<meta http-equiv="refresh" content="1; url=/nojavascript/"> Additive and Multiplicative Rules | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Probability and Statistics - Advanced Go to the latest version.

3.5: Additive and Multiplicative Rules

Created by: CK-12
0  0  0

Learning Objectives

• Calculate probabilities using the additive rule.
• Calculate probabilities using the multiplicative rule.
• Identify events that are not mutually exclusive and how to represent them in a Venn diagram.
• Understand the condition of independence.

When the probabilities of certain events are known, we can use those probabilities to calculate the probabilities of their respective unions and intersections. We use two rules: the additive and the multiplicative rules to find those probabilities. The examples that follow will illustrate how we can do so.

Example:

Suppose we have a loaded (unfair) die. We toss it several times and record the outcomes. If we define the following events:

$& \text{A:} \left \{ \text{observe an even number}\right \}\\& \text{B:} \left \{\text{observe a number less than 3}\right \}$

Let us suppose that we have come up with $P(A) = 0.4, P(B) = 0.3,$ and $P(A \cap B) = 0.1$. We want to find $P(A \cup B)$.

Solution:

It is probably best to draw the Venn diagram to illustrate the situation. As you can see, the probability of the events $A$ and $B$ occurring is the union of the individual probabilities in each event.

Therefore,

$P(A \cup B) = P(1) + P(2) + P(4) + P(6)$

Since

$P(A) & = P(2) + P(4) + P(6) = 0.4\\P(B) & = P(1) + P(2) = 0.3\\P(A \cap B) & = P(2) = 0.1$

If we add the probabilities of $P(A)$ and $P(B)$, we get

$P(A) + P(B) = P(2) + P(4) + P(6) + P(1) + P(2)$

But since

$P(A \cup B) = P(1) + P(2) + P(4) + P(6)$

Substituting, yields

$P(A) + P(B) = P(A \cup B) + P(2)$

However, $P(2) = P(A \cap B)$, thus

$P(A) + P(B) = P(A \cup B) + P(A \cap B)$

Or,

$P(A \cup B) & = P(A) + P(B) - P(A \cap B) && \\& = 0.4 + 0.3 - 0.1 = 0.6$

What we have demonstrated is that the probability of the union of two events, $A$ and $B$, can be obtained by adding the individual probabilities of $A$ and $B$ and subtracting $P(A \cap B)$. The Venn diagram above illustrates this union. Formula (1) above is called the Additive Rule of Probability.

Additive Rule of Probability

The union of two events, $A$ and $B$, can be obtained by adding the individual probabilities of $A$ and $B$ and subtracting $P(A \cap B)$. The Venn diagram above illustrates this union.

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

We can rephrase the definition as follows: The probability that either event $A$ or event $B$ occurs is equal to the probability that event $A$ occurs plus the probability that event $B$ occurs minus the probability that both occur.

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?

Solution:

Our event is

$\text{E} = \left \{\text{the card selected is either a spade or a face card} \right \}$

The event $E$ consists of $22\;\mathrm{cards}$; namely, $13$ spade cards and $9$ face cards that are not spade. Be careful, if we say that we have $12$ face cards, we would be over counting the face-spade cards!

To find $P(E)$ we use the additive rules of probability. First, let

$& \text{C} = \left \{\text{card selected is a spade}\right \}\\& \text{D} = \left \{\text{card selected is a face card}\right \}$

Note that $P(E) = P(C \cup D)$. Remember, event $C$ consists of $13\;\mathrm{cards}$ and event $D$ consists of $12$ face cards. Event $P(C \cap D)$ consists of the $3$ face-spade cards: The king, jack and, queen of spades cards. Using the additive rule of probability formula,

$P(A \cup B) & = P(A) + P(B) - P(A \cap B) \\& = \frac{13} {52} + \frac{12} {52} - \frac{3} {52} \\& = 0.250 + .231 - .058\\& = 0.423 \\& = 42.3 \%$

I hope that you have learned, through this example, the reason why we subtract $P(C \cap D)$. It is because we do not want to count the face-spade cards twice.

Example:

If you know that $84.2 \%$ of the people arrested in the mid 1990’s were males, $18.3 \%$ are under the age of $18$, and $14.1 \%$ were males under $18$. What is the probability, that a person selected at random from all those arrested, is either male or under $18$?

