7.1 Probability

Chapter Overview

In this chapter, you will learn to:

Write sample spaces.
Determine whether two events are mutually exclusive.
Use the Addition Rule.
Calculate probabilities using both tree diagrams and combinations.
Do problems involving conditional probability.
Determine whether two events are independent.

Sample Spaces and Probability

If two coins are tossed, what is the probability that both coins will fall heads? The problem seems simple enough, but it is not uncommon to hear the incorrect answer $1 / 3$ . A student may incorrectly reason that if two coins are tossed there are three possibilities, one head, two heads, or no heads. Therefore, the probability of two heads is one out of three. The answer is wrong because if we toss two coins there are four possibilities and not three. For clarity, assume that one coin is a penny and the other a nickel. Then we have the following four possibilities.

HH HT TH TT

The possibility HT, for example, indicates a head on the penny and a tail on the nickel, while TH represents a tail on the penny and a head on the nickel.

It is for this reason, we emphasize the need for understanding sample spaces.

An act of flipping coins, rolling dice, drawing cards, or surveying people are referred to as an experiment.

Example 3

Two dice are rolled. Write the sample space.

We assume one of the dice is red, and the other green. We have the following 36 possibilities.

Table 7.1

Green

Red

(1,1)

(1,2)

(1,3)

(1,4)

(1,5)

(1,6)

(2,1)

(2,2)

(2,3)

(2,4)

(2,5)

(2,6)

(3,1)

(3,2)

(3,3)

(3,4)

(3,5)

(3,6)

(4,1)

(4,2)

(4,3)

(4,4)

(4,5)

(4,6)

(5,1)

(5,2)

(5,3)

(5,4)

(5,5)

(5,6)

(6,1)

(6,2)

(6,3)

(6,4)

(6,5)

(6,6)

The entry (2, 5), for example, indicates that the red die shows a two, and the green a 5.

Now that we understand the concept of a sample space, we will define probability.

Probability For a sample space $S$ , and an outcome $A$ of $S$ , the following two properties are satisfied.

If $A$ is an outcome of a sample space, then the probability of $A$ , denoted by $P (A)$ , is between 0 and 1, inclusive.
$0 less than or equal to P open parenthesis A close parenthesis less than or equal to 1$
The sum of the probabilities of all the outcomes in $S$ equals 1.

The example we just considered consisted of only one outcome of the sample space. We are often interested in finding probabilities of several outcomes represented by an event.

An event is a subset of a sample space. If an event consists of only one outcome, it is called a simple event.

Example 6

A jar contains 3 red, 4 white, and 3 blue marbles. If a marble is chosen at random, what is the probability that the marble is a red marble or a blue marble?

We assume the marbles are $r sub 1$ , $r sub 2$ , $r sub 3$ , $w sub 1$ , $w sub 2$ , $w sub 3$ , $w sub 4$ , $b sub 1$ , $b sub 2$ , $b sub 3$ . Let the event $C$ represent that the marble is red or blue.

The sample space $S equals open brace r sub 1, r sub 2, r sub 3, w sub 1, w sub 2, w sub 3, w sub 4, b sub 1, b sub 2, b sub 3 close brace$

And the event $C equals open brace r sub 1, r sub 2, r sub 3, b sub 1, b sub 2, b sub 3 close brace$

Therefore, the probability of $C$ ,

$P open parenthesis C close parenthesis equals 6 / 10$ or $3 / 5$ .

Mutually Exclusive Events and the Addition Rule

In the the related section, we learned to find the union, intersection, and complement of a set. We will now use these set operations to describe events.

The union of two events $E$ and $F$ , $E \cup F$ , is the set of outcomes that are in $E$ or in $F$ or in both.

The intersection of two events $E$ and $F$ , $E \cap F$ , is the set of outcomes that are in both $E$ and $F$ .

The complement of an event $E$ , denoted by $E^{c}$ , is the set of outcomes in the sample space $S$ that are not in $E$ . It is worth noting that $P (E^{C}) = 1 - P (E)$ . This follows from the fact that if the sample space has $n$ elements and $E$ has $k$ elements, then $E^{c}$ has $n - k$ elements. Therefore,

$P (E^{C}) = \frac{n - k}{n} = 1 - \frac{k}{n} = 1 - P (E)$ .

Of particular interest to us are the events whose outcomes do not overlap. We call these events mutually exclusive.

Two events $E$ and $F$ are said to be mutually exclusive if they do not intersect. That is, $E intersection F equals ∅$ .

