3.5: Additive and Multiplicative Rules
Learning Objectives
- Calculate probabilities using the Additive Rule.
- Calculate probabilities using the Multiplicative Rule.
- Identify events that are not mutually exclusive and explain how to represent them in a Venn diagram.
- Understand the condition of independence.
Introduction
In this lesson, you will learn how to combine probabilities with the Additive Rule and the Multiplicative Rule. Through the examples in this lesson, it will become clear when to use which rule. You will also be presented with information about mutually exclusive events and independent events.
Venn Diagrams
When the probabilities of certain events are known, we can use these probabilities to calculate the probabilities of their respective unions and intersections. We use two rules, the Additive Rule and the Multiplicative Rule, to find these probabilities. The examples that follow will illustrate how we can do this.
Example: Suppose we have a loaded (unfair) die, and we toss it several times and record the outcomes. We will define the following events:
\begin{align*}& A: {\text{observe an even number}}\end{align*}
\begin{align*}& B: {\text{observe a number less than } 3}\end{align*}
Let us suppose that we have \begin{align*}P(A)=0.4, \ P(B)=0.3,\end{align*}
It is probably best to draw a Venn diagram to illustrate this situation. As you can see, the probability of events \begin{align*}A\end{align*}
Therefore, adding the probabilities together, we get the following:
\begin{align*}P(A \cup B) = P(1)+P(2)+P(4)+P(6)\end{align*}
We have also previously determined the probabilities below:
\begin{align*}P(A)& =P(2)+P(4)+P(6)=0.4\\
P(B)& =P(1)+P(2)=0.3\\
P(A \cap B)& =P(2)=0.1\end{align*}
If we add the probabilities \begin{align*}P(A)\end{align*}
\begin{align*}P(A)+P(B)=P(2)+P(4)+P(6)+P(1)+P(2)\end{align*}
Note that \begin{align*}P(2)\end{align*}
\begin{align*}P(A \cup B)& =P(1)+P(2)+P(4)+P(6)\\
P(A)& =P(2)+P(4)+P(6)\\
P(B)& =P(1)+P(2)\\
P(A \cap B)& =P(2)\\
P(A \cup B)& =P(A)+P(B)-P(A \cap B)\end{align*}
This is the Additive Rule of Probability, which is demonstrated below:
\begin{align*}P(A \cup B)=0.4+0.3-0.1=0.6\end{align*}
What we have shown is that the probability of the union of two events, \begin{align*}A\end{align*}
Additive Rule of Probability
The probability of the union of two events can be obtained by adding the individual probabilities and subtracting the probability of their intersection: \begin{align*}P(A \cup B)=P(A)+P(B)-P(A \cap B)\end{align*}
We can rephrase the definition as follows: The probability that either event \begin{align*}A\end{align*}
Example: Consider the experiment of randomly selecting a card from a deck of 52 playing cards. What is the probability that the card selected is either a spade or a face card?
Our event is defined as follows:
\begin{align*}& E: {\text{card selected is either a spade or a face card}}\end{align*}
There are 13 spades and 12 face cards, and of the 12 face cards, 3 are spades. Therefore, the number of cards that are either a spade or a face card or both is \begin{align*}13 + 9 = 22.\end{align*}
\begin{align*}& C: {\text{card selected is a spade}}\end{align*}
\begin{align*}& D: {\text{card selected is a face card}}\end{align*}
Note that \begin{align*}P(E)=P(C \cup D) = P(C) + P(D)-P(C \cap D)\end{align*}
\begin{align*}P(A \cup B) & = P(A)+P(B)-P(A \cap B)\\
& =\frac{13}{52}+\frac{12}{52}-\frac{3}{52}\\
& =0.250+0.231-0.058\\
& =0.423\\
& =42.3\%\end{align*}
Recall that we are subtracting 0.58 because we do not want to double-count the cards that are at the same time spades and face cards.
