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Probability

Construct probability models.

Constructing 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 \(\{1,2,3,4,5,6\}.\) An event is any subset of a sample space.

The likelihood of an event is known as probability. The probability of an event \(p\) is a number that always satisfies \(0\le p\le 1,\) 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 .

Outcome	Probability
Winning the raffle	1%
Losing the raffle	99%

The sum of the probabilities listed in a probability model must equal 1, or 100%.

Example

Try it.

Construct a probability model for rolling a single, fair die, with the event being the number shown on the die.

Solution

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 \(\frac{1}{6}.\)

Outcome	Roll of 1	Roll of 2	Roll of 3	Roll of 4	Roll of 5	Roll of 6
Probability	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)

Computing Probabilities of Equally Likely Outcomes

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 \(\frac{4}{6}=\frac{2}{3}.\)

Example

Try it.

A six-sided number cube is rolled. Find the probability of rolling an odd number.

Solution

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.

\[P(E)=\frac{3}{6}=\frac{1}{2}\]

Computing the Probability of the Union of Two Events

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 \(E\ \text{and }F,\text{written }E\cup F,\) is the event that occurs if either or both events occur.

\[P(E\cup F)=P(E)+P(F)-P(E\cap F)\]

Suppose the spinner in is spun. We want to find the probability of spinning orange or spinning a \(b.\)

There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is \(\frac{3}{6}=\frac{1}{2}.\) There are a total of 6 sections, and 2 of them have a \(b.\) So the probability of spinning a \(b\) is \(\frac{2}{6}=\frac{1}{3}.\) 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 \(b.\)

\[\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}\]

The probability of spinning orange or a \(b\) is \(\frac{2}{3}.\)

Example

Try it.

A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.

Solution

A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is \(\frac{1}{4}.\) There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is \(\frac{1}{13}.\)

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 \(\frac{1}{52}.\) Substitute \(P(H)=\frac{1}{4},P(7)=\frac{1}{13},\text{and}P(H\cap 7)=\frac{1}{52}\) into the formula.

\[\begin{array}{l}P(E{\cup }^{\text{}}F)=P(E)+P(F)-P(E{\cap }^{\text{}}F) \\ \ =\frac{1}{4}+\frac{1}{13}-\frac{1}{52} \\ \ =\frac{4}{13}\end{array}\]

The probability of drawing a heart or a 7 is \(\frac{4}{13}.\)

Computing the Probability of Mutually Exclusive Events

Suppose the spinner in 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 when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is

\[P(E\cup F)=P(E)+P(F)\]

Notice that with mutually exclusive events, the intersection of \(E\) and \(F\) is the empty set. The probability of spinning an orange is \(\frac{3}{6}=\frac{1}{2}\) and the probability of spinning a \(d\) is \(\frac{1}{6}.\) We can find the probability of spinning an orange or a \(d\) simply by adding the two probabilities.

\[\begin{array}{l}P(E{\cup }^{\text{}}F)=P(E)+P(F) \\ \ =\frac{1}{2}+\frac{1}{6} \\ \ =\frac{2}{3}\end{array}\]

The probability of spinning an orange or a \(d\) is \(\frac{2}{3}.\)

Example

Try it.

A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.

Solution

The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is \(\frac{1}{4},\) and the probability of drawing a spade is also \(\frac{1}{4},\) so the probability of drawing a heart or a spade is

\[\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\]

Using the Complement Rule to Compute Probabilities

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.

\[P({E}^{'})=1-P(E)\]

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 \(\frac{1}{9},\) the probability of the horse losing the race is simply

\[1-\frac{1}{9}=\frac{8}{9}\]

Example

Try it.

Two six-sided number cubes are rolled.

ⓐ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.

Solution

The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are \(6\times 6,\) or \(\ \text{36 }\) total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.

