| title | layout |
|---|---|
Probability |
page |
{: #CNX_Precalc_Figure_11_07_001}
Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one in [link]. 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.
Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment{: data-type="term"}, or an activity with an observable result. The numbers on the cube are possible results, or outcomes{: data-type="term"}, of this experiment. The set of all possible outcomes of an experiment is called the sample space{: data-type="term"} of the experiment. The sample space for this experiment is {1,2,3,4,5,6 }.
An event{: data-type="term"} is any subset of a sample space.
The likelihood of an event is known as probability{: data-type="term"}. The probability of an event p
is a number that always satisfies 0≤p≤1,
where 0 indicates an impossible event and 1 indicates a certain event. A probability model{: data-type="term"} 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 [link].
| Outcome | Probability | |---------- | Winning the raffle | 1% | | Losing the raffle | 99% | {: #Table_11_07_01 summary=".."}
The sum of the probabilities listed in a probability model must equal 1, or 100%.
- Identify every outcome.
- Determine the total number of possible outcomes.
- Compare each outcome to the total number of possible outcomes. {: type="1"}
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 1 6 .
| Outcome | Roll of 1 | Roll of 2 | Roll of 3 | Roll of 4 | Roll of 5 | Roll of 6 | | Probability | 1 6
| 1 6
| 1 6
| 1 6
| 1 6
| 1 6
| {: #Table_11_07_02 summary=".."}
**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.
| | Tails | 12
| {: .unnumbered summary=".." data-label=""}
Let S
be a sample space for an experiment. When investigating probability, an event is any subset of S.
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 S.
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 S,
so the probability of the event is 4 6 = 2 3 .
in an experiment with sample space S
with equally likely outcomes is given by
is a subset of S,
so it is always true that 0≤P(E)≤1.
We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events{: data-type="term"} E and F,written E∪F,
is the event that occurs if either or both events occur.
Suppose the spinner in [link] is spun. We want to find the probability of spinning orange or spinning a b.
{: #Figure_11_07_002}
There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is 3 6 = 1 2 .
There are a total of 6 sections, and 2 of them have a b.
So the probability of spinning a b
is 2 6 = 1 3 .
If we added these two probabilities, we would be counting the sector that is both orange and a b
twice. To find the probability of spinning an orange or a b,
we need to subtract the probability that the sector is both orange and has a b.
The probability of spinning orange or a b
is 2 3 .
and F
(written E∪F
) equals the sum of the probability of E
and the probability of F
minus the probability of E
and F
occurring together (
which is called the intersection{: data-type="term" .no-emphasis} of E
and F
and is written as E∩F
).
There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is 1 13 .
The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is 1 52 .
Substitute P(H)= 1 4 , P(7)= 1 13 , and P(H∩7)= 1 52
into the formula.
Suppose the spinner in [link] is spun again, but this time we are interested in the probability of spinning an orange or a d.
There are no sectors that are both orange and contain a d,
so these two events have no outcomes in common. Events are said to be mutually exclusive events{: data-type="term"} when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is
Notice that with mutually exclusive events, the intersection of E
and F
is the empty set. The probability of spinning an orange is 3 6 = 1 2
and the probability of spinning a d
is 1 6 .
We can find the probability of spinning an orange or a d
simply by adding the two probabilities.
The probability of spinning an orange or a d
is 2 3 .
is given by
- Determine the total number of outcomes for the first event.
- Find the probability of the first event.
- Determine the total number of outcomes for the second event.
- Find the probability of the second event.
- Add the probabilities. {: type="1"}
and the probability of drawing a spade is also 1 4 ,
so the probability of drawing a heart or a spade is
We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event E,
denoted E ′ ,
is the set of outcomes in the sample space that are not in E.
For example, suppose we are interested in the probability that a horse will lose a race. If event W
is the horse winning the race, then the complement of event W
is the horse losing the race.
To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1.
The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is 1 9 ,
the probability of the horse losing the race is simply
- Find the probability that the sum of the numbers rolled is less than or equal to 3.
- Find the probability that the sum of the numbers rolled is greater than 3. {: type="a"}
or 36
total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.
| 1-1
| 1-2
| 1-3
| 1-4
| 1-5
| 1-6
| | 2-1
| 2-2
| 2-3
|
2-4
| 2-5
| 2-6
| | 3-1
| 3-2
| 3-3
| 3-4
| 3-5
| 3-6
| | 4-1
| 4-2
| 4-3
| 4-4
| 4-5
| 4-6
| | 5-1
| 5-2
| 5-3
| 5-4
| 5-5
| 5-6
| | 6-1
| 6-2
| 6-3
| 6-4
| 6-5
| 6-6
| {: #eip-id1165135680167 summary=".."}
-
We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is
3 36 = 1 12 -
Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3.