Example: If you know that 84.2% of the people arrested in the mid 1990’s were males, 18.3% of those arrested were under the age of 18, and 14.1% were males under the age of 18, what is the probability that a person selected at random from all those arrested is either male or under the age of 18?
First, define the events:
\begin{align*}& A: {\text{person selected is male}}\end{align*}
\begin{align*}& B: {\text{person selected in under 18}}\end{align*}
Also, keep in mind that the following probabilities have been given to us:
\begin{align*}P(A)=0.842 && P(B)=0.183 && P(A \cap B)=0.141\end{align*}
Therefore, the probability of the person selected being male or under 18 is \begin{align*}P(A \cup B)\end{align*}
\begin{align*}P(A\cup B)& =P(A)+P(B)-P(A \cap B)\\
& = 0.842+0.183-0.141\\
& = 0.884\\
& = 88.4\%\end{align*}
This means that 88.4% of the people arrested in the mid 1990’s were either male or under 18. If \begin{align*}A \cap B\end{align*}
The figure below is a Venn diagram of mutually exclusive events. For example, set \begin{align*}A\end{align*} might represent all the outcomes of drawing a card, and set \begin{align*}B\end{align*} might represent all the outcomes of tossing three coins. These two sets have no elements in common.
If the events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are mutually exclusive, then the probability of the union of \begin{align*}A\end{align*} and \begin{align*}B\end{align*} is the sum of the probabilities of \begin{align*}A\end{align*} and \begin{align*}B\end{align*}: \begin{align*}P(A \cup B)=P(A)+P(B)\end{align*}.
Note that since the two events are mutually exclusive, there is no double-counting.
Example: If two coins are tossed, what is the probability of observing at least one head?
First, define the events as follows:
\begin{align*}& A: {\text{observe only one head}}\end{align*}
\begin{align*}& B: {\text{observe two heads}}\end{align*}
Now the probability of observing at least one head can be calculated as shown:
\begin{align*}P(A \cup B)=P(A)+P(B)=0.5+0.25=0.75=75\%\end{align*}
Multiplicative Rule of Probability
Recall from the previous section that conditional probability is used to compute the probability of an event, given that another event has already occurred:
\begin{align*}P(A|B)=\frac{P(A\cap B)}{P(B)}\end{align*}
This can be rewritten as \begin{align*}P(A \cap B) = P(A|B) \bullet P(B)\end{align*} and is known as the Multiplicative Rule of Probability.
The Multiplicative Rule of Probability says that the probability that both \begin{align*}A\end{align*} and \begin{align*}B\end{align*} occur equals the probability that \begin{align*}B\end{align*} occurs times the conditional probability that \begin{align*}A\end{align*} occurs, given that \begin{align*}B\end{align*} has occurred.
Example: In a certain city in the USA some time ago, 30.7% of all employed female workers were white-collar workers. If 10.3% of all workers employed at the city government were female, what is the probability that a randomly selected employed worker would have been a female white-collar worker?
We first define the following events:
\begin{align*}& F: {\text{randomly selected worker who is female}}\end{align*}
\begin{align*}& W: {\text{randomly selected white-collar worker}}\end{align*}
We are trying to find the probability of randomly selecting a female worker who is also a white-collar worker. This can be expressed as \begin{align*}P(F \cap W)\end{align*}.
According to the given data, we have:
\begin{align*}P(F)& =10.3\%=0.103\\ P(W|F)& =30.7\%=0.307\end{align*}
Now, using the Multiplicative Rule of Probability, we get:
\begin{align*}P(F \cap W) = P(F)P(W|F)=(0.103)(0.307)=0.0316=3.16\%\end{align*}
Thus, 3.16% of all employed workers were white-collar female workers.
Example: A college class has 42 students of which 17 are male and 25 are female. Suppose the teacher selects two students at random from the class. Assume that the first student who is selected is not returned to the class population. What is the probability that the first student selected is female and the second is male?