\(\text{1-1}\)	\(\text{1-2}\)	\(\text{1-3}\)	\(\text{1-4}\)	\(\text{1-5}\)	\(\text{1-6}\)
\(\text{2-1}\)	\(\text{2-2}\)	\(\text{2-3}\)	\(\\) \(\text{2-4}\)	\(\text{2-5}\)	\(\text{2-6}\)
\(\text{3-1}\)	\(\text{3-2}\)	\(\text{3-3}\)	\(\text{3-4}\)	\(\text{3-5}\)	\(\text{3-6}\)
\(\text{4-1}\)	\(\text{4-2}\)	\(\text{4-3}\)	\(\text{4-4}\)	\(\text{4-5}\)	\(\text{4-6}\)
\(\text{5-1}\)	\(\text{5-2}\)	\(\text{5-3}\)	\(\text{5-4}\)	\(\text{5-5}\)	\(\text{5-6}\)
\(\text{6-1}\)	\(\text{6-2}\)	\(\text{6-3}\)	\(\text{6-4}\)	\(\text{6-5}\)	\(\text{6-6}\)

ⓐ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\[\frac{3}{36}=\frac{1}{12}\]
ⓑ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. \[\begin{array}{l}P({E}^{'})=1-P(E) \\ \ =1-\frac{1}{12} \\ \ =\frac{11}{12}\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

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 \(C(5,2)\) 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:

\[\begin{array}{l}\frac{\text{ways to select 2 phones that are not defective}}{\text{ways to select 2 phones}}=\frac{C(5,2)}{C(8,2)} \\ =\frac{10}{28} \\ =\frac{5}{14}\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Key Equations

probability of an event with equally likely outcomes	\(P(E)=\frac{n(E)}{n(S)}\)
probability of the union of two events	\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
probability of the union of mutually exclusive events	\(P(E\cup F)=P(E)+P(F)\)
probability of the complement of an event	\(P(E')=1-P(E)\)

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. See .
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 .
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 .
To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See .
The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See .
In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See .

Constructing Probability Models

Outcome	Probability
Winning the raffle	1%
Losing the raffle	99%

The sum of the probabilities listed in a probability model must equal 1, or 100%.

Example

Try it.

Construct a probability model for rolling a single, fair die, with the event being the number shown on the die.

Solution

Outcome	Roll of 1	Roll of 2	Roll of 3	Roll of 4	Roll of 5	Roll of 6
Probability	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)	\(\frac{1}{6}\)

Computing Probabilities of Equally Likely Outcomes

Example

Try it.

A six-sided number cube is rolled. Find the probability of rolling an odd number.

Solution

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.

\[P(E)=\frac{3}{6}=\frac{1}{2}\]

Computing the Probability of the Union of Two Events

\[P(E\cup F)=P(E)+P(F)-P(E\cap F)\]

Suppose the spinner in is spun. We want to find the probability of spinning orange or spinning a \(b.\)

\[\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}\]

The probability of spinning orange or a \(b\) is \(\frac{2}{3}.\)

Example

Try it.

A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.

Solution

\[\begin{array}{l}P(E{\cup }^{\text{}}F)=P(E)+P(F)-P(E{\cap }^{\text{}}F) \\ \ =\frac{1}{4}+\frac{1}{13}-\frac{1}{52} \\ \ =\frac{4}{13}\end{array}\]

The probability of drawing a heart or a 7 is \(\frac{4}{13}.\)

Computing the Probability of Mutually Exclusive Events

\[P(E\cup F)=P(E)+P(F)\]

\[\begin{array}{l}P(E{\cup }^{\text{}}F)=P(E)+P(F) \\ \ =\frac{1}{2}+\frac{1}{6} \\ \ =\frac{2}{3}\end{array}\]

The probability of spinning an orange or a \(d\) is \(\frac{2}{3}.\)

Example

Try it.

A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.

Solution

\[\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\]

Using the Complement Rule to Compute Probabilities

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.

\[P({E}^{'})=1-P(E)\]

\[1-\frac{1}{9}=\frac{8}{9}\]

Example

Try it.

Two six-sided number cubes are rolled.

ⓐ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.