Solution:

Let

$& \text{A} = \left \{\text{person selected is male}\right \}\\& \text{B} = \left \{\text{person selected is under 18}\right \}$

From the percents given,

$P(A) = 0.842 && P(B) = 0.183 && P(A \cap B) = 0.141$

The probability of a person selected is male or under 18 $P(A \cup B)$:

$P(A \cup B) & = P(A) + P(B) - P(A \cap B)\\& = 0.842 + 0.183 - 0.141\\& = 0.884\\& = 88.4 \%$

This means that $88.4 \%$ of the people arrested in the mid 1990’s are either males or under $18$.

It happens sometimes that $A \cap B$ contains no simple events, i.e., $A \cap B = \left \{\phi \right \}$, the empty set. In this case, we say that the events $A$ and $B$ are mutually exclusive.

Definition
If $A \cap B$ contains no simple events, then $A$ and $B$ are mutually exclusive.

The figure below is the Venn diagram of mutually exclusive events, for example set $A$ might represent all the outcomes of drawing a card, and set $B$ might represent all the outcomes of tossing three coins.

This figure shows that the events $A$ and $B$ have no simple events in common, that is, events $A$ and $B$ can not occur simultaneously, and therefore, $P(A \cap B) = 0.$

If the events $A$ and $B$ are mutually exclusive, then the probability of the union of $A$ and $B$ is the sum of the probabilities of $A$ and $B$, that is

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

Notice that since the two events are mutually exclusive, there is no over-counting.

Example:

If two coins are tossed, what is the probability of observing at least one head?

Solution:

Let

$& \text{A:} \left \{\text{observe only one head}\right \}\\& \text{B:} \left \{\text{observe two heads}\right \}$

$P(A \cup B) = P(A) + P(B) = 0.5 + 0.25 = 0.75 = 75 \%$

Recall from previous section that the conditional probability rule is used to compute the probability of an event, given that another event had already occurred. The formula is

$P(A |B) = \frac{P(A \cap B)} {P(B)}$

Solving for $P(A \cap B)$, we get

$P(A \cap B) = P(A) P(A | B)$

This result is the Multiplicative Rule of Probability.

Multiplicative Rule of Probability

If $A$ and $B$ are two events, then

$P(A \cap B) = P(B) P(A | B)$

This says that the probability that both $A$ and $B$ occur equals to the probability that $B$ occurs times the conditional probability that $A$ occurs, given that $B$ occurs.

Keep in mind that the conditional probability and the multiplicative rule of probability are simply variations of the same thing.

Example:

In a certain city in the US some time ago, $30.7 \%$ of all employed female workers were white-collar workers. If $10.3 \%$ of all 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?

Solution:

We first define the following events

$\text{F} & = \left \{\text{randomly selected worker who is female}\right \}\\\text{W} & = \left \{\text{randomly selected white-collar worker}\right \}$

We are seeking to find the probability of randomly selecting a female worker who is also a white-collar worker. This can be expressed as $P(F \cap W)$.

According to the given data, we have

$P(F) & = 10.3 \% = 0.103\\P(W|F) & = 30.7 \% = 0.307$

Now using the multiplicative rule of probability we get,

$P(F \cap W) = P(F) P(W|F) = (0.103)(0.30) = 0.0316 = 3.16 \%$

Thus $3.16 \%$ of all employed workers were white-collar female workers.

Example:

A college class has $42$ students of which $17$ are males and $25$ are females. 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 a female and the second is male?

Solution:

Here we may define two events

$\text{F1} & = \left \{\text{first student selected is female} \right \}\\\text{M2} & = \left \{\text{second student selected is male}\right \}$

In this problem, we have a conditional probability situation. We want to determine the probability that the first student is female and the second student selected is male.

To do so we apply the multiplicative rule,

$P(F1 \cap M2) = P(F1) P(M2|F1)$

Before we use this formula, we need to calculate the probability of randomly selecting a female student from the population.

$P(F1) = \frac{25} {42} =0.595$

Now given that the first student is selected and not returned back to the population, the remaining number of students now is $41$, of which $24$ female students and $17$ male students. Thus the conditional probability that a male student is selected, given that the first student selected is a female,

$P(M2|F1) = P(M2) = \frac{17} {41} = 0.415$

Substituting these values into our equation, we get

$P(F1 \cap M2) = P(F1) P(M2|F1) = (0.595)(0.415) = 0.247 = 24.7 \%$

We conclude that there is a probability of $24.7 \%$ that the first student selected is a female and the second one is a male.