Next we'll determine whether a given pair of events are mutually exclusive.

We will now consider problems that involve the union of two events.

Example 12

If a die is rolled, what is the probability of obtaining an even number or a number greater than four?

Let $E$ be the event that the number shown on the die is an even number, and let $F$ be the event that the number shown is greater than four.

The sample space $S equals open brace 1,2,3,4,5,6 close brace$ . The event $E equals open brace 2,4,6 close brace$ , and the event $F equals open brace 5,6 close brace$

We need to find $P (E \cup F)$ .

Since $P open parenthesis E close parenthesis equals 3 / 6$ , and $P open parenthesis F close parenthesis equals 2 / 6$ , a student may say $P (E \cup F) = 3 / 6 + 2 / 6$ . This will be incorrect because the element 6, which is in both $E$ and $F$ has been counted twice, once as an element of $E$ and once as an element of $F$ . In other words, the set $E \cup F$ has only four elements and not five. Therefore, $P open parenthesis E union F close parenthesis equals 4 / 6$ and not $5 / 6$ .

This can be illustrated by a Venn diagram.

The sample space $S$ , the events $E$ and $F$ , and $E \cap F$ are listed below.

$S equals open brace 1,2,3,4,5,6 close brace$ , $E equals open brace 2,4,6 close brace$ , $F equals open brace 5,6 close brace$ , and $E intersection F equals open brace 6 close brace$ .

Figure 7.2

The above figure shows $S$ , $E$ , $F$ , and $E \cap F$ .

Finding the probability of $E \cup F$ , is the same as finding the probability that $E$ will happen, or $F$ will happen, or both will happen. If we count the number of elements $n (E)$ in $E$ , and add to it the number of elements $n (F)$ in $F$ , the points in both $E$ and $F$ are counted twice, once as elements of $E$ and once as elements of $F$ . Now if we subtract from the sum, $n (E) + n (F)$ , the number $n (E \cap F)$ , we remove the duplicity and get the correct answer. So as a rule,

$n (E \cup F) = n (E) + n (F) - n (E \cap F)$

By dividing the entire equation by $n (S)$ , we get

$\frac{n (E \cup F)}{n (S)} = \frac{n (E)}{n (S)} + \frac{n (F)}{n (S)} - \frac{n (E \cap F)}{n (S)}$

Since the probability of an event is the number of elements in that event divided by the number of all possible outcomes, we have

$P (E \cup F) = P (E) + P (F) - P (E \cap F)$

Applying the above for this example, we get

$P (E \cup F) = 3 / 6 + 2 / 6 - 1 / 6 = 4 / 6$

This is because, when we add $P (E)$ and $P (F)$ , we have added $P (E \cap F)$ twice. Therefore, we must subtract $P (E \cap F)$ , once.

This gives us the general formula, called the Addition Rule, for finding the probability of the union of two events. It states

$P (E \cup F) = P (E) + P (F) - P (E \cap F)$

If two events E and F are mutually exclusive, then $E \cap F = \emptyset$ and $P open parenthesis E intersection F close parenthesis equals 0$ , and we get

$P (E \cup F) = P (E) + P (F)$

Example 14

Two dice are rolled, and the events $F$ and $T$ are as follows:

$F = {The sum of the dice is four}$ and $T = {At least one die shows a three}$

Find $P (F \cup T)$ .

We list $F$ and $T$ , and $F \cap T$ as follows:

$F equals open brace open parenthesis 1,3 close parenthesis, open parenthesis 2,2 close parenthesis, open parenthesis 3,1 close parenthesis close brace$

$T equals open brace open parenthesis 3,1 close parenthesis, open parenthesis 3,2 close parenthesis, open parenthesis 3,3 close parenthesis, open parenthesis 3,4 close parenthesis, open parenthesis 3,5 close parenthesis, open parenthesis 3,6 close parenthesis, open parenthesis 1,3 close parenthesis, open parenthesis 2,3 close parenthesis, open parenthesis 4,3 close parenthesis, open parenthesis 5,3 close parenthesis, open parenthesis 6,3 close parenthesis close brace$

$F intersection T equals open brace open parenthesis 1,3 close parenthesis, open parenthesis 3,1 close parenthesis close brace$

Since $P (F \cup T) = P (F) + P (T) - P (F \cap T)$

We have $P (F \cup T) = 3 / 36 + 11 / 36 - 2 / 36 = 12 / 36$ .

