Common Core: High School - Statistics and Probability : Apply and Interpret the Addition Rule: CCSS.Math.Content.HSS-CP.B.7

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Example Questions

Example Question #1 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

What is the probability of choosing a king OR a spade from a regular deck of cards?

Possible Answers:

\(\displaystyle \frac{13}{52}=\frac{1}{4}\)

\(\displaystyle \frac{7}{52}\)

\(\displaystyle \frac{17}{52}\)

\(\displaystyle \frac{4}{52}=\frac{1}{13}\)

\(\displaystyle \frac{16}{52}=\frac{4}{13}\)

Correct answer:

\(\displaystyle \frac{16}{52}=\frac{4}{13}\)

Explanation:

 

 

 

Example Question #1 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

Screen shot 2016 03 29 at 8.35.06 pm

Possible Answers:

\(\displaystyle 70.06\%\)

\(\displaystyle 110.71\%\)

\(\displaystyle 76.07\%\)

\(\displaystyle 60.71\%\)

Cannot be determined

Correct answer:

\(\displaystyle 76.07\%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events, we can simply add the probabilities, which can be illustrated by the following figure. In this figure, each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die. 

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and, at times, intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(V8\cup Manual)=P(V8)+P(Manual)-P(V8\cap Manual)\)

\(\displaystyle P(V8\cup Manual)=\frac{170}{280}+\frac{140}{280}-\frac{97}{280}\)

\(\displaystyle P(V8\cup Manual)=0.6071+0.5000-0.3464\)

\(\displaystyle P(V8\cup Manual)=0.7607\)

\(\displaystyle P(V8\cup Manual)=0.7607\times 100\%=76.07\%\)

Example Question #2 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

Screen shot 2016 03 29 at 8.35.06 pm

Possible Answers:

\(\displaystyle 70.06\%\)

Cannot be determined

\(\displaystyle 60.71\%\)

\(\displaystyle 76.07\%\)

\(\displaystyle 110.71\%\)

Correct answer:

\(\displaystyle 76.07\%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(V8\cup Manual)=P(V8)+P(Manual)-P(V8\cap Manual)\)

\(\displaystyle P(V8\cup Manual)=\frac{170}{280}+\frac{140}{280}-\frac{97}{280}\)

\(\displaystyle P(V8\cup Manual)=0.6071+0.5000-0.3464\)

\(\displaystyle P(V8\cup Manual)=0.7607\)

\(\displaystyle P(V8\cup Manual)=0.7607\times 100\%=76.07\%\)

Example Question #3 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 71 & 190 & 261 \\ \hline V8 & 88 & 66 & 154 \\ \hline Total & 154 & 261 & 415 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 24.7 \%\)

\(\displaystyle 98.8 \%\)

\(\displaystyle 49.4 \%\)

\(\displaystyle 9.88 \%\)

\(\displaystyle 74.1 \%\)

Correct answer:

\(\displaystyle 98.8 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{154}{415} + \frac{256}{415} - \frac{66}{415}\)

\(\displaystyle P(V8 \cup Manual)= 0.3711 + 0.6169 - 0.159\)

\(\displaystyle P(V8 \cup Manual)= 0.988\)

\(\displaystyle P(V8 \cup Manual)= 0.988 \times 100 \%= 98.8 \%\)

 

Example Question #73 : Conditional Probability & The Rules Of Probability

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 118 & 59 & 177 \\ \hline V8 & 95 & 167 & 262 \\ \hline Total & 262 & 177 & 439 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 11.12 \%\)

\(\displaystyle 83.37 \%\)

\(\displaystyle 27.79 \%\)

\(\displaystyle 111.16 \%\)

\(\displaystyle 55.58 \%\)

Correct answer:

\(\displaystyle 111.16 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{262}{439} + \frac{226}{439} - \frac{167}{439}\)

\(\displaystyle P(V8 \cup Manual)= 0.5968 + 0.5148 - 0.3804\)

\(\displaystyle P(V8 \cup Manual)= 1.1116\)

\(\displaystyle P(V8 \cup Manual)= 1.1116 \times 100 \%= 111.16 \%\)

 

 

Example Question #74 : Conditional Probability & The Rules Of Probability

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 80 & 113 & 193 \\ \hline V8 & 74 & 107 & 181 \\ \hline Total & 181 & 193 & 374 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 26.8 \%\)

\(\displaystyle 10.72 \%\)

\(\displaystyle 107.22 \%\)

\(\displaystyle 80.42 \%\)

\(\displaystyle 53.61 \%\)

Correct answer:

\(\displaystyle 107.22 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{181}{374} + \frac{10}{17} - \frac{107}{374}\)

\(\displaystyle P(V8 \cup Manual)= 0.484 + 0.5882 - 0.2861\)

\(\displaystyle P(V8 \cup Manual)= 1.0722\)