Example:

Suppose a coin was tossed twice and the observed face was recorded on each toss. The following events are defined

$\text{A} & = \left \{\text{first toss is head}\right \}\\\text{B} & = \left \{\text{second toss is also head}\right \}$

Does knowing that event $A$ has occurred affect the probability of the occurrence of $B$?

Solution:

You would probably say no. Let’s see if this is so. The sample space of this experiment is

$S = \left \{HH, HT, TH, TT \right \}$

Each of these simple events has a probability of $1/4 = 25 \%$. Looking back at the problem, we have events $A$ and $B$.

Since the first toss is a head, we have

$P(A) = P(HH) + P(HT) = 1/4 + 1/4 = 1/2$

And since the second toss is also a head,

$P(B) = P(HH) + P(TH) 1/4 + 1/4 = 1/2$

Now, what is the conditional probability? Here it is,

$P(B|A) & = \frac{P(A \cap B)} {P(A)} \\& = \frac{1/4} {1/2} \\& = \frac{1} {2}$

What does this tell us? It tells us that $P(B) = 1/2$ and $P(B|A) = 1/2$ also. Which means knowing that the first toss resulted in a head does not affect the probability of the second toss. In other words,

$P(B) = P(B|A)$

When this occurs, we say that events $A$ and $B$ are independent.

Condition of Independence

If event $B$ is independent of event $A$, then the occurrence of $A$ does not affect the probability of the occurrence of event $B$. So we write,

$P(B) = P(B|A)$

Example:

The table below gives the number of physicists (in thousands) in the US cross classified by specialties $(P1, P2, P3, P4)$ and base of practice $(B1, B2, B3)$. (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 the event $B1$ independent of $P3$?

Industry

$(B1)$

$(B2)$

Government

$(B3)$

Total

General Physics

$(P1)$

$10.3$ $72.3$ $11.2$ $93.8$

Semiconductors

$(P2)$

$11.4$ $0.82$ $5.2$ $17.42$

Nuclear Physics

(P3)

$1.25$ $0.32$ $34.3$ $35.87$

Astrophysics

$(P4)$

$0.42$ $31.1$ $35.2$ $66.72$
Total $23.37$ $104.54$ $85.9$ $213.81$

Figure: A table showing the number of physicists in each specialty (thousands). This data is hypothetical.

Solution:

The problem may appear a little difficult at first but it is actually much simpler, especially, if we make use of the condition of independence. All we need to do is to calculate $P(B1|P3)$ and $P(B1)$. If those two probabilities are equal, then the two events $B1$ and $P3$ are indeed independent. Otherwise, they are dependent. From the table we find,

$P(B1) = \frac{23.37} {213.81} = 0.109$

And

$P(B1|P3) & = \frac{P(P3 \cap B1)} {P(P3)} \\& = \frac{1.25} {35.87} \\& = 0.035$

Thus, $P(B1|P3) \neq P(B1)$ and so the event $B1$ is dependent on the event $P3$. This lack of independence results from the fact that the percentage of nuclear physicists who are working in the industry $(3.5 \%)$ is not the same as the percentage of all physicists who are in the industry $(10.9 \%)$.

Lesson Summary

1. 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 $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
2. If $A \cap B$ contains no simple events, then $A$ and $B$ are mutually exclusive. Mathematically, this means $P(A \cup B) = P(A) + P(B)$.
3. The Multiplicative Rule of Probability states $P(A \cap B) = P(B) P(A | B)$.
4. If event $B$ is independent of event $A$, then the occurrence of $A$ does not affect the probability of the occurrence of event $B$. Mathematically, $P(B) = P(B|A)$.

Review Questions

1. Two fair dice are tossed and the following events are identified: $& \text{A:} \left \{\text{Sum of the numbers is odd} \right \} \\& \text{B:} \left \{\text{Sum of the numbers is}\ 9, 11,\ \text{or}\ 12\right \}$
1. Are events $A$ and $B$ independent? Why?
2. Are events $A$ and $B$ mutually exclusive? Why?
2. 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 $1\;\mathrm{year}$ is $0.99.$ What is the probability that a new television of the same brand will last $1\;\mathrm{year}$?

1. No; $P(A) \neq P(A|B)$
2. No; $P(A \cap B) \neq 0$
1. $0.891$

Date Created:

Feb 23, 2012

Jul 03, 2014
Files can only be attached to the latest version of None