Example 15

Mr. Washington is seeking a mathematics instructor's position at his favorite community college in Cupertino. His employment depends on two conditions: whether the board approves the position, and whether the hiring committee selects him. There is a 80% chance that the board will approve the position, and there is a 70% chance that the hiring committee will select him. If there is a 90% chance that at least one of the two conditions, the board approval or his selection, will be met, what is the probability that Mr. Washington will be hired?

Let $A$ be the event that the board approves the position, and S be the event that Mr. Washington gets selected. We have,

$P open parenthesis A close parenthesis equals .80$ , $P open parenthesis S close parenthesis equals .70$ , and $P open parenthesis A union S close parenthesis equals .90$ .

We need to find, $P (A \cap S)$ .

The addition formula states that,

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

Substituting the known values, we get

$. 90 = . 80 + . 70 - P (A \cap S)$

Therefore, $P open parenthesis A intersection S close parenthesis equals .60$ .

Example 16

The probability that this weekend will be cold is $.6$ , the probability that it will be rainy is $.7$ , and probability that it will be both cold and rainy is $.5$ . What is the probability that it will be neither cold nor rainy?

Let $C$ be the event that the weekend will be cold, and $R$ be event that it will be rainy. We are given that

$P open parenthesis C close parenthesis equals .6$ , $P open parenthesis R close parenthesis equals .7$ , $P open parenthesis C intersection R close parenthesis equals .5$

$P (C \cup R) = P (C) + P (R) - P (C \cap R) = . 6 + . 7 - . 5 = . 8$

We want to find $P ({(C \cup R)}^{c})$ .

$P ({(C \cup R)}^{c}) = 1 - P (C \cup R) = 1 - . 8 = . 2$

We summarize this section by listing the important rules.

Probability Using Tree Diagrams and Combinations

In this section, we will apply previously learnt counting techniques in calculating probabilities, and use tree diagrams to help us gain a better understanding of what is involved.

We begin with an example.

Example 18

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn with replacement, what is the probability that both marbles are red?

Let $E$ be the event that the first marble drawn is red, and let $F$ be the event that the second marble drawn is red.

We need to find $P (E \cap F)$ .

By the statement, "two marbles are drawn with replacement," we mean that the first marble is replaced before the second marble is drawn.

There are 7 choices for the first draw. And since the first marble is replaced before the second is drawn, there are, again, seven choices for the second draw. Using the multiplication axiom, we conclude that the sample space $S$ consists of 49 ordered pairs. Of the 49 ordered pairs, there are $3 \times 3 = 9$ ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

$P (E \cap F) = \frac{9}{49} = \frac{3}{7} \cdot \frac{3}{7}$

Further note that in this particular case

$P (E \cap F) = P (E) \cdot P (F)$

Example 19

If in the Example 18, the two marbles are drawn without replacement, then what is the probability that both marbles are red?

By the statement, "two marbles are drawn without replacement," we mean that the first marble is not replaced before the second marble is drawn.

Again, we need to find $P (E \cap F)$ .

There are, again, 7 choices for the first draw. And since the first marble is not replaced before the second is drawn, there are only six choices for the second draw. Using the multiplication axiom, we conclude that the sample space $S$ consists of 42 ordered pairs. Of the 42 ordered pairs, there are $3 \times 2 = 6$ ordered pairs that show red on the first draw and red on the second draw. Therefore,

$P (E \cap F) = \frac{6}{42} = \frac{3}{7} \cdot \frac{2}{6}$

Here $3 / 7$ represents $P (E)$ , and $2 / 6$ represents the probability of drawing a red on the second draw, given that the first draw resulted in a red. We write the latter as $P open parenthesis Red on the second ∣ red on first close parenthesis$ or $P open parenthesis F ∣ E close parenthesis$ . The "|" represents the word "given." Therefore,

$P (E \cap F) = P (E) \cdot P (F ∣ E)$

The above result is an important one and will appear again in later sections.

We now demonstrate the above results with a tree diagram.

Example 20

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn without replacement, find the following probabilities using a tree diagram.

The probability that both marbles are white.
The probability that the first marble is red and the second white.
The probability that one marble is red and the other white.

Let $R$ be the event that the marble drawn is red, and let $W$ be the event that the marble drawn is white.

We draw the following tree diagram.

Figure 7.3

Although the tree diagrams give us better insight into a problem, they are not practical for problems where more than two or three things are chosen. In such cases, we use the concept of combinations that we learned in the related section. This method is best suited for problems where the order in which the objects are chosen is not important, and the objects are chosen without replacement.