Solution

\(\text{1-1}\)	\(\text{1-2}\)	\(\text{1-3}\)	\(\text{1-4}\)	\(\text{1-5}\)	\(\text{1-6}\)
\(\text{2-1}\)	\(\text{2-2}\)	\(\text{2-3}\)	\(\\) \(\text{2-4}\)	\(\text{2-5}\)	\(\text{2-6}\)
\(\text{3-1}\)	\(\text{3-2}\)	\(\text{3-3}\)	\(\text{3-4}\)	\(\text{3-5}\)	\(\text{3-6}\)
\(\text{4-1}\)	\(\text{4-2}\)	\(\text{4-3}\)	\(\text{4-4}\)	\(\text{4-5}\)	\(\text{4-6}\)
\(\text{5-1}\)	\(\text{5-2}\)	\(\text{5-3}\)	\(\text{5-4}\)	\(\text{5-5}\)	\(\text{5-6}\)
\(\text{6-1}\)	\(\text{6-2}\)	\(\text{6-3}\)	\(\text{6-4}\)	\(\text{6-5}\)	\(\text{6-6}\)

ⓐ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\[\frac{3}{36}=\frac{1}{12}\]
ⓑ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. \[\begin{array}{l}P({E}^{'})=1-P(E) \\ \ =1-\frac{1}{12} \\ \ =\frac{11}{12}\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Computing Probability Using Counting Theory

\[\begin{array}{l}\frac{\text{ways to select 2 phones that are not defective}}{\text{ways to select 2 phones}}=\frac{C(5,2)}{C(8,2)} \\ =\frac{10}{28} \\ =\frac{5}{14}\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

Key Equations

probability of an event with equally likely outcomes	\(P(E)=\frac{n(E)}{n(S)}\)
probability of the union of two events	\(P(E\cup F)=P(E)+P(F)-P(E\cap F)\)
probability of the union of mutually exclusive events	\(P(E\cup F)=P(E)+P(F)\)
probability of the complement of an event	\(P(E')=1-P(E)\)

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. See .
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 .
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 .
To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See .
The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See .
In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Tossing a coin:
- ⓐ Describe in set notation the sample space of tossing a coin.
- ⓑ Find the probability of “Coin lands on heads.”
Rolling a die:
- ⓐ Describe in set notation the event “Rolling an odd number.”
- ⓑ Find the probability of “Rolling an odd number.”
Drawing a card:
- ⓐ Describe in set notation the event “Drawing an Ace.”
- ⓑ Find the probability of “Drawing an Ace.”
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Tossing a coin:
- ⓐ When you toss a coin, there are two outcomes. The sample space is: Heads, Tails.
- ⓑ There is only one outcome for the coin landing on heads, so P(Lands on Heads)=12.
Rolling a die:
- ⓐ When you roll a dice, there are six outcomes. The event "Rolling an odd number" has three outcomes. The event is set notation is: 1,3,5.
- ⓑ P(rolling an odd number)=3/6=1/2.
Drawing a card:
- ⓐ When you draw a card, there are 52 outcomes. The event "Drawing an Ace" has four outcomes. The event in set notation is:Ace of hearts, Ace of diamonds, Ace of clubs, Ace of spades.
- ⓑ P(Drawing an Ace)=4/52=1/13.
Spinning a dial:
- ⓐ Describe the sample space in set notation.
- ⓑ Find the probability of “Dial stops on a yellow slice.”
- ⓒ Find the probability of “Dial stops on a red slice.”
- ⓓ Find the probability of “Dial stops on a blue slice.”
- Draw a diagram showing the sample space of a standard deck of 52 cards. Begin by distinguishing between red and black cards showing the number of each. Next show the suits: diamonds, hearts, clubs and spades. Below this list the number or face card appearing in each suit. Use your diagram to help you find the following.
- ⓐ Describe in set notation the event “Drawing a king.”
- ⓑ Find the probability of “Drawing a king.”
- ⓒ Describe in set notation the event “Drawing a club.”
- ⓓ Find the probability of “Drawing a club.”
- ⓔ Find the probability of “Drawing a red six.”
- ⓕ Find the probability of “Drawing a black queen.”
Construct a probability model for rolling a single, fair die, with the event being the number shown on the die.

