RandomUnable to be predicted with certainty. situations are those that cannot be predicted with certainty. However, by using probabilityA measure of how likely it is that something will occur., a measure of how likely it is that a random situation will turn out a particular way, we may still be able to make some predictions about those situations. For example, many games use dice or spinners to generate numbers randomly. If we understand how to calculate probabilities, we can make thoughtful decisions about how to play these games by knowing the likelihood of various outcomes.
First we need to introduce some terms. When working with probability, a random action or series of actions is called a trialA random action or series of actions.. An outcomeA result of a trial. is the result of a trial, and an eventA collection of possible outcomes, often describable using a common characteristic, such as rolling an even number with a die or picking a card from a specific suit. 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 |
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 is impossible. The event has no outcomes in it.) |
Notice that a collection of outcomes is put in brackets and separated by commas.
A simple eventAn event with only one outcome. 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 eventAn event with more than one outcome. 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 we roll a six-sided die many times, we should not expect any outcome to happen more often than another. The outcomes in a situation like this are said to be equally likelyHaving the same likelihood of occurring, such that in a large number of trials, two equally likely outcomes would happen roughly the same number of times.. 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, we would expect to get each outcome `1/6` of the rolls. That is, we'd expect `1/6` of the rolls to be `1`, `1/6` of the rolls to be `2`, `1/6` of the rolls to be `3`, and so on.
|
Example |
|||||||||||||||||
|
Problem |
Tori is flipping a pair of coins and noting how many heads she gets. What are the outcomes in this trial? Are they equally likely? |
||||||||||||||||
|
|
A single coin can turn up heads (H) or tails (T). Tori can flip two heads, two tails, or one of each. There are `3` outcomes: `0` heads, `1` head, or `2` heads.
These outcomes are not equally likely. This might be surprising, but think about it this way: Imagine one coin is a nickel and the other is a dime. The possible ways to flip the coins are:
Notice that there are two ways to get only `1` head, but only one way to get `2` heads and one way to get `0` heads. Tori should expect to get `1` head on `1/2` of the flips, `0` heads on `1/4` of the flips, and `2` heads on `1/4` of the flips. |
|
|||||||||||||||
|
Answer |
There are `3` outcomes, but they are not equally likely. |
|
|||||||||||||||
|
When this spinner is spun, the arrow points to one of the colors. Are the outcomes equally likely?
A) Yes, they are equally likely.
B) No, they are not equally likely.
|
The probability of an event is how often it is expected to occur. When all the possible outcomes for a trial are equally likely, the probability is the ratio of the size of the event spaceThe set of possible outcomes in an event: for example, the event “rolling an even number” on a die has the event space of `2`, `4`, and `6`. (the outcomes in the event) to the sample spaceThe set of all outcomes. (all the possible outcomes for the trial). The probability of an event `E` is usually written `P(E)`.
|
`P(E)=("size of the event space")/("size of the sample space")=("number of outcomes in the event")/("total number of possible outcomes")` |
Notice that an event is often described using common characteristics of the outcomes, if possible—such as rolling an even number on a die. The event space, however, is a list of all the outcomes in the event—such as `{2,4,6}`. The sample space is all possible outcomes, not just the ones in the event.
|
Example |
||||
|
Problem |
A game requires rolling a six-sided die numbered from `1` to `6`. What is the probability of rolling an even number? |
|||
|
|
`"Sample space"={1, 2, 3, 4, 5, 6}`
`"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.”
|
|
|
|
|
|
Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. |
|
|
Answer |
`P"(even number)"=1/2` |
|
|
|
Sometimes the outcomes are not equally likely. In that case, we need to find a way to represent the outcomes so that they are equally likely.
|
Example |
|||||||||||||||||||
|
Problem |
Tori is flipping a pair of coins and noting how many heads she gets. What is the probability of getting `2` heads? What is the probability of getting `1` head? |
||||||||||||||||||
|
|
|
|
First, find the sample space and the event space. Remember that we want the outcomes to be equally likely. In this case, since there are more ways to get `1` head, we can't just list `0`, `1`, and `2` as the outcomes. We have to represent the outcomes so that they are equally likely. |
||||||||||||||||
|
|
Outcomes:
sample space: `{`HH, HT, TH, TT`}` event space for `2` heads: `{`HH`}` event space for `1` head: `{`HT, TH`}` |
|
We can do this by treating each coin differently. For example, if we always toss one coin first, then the other, we can write the result of each coin flip.