Here we can define two events:
\begin{align*}& F1: {\text{first student selected is female}}\end{align*}
\begin{align*}& M2: {\text{second student selected is male}}\end{align*}
In this problem, we have a conditional probability situation. We want to determine the probability that the first student selected is female and the second student selected is male. To do so, we apply the Multiplicative Rule:
\begin{align*}P(F1 \cap M2) = P(F1)P(M2|F1)\end{align*}
Before we use this formula, we need to calculate the probability of randomly selecting a female student from the population. This can be done as follows:
\begin{align*}P(F1)=\frac{25}{42}=0.595\end{align*}
Now, given that the first student selected is not returned back to the population, the remaining number of students is 41, of which 24 are female and 17 are male.
Thus, the conditional probability that a male student is selected, given that the first student selected was a female, can be calculated as shown below:
\begin{align*}P(M2|F1)=P(M2)=\frac{17}{41}=0.415\end{align*}
Substituting these values into our equation, we get:
\begin{align*}P(F1 \cap M2)=P(F1)P(M2|F1)=(0.595)(0.415)=0.247=24.7\%\end{align*}
We conclude that there is a probability of 24.7% that the first student selected is female and the second student selected is male.
Example: Suppose a coin was tossed twice, and the observed face was recorded on each toss. The following events are defined:
\begin{align*}& A: {\text{first toss is a head}}\end{align*}
\begin{align*}& B: {\text{second toss is a head}}\end{align*}
Does knowing that event \begin{align*}A\end{align*} has occurred affect the probability of the occurrence of \begin{align*}B\end{align*}?
The sample space of this experiment is \begin{align*}S=\left \{HH, HT, TH, TT\right \}\end{align*}, and each of these simple events has a probability of 0.25. So far we know the following information:
\begin{align*}P(A) & = P(HT) + P(HH) = \frac{1}{4} + \frac{1}{4} = 0.5\\ P(B) & = P(TH) + P(HH) = \frac{1}{4} + \frac{1}{4} = 0.5\\ A \cap B & = \left \{\text{HH}\right \}\\ P(A \cap B) & = 0.25\end{align*}
Now, what is the conditional probability? It is as follows:
\begin{align*}P(B|A)& =\frac{P(A \cap B)}{P(A)}\\ & =\frac{\frac{1}{4}}{\frac{1}{2}}\\ & =\frac{1}{2}\end{align*}
What does this tell us? It tells us that \begin{align*}P(B)=\frac{1}{2}\end{align*} and also that \begin{align*}P(B|A)=\frac{1}{2}\end{align*}. This means knowing that the first toss resulted in heads does not affect the probability of the second toss being heads. In other words, \begin{align*}P(B|A)=P(B)\end{align*}.
When this occurs, we say that events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent events.
Independence
If event \begin{align*}B\end{align*} is independent of event \begin{align*}A\end{align*}, then the occurrence of event \begin{align*}A\end{align*} does not affect the probability of the occurrence of event \begin{align*}B\end{align*}. Therefore, we can write \begin{align*}P(B)=P(B|A)\end{align*}.
Recall that \begin{align*}P(B|A)=\frac{P(B \cap A)}{P(A)}\end{align*}. Therefore, if \begin{align*}B\end{align*} and \begin{align*}A\end{align*} are independent, the following must be true:
\begin{align*}P(B|A)=\frac{P(A \cap B)}{P(A)}=P(B)\end{align*}
\begin{align*}P(A \cap B)=P(A) \bullet P(B)\end{align*}
That is, if two events are independent, \begin{align*}P(A \cap B)=P(A) \bullet P(B)\end{align*}.
Example: The table below gives the number of physicists (in thousands) in the US cross-classified by specialty \begin{align*}(P1, P2, P3, P4)\end{align*} and base of practice \begin{align*}(B1, B2, B3)\end{align*}. (Remark: The numbers are absolutely hypothetical and do not reflect the actual numbers in the three bases.) Suppose a physicist is selected at random. Is the event that the physicist selected is based in academia independent of the event that the physicist selected is a nuclear physicist? In other words, is event \begin{align*}B1\end{align*} independent of event \begin{align*}P3\end{align*}?