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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 \(\frac{1}{6}.\)
Outcome Roll of 1 Roll of 2 Roll of 3 Roll of 4 Roll of 5 Roll of 6
Probability \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\)
Construct a probability model for tossing a fair coin.

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Outcome Probability
Heads \(\frac{1}{2}\)
Tails \(\frac{1}{2}\)
A six-sided number cube is rolled. Find the probability of rolling an odd number.

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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.
\[P(E)=\frac{3}{6}=\frac{1}{2}\]
A number cube is rolled. Find the probability of rolling a number greater than 2.

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\(\frac{2}{3}\)
A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.

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A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is \(\frac{1}{4}.\) There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is \(\frac{1}{13}.\)
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 \(\frac{1}{52}.\) Substitute \(P(H)=\frac{1}{4},P(7)=\frac{1}{13},\text{and}P(H\cap 7)=\frac{1}{52}\) into the formula.
\[\begin{array}{l}P(E{\cup }^{\text{}}F)=P(E)+P(F)-P(E{\cap }^{\text{}}F) \\ \ =\frac{1}{4}+\frac{1}{13}-\frac{1}{52} \\ \ =\frac{4}{13}\end{array}\]
The probability of drawing a heart or a 7 is \(\frac{4}{13}.\)
A card is drawn from a standard deck. Find the probability of drawing a red card or an ace.

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\(\frac{7}{13}\)
A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.

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The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is \(\frac{1}{4},\) and the probability of drawing a spade is also \(\frac{1}{4},\) so the probability of drawing a heart or a spade is
\[\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\]
A card is drawn from a standard deck. Find the probability of drawing an ace or a king.

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\(\frac{2}{13}\)

Two six-sided number cubes are rolled.

ⓐ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.

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\(\text{1-1}\)	\(\text{1-2}\)	\(\text{1-3}\)	\(\text{1-4}\)	\(\text{1-5}\)	\(\text{1-6}\)
\(\text{2-1}\)	\(\text{2-2}\)	\(\text{2-3}\)	\(\\) \(\text{2-4}\)	\(\text{2-5}\)	\(\text{2-6}\)
\(\text{3-1}\)	\(\text{3-2}\)	\(\text{3-3}\)	\(\text{3-4}\)	\(\text{3-5}\)	\(\text{3-6}\)
\(\text{4-1}\)	\(\text{4-2}\)	\(\text{4-3}\)	\(\text{4-4}\)	\(\text{4-5}\)	\(\text{4-6}\)
\(\text{5-1}\)	\(\text{5-2}\)	\(\text{5-3}\)	\(\text{5-4}\)	\(\text{5-5}\)	\(\text{5-6}\)
\(\text{6-1}\)	\(\text{6-2}\)	\(\text{6-3}\)	\(\text{6-4}\)	\(\text{6-5}\)	\(\text{6-6}\)

ⓐ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\[\frac{3}{36}=\frac{1}{12}\]
ⓑ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. \[\begin{array}{l}P({E}^{'})=1-P(E) \\ \ =1-\frac{1}{12} \\ \ =\frac{11}{12}\end{array}\]

Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10.