P( E ′ )=1−P(E) =1− 1 12 = 11 12
{: type="a"}
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 C(5,2)
ways to select 2 phones that are not defective. There are 8 phones, so there are C(8,2)
ways to select 2 phones. The probability of selecting 2 phones that are not defective is:
- 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. {: type="a"}
ways to choose 5 bears. There are 14 toys, so there are
<math xmlns="http://www.w3.org/1998/Math/MathML"> <mrow> <mtext> </mtext><mi>C</mi><mo stretchy="false">(</mo><mn>14</mn><mo>,</mo><mn>5</mn><mo stretchy="false">)</mo><mtext> </mtext> </mrow> </math>
ways to choose any 5 toys.
<div data-type="equation" id="eip-id1165135359779" class="unnumbered" data-label="">
<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"> <mrow> <mtext> </mtext><mfrac> <mrow> <mi>C</mi><mo stretchy="false">(</mo><mn>6</mn><mtext>,</mtext><mn>5</mn><mo stretchy="false">)</mo> </mrow> <mrow> <mi>C</mi><mo stretchy="false">(</mo><mn>14</mn><mtext>,</mtext><mn>5</mn><mo stretchy="false">)</mo> </mrow> </mfrac> <mo>=</mo><mfrac> <mn>6</mn> <mrow> <mn>2</mn><mtext>,</mtext><mn>002</mn> </mrow> </mfrac> <mo>=</mo><mfrac> <mn>3</mn> <mrow> <mn>1</mn><mtext>,</mtext><mn>001</mn> </mrow> </mfrac> <mtext> </mtext> </mrow> </math>
</div>
-
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 C(6,2)
ways to choose 2 bears. There are 5 dogs, so there are C(5,3)
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 C(6,2)⋅C(5,3)
ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
C(6,2)C(5,3) C(14,5) = 15⋅10 2,002 = 75 1,001 -
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 C(9,5)
ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are C(14,5)
ways to choose the 5 toys from all of the toys.
C(9,5) C(14,5) = 63 1,001If 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 C(5,1)⋅C(9,4)
ways to choose 1 dog and 1 other toy.
C(5,1)C(9,4) C(14,5) = 5⋅126 2,002 = 315 1,001Because 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.
63 1,001 + 315 1,001 = 378 1,001We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.
1− 378 1,001 = 623 1,001
{: type="a"}
- 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. {: type="a"}
Visit this website for additional practice questions from Learningpod.
| probability of an event with equally likely outcomes | P(E)= n(E) n(S)
| | probability of the union of two events | P(E∪F)=P(E)+P(F)−P(E∩F)
| | probability of the union of mutually exclusive events | P(E∪F)=P(E)+P(F)
| | probability of the complement of an event | P(E')=1−P(E)
| {: #eip-id1165134166592 summary=".."}
- 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. See [link].
- 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. See [link].
- 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. See [link].
- To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See [link].
- The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See [link].
- In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See [link].
and 1,
inclusive of 0
and 1.
and a union of events A and B,
the union includes either A or B
or both. The difference is that a union of sets results in another set, while the union of events is a probability, so it is always a numerical value between 0
and 1.
For the following exercises, use the spinner shown in [link] to find the probabilities indicated.
{: #CNX_Precalc_Figure_11_07_201}
For the following exercises, two coins are tossed.
For the following exercises, four coins are tossed.
For the following exercises, one card is drawn from a standard deck of 52
cards. Find the probability of drawing the following:
For the following exercises, two dice are rolled, and the results are summed.
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | (1, 1) 2 |
(1, 2) 3 |
(1, 3) 4 |
(1, 4) 5 |
(1, 5) 6 |
(1, 6) 7 |
| 2 | (2, 1) 3 |
(2, 2) 4 |
(2, 3) 5 |
(2, 4) 6 |
(2, 5) 7 |
(2, 6) 8 |
| 3 | (3, 1) 4 |
(3, 2) 5 |
(3, 3) 6 |
(3, 4) 7 |
(3, 5) 8 |
(3, 6) 9 |
| 4 | (4, 1) 5 |
(4, 2) 6 |
(4, 3) 7 |
(4, 4) 8 |
(4, 5) 9 |
(4, 6) 10 |
| 5 | (5, 1) 6 |
(5, 2) 7 |
(5, 3) 8 |
(5, 4) 9 |
(5, 5) 10 |
(5, 6) 11 |
| 6 | (6, 1) 7 |
(6, 2) 8 |
(6, 3) 9 |
(6, 4) 10 |
(6, 5) 11 |
(6, 6) 12 |
or greater than 9.
and 9,
inclusive.
or 6.
or 6.