Academia \begin{align*}(B1)\end{align*} | Industry \begin{align*}(B2)\end{align*} | Government \begin{align*}(B3)\end{align*} | Total | |
---|---|---|---|---|
General Physics \begin{align*}(P1)\end{align*} | 10.3 | 72.3 | 11.2 | 93.8 |
Semiconductors \begin{align*}(P2)\end{align*} | 11.4 | 0.82 | 5.2 | 17.42 |
Nuclear Physics \begin{align*}(P3)\end{align*} | 1.25 | 0.32 | 34.3 | 35.87 |
Astrophysics \begin{align*}(P4)\end{align*} | 0.42 | 31.1 | 35.2 | 66.72 |
Total | 23.37 | 104.54 | 85.9 | 213.81 |
Figure: A table showing the number of physicists in each specialty (thousands). These data are hypothetical.
We need to calculate \begin{align*}P(B1|P3)\end{align*} and \begin{align*}P(B1)\end{align*}. If these two probabilities are equal, then the two events \begin{align*}B1\end{align*} and \begin{align*}P3\end{align*} are indeed independent. From the table, we find the following:
\begin{align*}P(B1)=\frac{23.37}{213.81}=0.109\end{align*}
and
\begin{align*}P(B1|P3)=\frac{P(B1 \cap P3)}{P(P3)}=\frac{1.25}{35.87}=0.035\end{align*}
Thus, \begin{align*}P(B1|P3)\neq P(B1)\end{align*}, and so events \begin{align*}B1\end{align*} and \begin{align*}P3\end{align*} are not independent.
Caution! If two events are mutually exclusive (they have no overlap), they are not independent. If you know that events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} do not overlap, then knowing that \begin{align*}B\end{align*} has occurred gives you information about \begin{align*}A\end{align*} (specifically that \begin{align*}A\end{align*} has not occurred, since there is no overlap between the two events). Therefore, \begin{align*}P(A|B)\neq P(A)\end{align*}.
Lesson Summary
The Additive Rule of Probability states that the union of two events can be found by adding the probabilities of each event and subtracting the intersection of the two events, or \begin{align*}P(A \cup B)= P(A)+P(B)-P(A \cap B)\end{align*}.
If \begin{align*}A \cap B\end{align*} contains no simple events, then \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are mutually exclusive. Mathematically, this means \begin{align*}P(A \cup B)=P(A)+P(B)\end{align*}.
The Multiplicative Rule of Probability states \begin{align*}P(A \cap B)=P(B) \bullet P(A|B)\end{align*}.
If event \begin{align*}B\end{align*} is independent of event \begin{align*}A\end{align*}, then the occurrence of event \begin{align*}A\end{align*} does not affect the probability of the occurrence of event \begin{align*}B\end{align*}. Mathematically, this means \begin{align*}P(B)=P(B|A)\end{align*}. Another formulation of independence is that if the two events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are independent, then \begin{align*}P(A \cap B)=P(A) \bullet P(B)\end{align*}.
Multimedia Links
For an explanation of how to find probabilities using the Multiplicative and Additive Rules with combination notation (1.0), see bullcleo1, Determining Probability (9:42).
For an explanation of how to find the probability of 'and' statements and independent events (1.0), see patrickJMT, Calculating Probability - "And" Statements, Independent Events (8:04).
Review Questions
- Two fair dice are tossed, and the following events are identified: \begin{align*}& A: {\text{sum of the numbers is odd}}\\
& B: {\text{sum of the numbers is 9, 11, or 12}}\end{align*}
- Are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} independent? Why or why not?
- Are events \begin{align*}A\end{align*} and \begin{align*}B\end{align*} mutually exclusive? Why or why not?
- The probability that a certain brand of television fails when first used is 0.1. If it does not fail immediately, the probability that it will work properly for 1 year is 0.99. What is the probability that a new television of the same brand will last 1 year?