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\(\frac{5}{6}\)
A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears.
1. ⓐFind the probability that only bears are chosen.
2. ⓑFind the probability that 2 bears and 3 dogs are chosen.
3. ⓒFind the probability that at least 2 dogs are chosen.
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1. ⓐ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 \(C(6,5)\) ways to choose 5 bears. There are 14 toys, so there are \(C(14,5)\) ways to choose any 5 toys. \[\frac{C(6\text{,}5)}{C(14\text{,}5)}=\frac{6}{2\text{,}002}=\frac{3}{1\text{,}001}\]
2. ⓑ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. \[\frac{C(6\text{,}2)C(5\text{,}3)}{C(14\text{,}5)}=\frac{15⋅10}{2\text{,}002}=\frac{75}{1\text{,}001}\]
3. ⓒ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.
  \[\frac{C(9\text{,}5)}{C(14\text{,}5)}=\frac{63}{1\text{,}001}\]
  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 \(C(5,1)⋅C(9,4)\) ways to choose 1 dog and 1 other toy.
  \[\frac{C(5\text{,}1)C(9\text{,}4)}{C(14\text{,}5)}=\frac{5⋅126}{2\text{,}002}=\frac{315}{1\text{,}001}\]
  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.
  \[\frac{63}{1\text{,}001}+\frac{315}{1\text{,}001}=\frac{378}{1\text{,}001}\]
  We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.
  \[1-\frac{378}{1\text{,}001}=\frac{623}{1\text{,}001}\]
A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs.
1. ⓐFind the probability that all 3 gumballs selected are purple.
2. ⓑFind the probability that no yellow gumballs are selected.
3. ⓒFind the probability that at least 1 yellow gumball is selected.
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\(\begin{array}{lll}\text{a}\text{. }\frac{1}{91}; & \text{b}\text{. }\frac{\text{5}}{\text{91}}; & \text{c}\text{. }\frac{86}{91}\end{array}\)
What term is used to express the likelihood of an event occurring? Are there restrictions on its values? If so, what are they? If not, explain.

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probability; The probability of an event is restricted to values between \(0\) and \(1,\) inclusive of \(0\) and \(1.\)
What is a sample space?
What is an experiment?

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An experiment is an activity with an observable result.
What is the difference between events and outcomes? Give an example of both using the sample space of tossing a coin 50 times.
The union of two sets is defined as a set of elements that are present in at least one of the sets. How is this similar to the definition used for the union of two events from a probability model? How is it different?

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The probability of the union of two events occurring is a number that describes the likelihood that at least one of the events from a probability model occurs. In both a union of sets \(A\ \text{and }B\) and a union of events \(A\text{and}B,\) the union includes either \(A\text{or}B\) 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 \(1.\)
Landing on red
Landing on a vowel

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\(\frac{1}{2}.\)
Not landing on blue
Landing on purple or a vowel

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\(\frac{5}{8}.\)
Landing on blue or a vowel
Landing on green or blue

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\(\frac{1}{2}.\)
Landing on yellow or a consonant
Not landing on yellow or a consonant

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\(\frac{3}{8}.\)
What is the sample space?
Find the probability of tossing two heads.

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\(\frac{1}{4}.\)
Find the probability of tossing exactly one tail.
Find the probability of tossing at least one tail.

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\(\frac{3}{4}.\)
What is the sample space?
Find the probability of tossing exactly two heads.

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\(\frac{3}{8}.\)
Find the probability of tossing exactly three heads.
Find the probability of tossing four heads or four tails.

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\(\frac{1}{8}.\)
Find the probability of tossing all tails.
Find the probability of tossing not all tails.

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\(\frac{15}{16}.\)
Find the probability of tossing exactly two heads or at least two tails.
Find the probability of tossing either two heads or three heads.

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\(\frac{5}{8}.\)

Symbols used here

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Probability

Construct probability models.
Compute probabilities of equally likely outcomes.
Compute probabilities of the union of two events.
Use the complement rule to find probabilities.
Compute probability using counting theory.
Introduction to Sample Spaces and Computing Basic Probabilities.
Draw a diagram showing the sample space of a standard deck of 52 cards. Begin by distinguishing between red and black cards showing the number of each. Next show the suits: diamonds, hearts, clubs and spades. Below this list the number or face card appearing in each suit. Use your diagram to help you find the following.
Identify every outcome.

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

بیشتر در Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates

Outcome	Probability
Heads	\(\frac{1}{2}\)
Tails	\(\frac{1}{2}\)

Outcome	Probability
Heads	\(\frac{1}{2}\)
Tails	\(\frac{1}{2}\)

Outcome	Probability
Heads	\(\frac{1}{2}\)
Tails	\(\frac{1}{2}\)