For the following exercises, a coin is tossed, and a card is pulled from a standard deck. Find the probability of the following:
For the following exercises, use this scenario: a bag of M&Ms contains 12
blue, 6
brown, 10
orange, 8
yellow, 8
red, and 4
green M&Ms. Reaching into the bag, a person grabs 5 M&Ms.
blue M&Ms?
blue M&Ms?
Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20
numbers from the numbers 1
to 80.
After the player makes his selections, 20
winning numbers are randomly selected from numbers 1
to 80.
A win occurs if the player has correctly selected 3,4,
or 5
of the 20
winning numbers. (Round all answers to the nearest hundredth of a percent.)
Use this data for the exercises that follow: In 2013, there were roughly 317 million citizens in the United States, and about 40 million were elderly (aged 65 and over).2{: data-type="footnote-link"}
Sequences and Their Notation{: .target-chapter}
Arithmetic Sequences{: .target-chapter}
arithmetic? If so, find the common difference.
arithmetic? If so, find the common difference.
and common difference d=−8.
What are the first five terms?
and a 8 =−14.6.
What is the first term?
and then find the 31st term.
Geometric Sequences{: .target-chapter}
geometric? If so find the common ratio. If not, explain why.
and a 9 =262,144 .
What are the first five terms?
and common ratio r= 1 2 .
What is the 8th term?
Series and Their Notation{: .target-chapter}
from m=0
to m=5.
twenty times.
terms of an arithmetic series to find the sum of the first eleven terms of the arithmetic series 2.5, 4, 5.5, … .
tapered rungs, the lengths of which increase by a common difference. The first rung is 5 inches long, and the last rung is 20 inches long. What is the sum of the lengths of the rungs?
for the series 12, 6, 3, 3 2 ,…
| Year | Membership Fees |
|---|---|
| 1 | $1500 |
| 2 | $1950 |
| 3 | $2535 |
the height of the previous bounce. Write a series representing the total distance traveled by the ball, assuming it was initially dropped from a height of 5 feet. What is the total distance? (Hint: the total distance the ball travels on each bounce is the sum of the heights of the rise and the fall.)
years old? How much more?
Counting Principles{: .target-chapter}
that is divisible by either 4
or 6?
musicians, 12
play piano, 7
play trumpet, and 2
play both piano and trumpet. How many musicians play either piano or trumpet?
freshman, 10
sophomores, 3
juniors, and 2
seniors, how many ways can a president, vice president, and treasurer be elected?
have?
Binomial Theorem{: .target-chapter}
without fully expanding the binomial.
Probability{: .target-chapter}
For the following exercises, assume two die are rolled.
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 1, 1 | 1, 2 | 1, 3 | 1, 4 | 1, 5 | 1, 6 |
| 2 | 2, 1 | 2, 2 | 2, 3 | 2, 4 | 2, 5 | 2, 6 |
| 3 | 3, 1 | 3, 2 | 3, 3 | 3, 4 | 3, 5 | 3, 6 |
| 4 | 4, 1 | 4, 2 | 4, 3 | 4, 4 | 4, 5 | 4, 6 |
| 5 | 5, 1 | 5, 2 | 5, 3 | 5, 4 | 5, 5 | 5, 6 |
| 6 | 6, 1 | 6, 2 | 6, 3 | 6, 4 | 6, 5 | 6, 6 |
For the following exercises, use the following data: An elementary school survey found that 350 of the 500 students preferred soda to milk. Suppose 8 children from the school are attending a birthday party. (Show calculations and round to the nearest tenth of a percent.)
arithmetic? If so find the common difference.
and common difference d=– 4 3 .
What is the 6th term?
and then find the 22nd term.
and then find the 32nd term.
geometric? If so find the common ratio. If not, explain why.
from k=−3
to k=15.
terms of a geometric series to find ∑ k=1 7 −0.2⋅ ( −5 ) k−1 .
Interest earned: $14,355.75
without fully expanding the binomial.
For the following exercises, use the spinner in [link].
{: #CNX_Precalc_Figure_11_07_202}
- {: data-type="footnote-ref" #footnote1} 1{: data-type="footnote-ref-link"} The figure is for illustrative purposes only and does not model any particular storm.
- {: data-type="footnote-ref" #footnote2} 2{: data-type="footnote-ref-link"} United States Census Bureau. http://www.census.gov {: data-list-type="bulleted" data-bullet-style="none"}
complement of an event : the set of outcomes in the sample space that are not in the event E ^
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 occur