|
||||||||||||||||
|
|
`P(2" heads")=P({"HH"})=1/4` `P(1" head")=P({"HT,TH"})=2/4=1/2` |
|
Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. |
||||||||||||||||
|
Answer |
`P(2" heads")=1/4` `P(1" head")=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 big it is. Unless there is reason not to do so, we will express all final probabilities in lowest terms.
|
Find the probability of spinning blue or green on this spinner:
A) `1/6`
B) `1/3`
C) `2`
D) `6`
|
So far we've been looking at trials with a small number of outcomes and events. But trials and events can get complex fast. The following example has a lot more outcomes than we've seen before:
|
Example |
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Problem |
Tori rolled a die and wanted to get either `1` or `3`. James rolled two dice, one blue and one red, and wanted to get both a `1` and a `3`, at the same time. Which event has a greater probability? |
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
|
Tori's sample space: `{1, 2, 3, 4, 5, 6}`
Tori's event space: `{1, 3}`
|
|
First, find the sample space and the event space for the two trials. For Tori's trial, this is easy. |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
|
James' sample space has `36` outcomes
James' event space has `2` outcomes |
|
It's not so obvious for James' trial, since he is rolling two dice. We'll use a chart to find the possibilities. There are `36` outcomes. Of these, there are `2` that have both `1` and `3`. |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
|
Tori: `P(1" or "3)=2/6=1/3`
James: `P(1" or "3)=2/36=1/18` |
|
Since the outcomes are equally likely, the probability of the event is the ratio of event space to sample space. |
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Answer |
Tori's event has a greater probability. |
|
|
||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
As we've seen, when a trial involves more than one random element, such as flipping more than one coin or rolling more than one die, finding the sample space of equally likely outcomes can get complicated. However, we don't need to identify every outcome in the sample space to calculate a probability. We only need the number of outcomes.
The Fundamental Counting PrincipleA way to find the number of outcomes in a sample space by finding the product of the number of outcomes for each element. is a way to find the number of outcomes without listing and counting every one of them.
|
The Fundamental Counting Principle
When a trial consists of more than one random element, the number of outcomes in the sample space is equal to the product of the number of outcomes for each random element.
Examples
|
The Fundamental Counting Principle also lets us see how many choices there are when we are not choosing randomly. 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. The Fundamental Counting Principle says they can make `4*3*2` or `24` different outfits.
Let's check that answer using a tree diagramA diagram that shows the choices or random outcomes from multiple elements, using branches for each new element.. A tree diagram has a branch for every possible outcome for each event. To save space, let's use B for black, W for white, G for green, P for purple, and Y for yellow.
Look at the diagram for a moment. There are three choices of pants. For each of those, there are four choices of shirt. That means there are `3*4` pants-shirt combinations. For each of those `12` combinations, there are two choices of shoes. That gives `3*4*2` pants-shirt-shoes combinations, so there are `24` possible outfits.
Suppose we choose pants, shirt, and shoes completely randomly from those choices—that is, we're as likely to choose any color pants, any color shirt, and either color shoes. There are `8` outfits in which the pants and shoes are the same color (black pants and shoes with any of the four shirts, or white pants and shoes with any of the four shirts). The probability that the pants and shoes are the same color would be:
`P("pants and shoes same color")=8/24=1/3`
|
Example |
|||
|
Problem |
Barry volunteers at a charity walk to make lunches for all the other volunteers. In each bag he puts:
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? |
||
|
|
Size of sample space:
(number of sandwich choices) `*` (number of chip choices) `*` (number of fruit choices) `= 2*3*2=12`
|
|
First, use the Fundamental Counting Principle to find the size of the sample space. We don't need to find all the outcomes, just how many there are. |
|
|
Size of event space:
(number of sandwich choices in event) `*` (number of chip choices in event) `*` (number of fruit choices in event) `= 1*3*1=3` |
|
For the event space, we 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. |
|
Answer |
`P("peanut butter and jelly and apple")=` `"size of event space"/"size of sample space"=3/12=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?
A) `1/16`
B) `1/8`
C) `1/4`
D) `1/2`
|
Probability helps us 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)="size of the event space"/"size of the sample space"="number of outcomes in the event"/"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:
When a trial consists of more than one random element, the number of outcomes in the sample space is equal to the product of the number of outcomes for each random element.