\(\displaystyle P(V8 \cup Manual)= 1.0722 \times 100 \%= 107.22 \%\)

 

 

Example Question #5 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 62 & 190 & 252 \\ \hline V8 & 60 & 58 & 118 \\ \hline Total & 118 & 252 & 370 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 74.19 \%\)

\(\displaystyle 24.73 \%\)

\(\displaystyle 9.89 \%\)

\(\displaystyle 49.46 \%\)

\(\displaystyle 98.92 \%\)

Correct answer:

\(\displaystyle 98.92 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{59}{185} + \frac{124}{185} - \frac{29}{185}\)

\(\displaystyle P(V8 \cup Manual)= 0.3189 + 0.6703 - 0.1568\)

\(\displaystyle P(V8 \cup Manual)= 0.9892\)

\(\displaystyle P(V8 \cup Manual)= 0.9892 \times 100 \%= 98.92 \%\)

 

Example Question #6 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 194 & 110 & 304 \\ \hline V8 & 98 & 157 & 255 \\ \hline Total & 255 & 304 & 559 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 23.34 \%\)

\(\displaystyle 9.34 \%\)

\(\displaystyle 70.03 \%\)

\(\displaystyle 46.69 \%\)

\(\displaystyle 93.38 \%\)

Correct answer:

\(\displaystyle 93.38 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{255}{559} + \frac{267}{559} - \frac{157}{559}\)

\(\displaystyle P(V8 \cup Manual)= 0.4562 + 0.4776 - 0.2809\)

\(\displaystyle P(V8 \cup Manual)= 0.9338\)

\(\displaystyle P(V8 \cup Manual)= 0.9338 \times 100 \%= 93.38 \%\)

 

Example Question #7 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 175 & 179 & 354 \\ \hline V8 & 137 & 90 & 227 \\ \hline Total & 227 & 354 & 581 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 42.69 \%\)

\(\displaystyle 85.37 \%\)

\(\displaystyle 64.03 \%\)

\(\displaystyle 8.54 \%\)

\(\displaystyle 21.34 \%\)

Correct answer:

\(\displaystyle 85.37 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{227}{581} + \frac{269}{581} - \frac{90}{581}\)

\(\displaystyle P(V8 \cup Manual)= 0.3907 + 0.463 - 0.1549\)

\(\displaystyle P(V8 \cup Manual)= 0.8537\)

\(\displaystyle P(V8 \cup Manual)= 0.8537 \times 100 \%= 85.37 \%\)

 

Example Question #8 : Apply And Interpret The Addition Rule: Ccss.Math.Content.Hss Cp.B.7

Given the following data, what is the probability that a car will have a V8 or a manual transmission?

\(\displaystyle \begin{tabular}{|c|c|c|c|} \hline ~ & Automatic & Manual & Total \\ \hline V6 & 80 & 108 & 188 \\ \hline V8 & 166 & 65 & 231 \\ \hline Total & 231 & 188 & 419 \\ \hline \end{tabular}\)

Possible Answers:

\(\displaystyle 9.64 \%\)

\(\displaystyle 96.42 \%\)

\(\displaystyle 24.1 \%\)

\(\displaystyle 48.21 \%\)

\(\displaystyle 72.31 \%\)

Correct answer:

\(\displaystyle 96.42 \%\)

Explanation:

In order to solve this problem, we need to discuss probabilities and more specifically probabilities of disjoint and non-disjoint events. We will start with discussing probabilities in a general sense. A probability is generally defined as the chances or likelihood of an event occurring. It is calculated by identifying two components: the event and the sample space. The event is defined as the favorable outcome or success that we wish to observe. On the other hand, the sample space is defined as the set of all possible outcomes for the event. Mathematically we calculate probabilities by dividing the event by the sample space:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

Let's use a simple example: the rolling of a die. We want to know the probability of rolling a one. We know that the sample space is six because there are six sides or outcomes to the die. Also, we know that there is only a single side with a value of one; therefore,

\(\displaystyle P=\frac{1}{6}\)

Now, let's convert this into a percentage:

\(\displaystyle \frac{1}{6}=0.1666\)

\(\displaystyle 0.1666\times100\%=16.66\%\)

Probabilities expressed in fraction form will have values between zero and one. One indicates that an event will definitely occur, while zero indicates that an event will not occur. Likewise, probabilities expressed as percentages possess values between zero and one hundred percent where probabilities closer to zero are unlikely to occur and those close to one hundred percent are more likely to occur. 

\(\displaystyle 0< P< 1 \textup{ or }0\%< P< 100\%\)

Now that we understand the definition of a probability in its most general sense, we can investigate disjoint and non-disjoint probabilities. We will begin by investigating how we calculate the probability of disjoint or mutually exclusive events. If events are referred to as disjoint or mutually exclusive, then they are independent of one another. In order to calculate the probability of two disjoint events, A and B, we need to calculate the probability of event A occurring and add it to the probability that event B will occur. This is formally written using the following equation:

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

In the case of disjointed events we can simply add the probabilities which can be illustrated by the following figure. In this figure each event is independent of the other.