Example 21

Suppose a jar contains 3 red, 2 white, and 3 blue marbles. If three marbles are drawn without replacement, find the following probabilities.

$P (Two red and one white)$
$P (One of each color)$
$P (None blue)$
$P (At least one blue)$

Let us suppose the marbles are labeled as $R sub 1$ , $R sub 2$ , $R sub 3$ , $W sub 1$ , $W sub 2$ , $B sub 1$ , $B sub 2$ , $B sub 3$ .

$P (Two red and one white)$
We analyze the problem in the following manner.
Since we are choosing 3 marbles from a total of 8, there are $8 C 3 equals 56$ possible combinations. Of these 56 combinations, there are $3 C 2 \times 2 C 1 = 6$ combinations consisting of 2 red and one white. Therefore,
$P (Two red and one white) = \frac{3 C 2 \times 2 C 1}{8} = \frac{6}{56}$ .
$P (One of each color)$
Again, there are $8 C 3 equals 56$ possible combinations. Of these 56 combinations, there are $3 C 1 \times 2 C 1 \times 3 C 1 = 18$ combinations consisting of one red, one white, and one blue. Therefore,
$P (One of each color) = \frac{3 C 1 \times 2 C 1 \times 3 C 1}{8} = \frac{18}{56}$ .
$P (None blue)$
There are 5 non-blue marbles, therefore
$P (None blue) = \frac{5}{C} = \frac{10}{56} = \frac{5}{28}$ .
$P (At least one blue)$
By "at least one blue marble," we mean the following: one blue marble and two non-blue marbles, or two blue marbles and one non-blue marble, or all three blue marbles. So we have to find the sum of the probabilities of all three cases.
$P (At least one blue) = P (one blue, two non-blue) + P (two blue, one non-blue) + P (three blue)$
$P (At least one blue) = \frac{3 C 1 \times 5 C 2}{8} + \frac{3 C 2 \times 5 C 1}{8} + \frac{3}{C}$
$P (At least one blue) = 30 / 56 + 15 / 56 + 1 / 56 = 46 / 56 = 23 / 28$ .
Alternately,
we use the fact that $P (E) = 1 - P (E^{c})$ .
If the event $E = At least one blue$ , then $E^{c} = None blue$ .
But from part c of this example, we have $(E^{c}) = 5 / 28$
Therefore, $P (E) = 1 - 5 / 28 = 23 / 28$ .

Example 22

Five cards are drawn from a deck. Find the probability of obtaining two pairs, that is, two cards of one value, two of another value, and one other card.

Let us first do an easier problem–the probability of obtaining a pair of kings and queens.

Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is

$P (A pair of kings and queens) = \frac{4 C 2 \times 4 C 2 \times 44 C1}{52 C5}$

To find the probability of obtaining two pairs, we have to consider all possible pairs.

Since there are altogether 13 values, that is, aces, deuces, and so on, there are $13 C2$ different combinations of pairs.

$P (Two pairs) = 13 C2 \cdot \frac{4 C 2 \times 4 C 2 \times 44 C1}{52 C5} = . 04754$

We end the section by solving a problem called the Birthday Problem.

Example 23

If there are 25 people in a room, what is the probability that at least two people have the same birthday?

Let event $E$ represent that at least two people have the same birthday.

We first find the probability that no two people have the same birthday.

We analyze as follows.

Suppose there are 365 days to every year. According to the multiplication axiom, there are $365 to the power 25$ possible birthdays for 25 people. Therefore, the sample space has $365 to the power 25$ elements. We are interested in the probability that no two people have the same birthday. There are 365 possible choices for the first person and since the second person must have a different birthday, there are 364 choices for the second, 363 for the third, and so on. Therefore,

$P (No two have the same birthday) = \frac{365 \cdot 364 \cdot 363 \dots 341}{365^{25}} = \frac{365 P 25}{365^{25}}$

Since $P (at least two people have the same birthday) = 1 - P (No two have the same birthday),$

$P (at least two people have the same birthday) = 1 - \frac{365 P 25}{365^{25}} = . 5687$

Conditional Probability

Suppose you and a friend wish to play a game that involves choosing a single card from a well-shuffled deck. Your friend deals you one card, face down, from the deck and offers you the following deal: If the card is a king, he will pay you $5, otherwise, you pay him $1. Should you play the game?

You reason in the following manner. Since there are four kings in the deck, the probability of obtaining a king is $4 / 52$ or $1 / 13$ . And, probability of not obtaining a king is $12 / 13$ . This implies that the ratio of your winning to losing is 1 to 12, while the payoff ratio is only $1 to $5. Therefore, you determine that you should not play.