Disjoint

Let's look at an example to illustrate this concept. Suppose someone wants to calculate rolling a five or a six on a die. In this problem, the word "or" indicates that we need to compute the probability of both; however, we need to know if the events are disjointed or non-disjointed. Disjointed events are independent of one another or mutually exclusive such as the dice rolls in this example. Let's calculate the probabilities of rolling any particular number on a die.  

Screen shot 2016 03 29 at 3.19.43 pm

We can see that the probability of rolling any value is one out of six. Now, let's use the formula to solve for the probability of rolling a five or a six.

\(\displaystyle P (A\cup B) = P(A)+P(B)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = P(Rolling\ a\ 5)+P(Rolling\ a\ 6)\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{6}+\frac{1}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{2}{6}\)

\(\displaystyle P (Rolling\ a\ 5\cup Rolling\ a\ 6) = \frac{1}{3} \textup{ or }33.{} \bar{3} \%\)

Next, we need to discuss how to calculate the probabilities of non-disjoint or non-mutually exclusive events. If events are described in this way, then they can intersect at some point. For example, physical traits are not exclusive of each other. A person can have brown eyes, brown hair, or they may have brown eyes and brown hair at the same time. In other words, just because someone has brown eyes does not mean that they cannot have brown hair (i.e. it could be a variety of colors). Likewise, brown hair does not limit a person's eye color to a shade other than brown; thus, the traits may intersect. This example can be illustrated in the following figure. In this figure, events A and B are not exclusive of one another and at times intersect. 

 

Non disjoint

When events, A and B, are non-disjoint we can calculate their probability by adding the probability that event A will occur to the probability that event B will occur and then subtract the intersection of both events A and B. This is formally written using the following formula:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

Now let's use this information to solve an example problem. Given the following information what is the probability of a truck being royal blue or having a V8 engine?

Screen shot 2016 03 29 at 7.53.59 pm

First, we know that the key word "or" indicates that we are dealing with disjoint or non-disjoint probabilities; however, we need to determine whether the events of our scenario are mutually exclusive or not. We know that they are non-disjoint because a truck can be both royal blue and have a V8 engine, which represents an intersection of the two events. Let's begin by calculating the probability that a truck is royal blue:

\(\displaystyle P=\frac{\textup{event: particular phenomenon we wish to observe}}{\textup{sample space: total number of possible outcomes}}\)

\(\displaystyle P(Royal\ Blue)=\frac{150}{280}\)

\(\displaystyle P(Royal\ Blue)=0.5357\)

Now, we can calculate the probability that a truck will have a V8 engine.

\(\displaystyle P(V8)=\frac{170}{280}\)

\(\displaystyle P(V8)=0.6071\)

At this point, an example of a common mistake can be illustrated. If we were to treat these events as mutually exclusive then we would follow the formula that simply adds together the probabilities of the two events. If we add together these probabilities, then we would obtain the following value:

\(\displaystyle P(Royal\ Blue\ or\ V8)=0.5357+0.6071\)

\(\displaystyle P(Royal\ Blue\ or\ V8)=1.1428\)

This answer would obviously be incorrect because a probability cannot be greater than one; therefore, we need to subtract the probability of the intersection of the two events. Now, we need to calculate the value of this intersection.

\(\displaystyle P(Royal\ Blue\cap V8)=\frac{84}{280}\)

\(\displaystyle P(Royal\ Blue\cap V8)=0.3000\)

Now, we can create an equation to calculate the probability of the non-mutually exclusive events:

\(\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

\(\displaystyle P(Royal\ Blue\cup V8)=P(Royal\ Blue)+P(V8)-P(Royal\ Blue\cap V8)\)

Substitute in values and solve.

\(\displaystyle P(Royal\ Blue\cup V8)=0.5357+0.6071-0.3000\)

\(\displaystyle P(Royal\ Blue\cup V8)=0.8428\)

Let's use this information to solve the question. We need to find the probability that a car will have a V8 or a manual transmission. First, we need to determine whether or not the events are mutually exclusive. We know that the events are non-disjoint because they intersect (i.e. a car may have both a manual and a V8 engine). Next, we need to create a formula to solve for the probability.

\(\displaystyle \\ \\ P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ P(V8 \cup Manual)=P(V8)+P(Manual)-P(V8 \cap Manual) \\ \\ P(V8 \cup Manual)= \frac{231}{419} + \frac{173}{419} - \frac{65}{419}\)

\(\displaystyle P(V8 \cup Manual)= 0.5513 + 0.4129 - 0.1551\)

\(\displaystyle P(V8 \cup Manual)= 0.9642\)

\(\displaystyle P(V8 \cup Manual)= 0.9642 \times 100 \%= 96.42 \%\) 

 

 

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