Now consider the following scenario. While your friend was dealing the card, you happened to get a glance of it and noticed that the card was a face card. Should you, now, play the game?

Since there are 12 face cards in the deck, the total elements in the sample space are no longer 52, but just 12. This means the chance of obtaining a king is $4 / 12$ or $1 / 3$ . So your chance of winning is $1 / 3$ and of losing $2 / 3$ . This makes your winning to losing ratio 1 to 2 which fares much better with the payoff ratio of $1 to $5. This time, you determine that you should play.

In the second part of the above example, we were finding the probability of obtaining a king knowing that a face card had shown. This is an example of conditional probability. Whenever we are finding the probability of an event E under the condition that another event F has happened, we are finding conditional probability.

The symbol $P open parenthesis E ∣ F close parenthesis$ denotes the problem of finding the probability of $E$ given that $F$ has occurred. We read $P open parenthesis E ∣ F close parenthesis$ as "the probability of $E$ , given $F$ ."

Example 24

A family has three children. Find the conditional probability of having two boys and a girl given that the first born is a boy.

Let event $E$ be that the family has two boys and a girl, and $F$ the event that the first born is a boy.

First, we list the sample space for a family of three children as follows.

$S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}$

Since we know that the first born is a boy, our possibilities narrow down to four outcomes, $BBB$ , $BBG$ , $BGB$ , and $BGG$ .

Among the four, $BBG$ and $BGB$ represent two boys and a girl.

Therefore, $P open parenthesis E ∣ F; equals 2 / 4$ or $1 / 2$ .

Let us now develop a formula for the conditional probability $P open parenthesis E ∣ F close parenthesis$ .

Suppose an experiment consists of $n$ equally likely events. Further suppose that there are $m$ elements in $F$ , and $c$ elements in $E \cap F$ , as shown in the following Venn diagram.

The figure shows that everything within the square is equal to S. In the Venn diagram m-c is equal to F where c is equal to both E and F. — **Figure 7.4**

If the event $F$ has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset $F$ . Therefore, we only look at the set $F$ and at nothing outside of $F$ . Since $F$ has $m$ elements, the denominator in the calculation of $P open parenthesis E ∣ F close parenthesis$ is m. We may think that the numerator for our conditional probability is the number of elements in $E$ . But clearly we cannot consider the elements of $E$ that are not in $F$ . We can only count the elements of $E$ that are in $F$ , that is, the elements in $E \cap F$ . Therefore,

$P (E ∣ F) = \frac{c}{m}$

Dividing both the numerator and the denominator by $n$ , we get

$P (E ∣ F) = \frac{c / n}{m / n}$

But $c / n equals P open parenthesis E intersection F close parenthesis$ , and $m / n equals P open parenthesis F close parenthesis$ .

Substituting, we derive the following formula for $P open parenthesis E ∣ F close parenthesis$ .

Example 25

For Two Events $E$ and $F$ , the Probability of $E$ Given $F$ is

$P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$

Example 26

A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.

Let $E$ be the event that an even number shows, and $F$ be the event that a number greater than three shows. We want $P open parenthesis E ∣ F close parenthesis$ .

$E equals open brace 2,4,6 close brace$ and $F equals open brace 4,5,6 close brace$ . Which implies, $E intersection F equals open brace 4,6 close brace$

Therefore, $P open parenthesis F close parenthesis equals 3 / 6$ , and $P open parenthesis E intersection F close parenthesis equals 2 / 6$

$P (E ∣ F) = \frac{P (E \cap F)}{P (F)} = \frac{2 / 6}{3 / 6} = \frac{2}{3}$ .

Example 27

The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.

**Table 7.2**
	Male(M)	Female(F)	Total
Public Transportation(P)	8	13	21
Drive(D)	39	40	79
Total	47	53	100

The events $M$ , $F$ , $P$ , and $D$ are self explanatory. Find the following probabilities.

$P open parenthesis D ∣ M close parenthesis$
$P open parenthesis F ∣ D close parenthesis$
$P open parenthesis M ∣ P close parenthesis$

We use the conditional probability formula $P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$ .

$P (D ∣ M) = \frac{P (D \cap M)}{P (M)} = \frac{39 / 100}{47 / 100} = \frac{39}{47}$ .
$P (F ∣ D) = \frac{P (F \cap D)}{P (D)} = \frac{40 / 100}{79 / 100} = \frac{40}{79}$ .
$P (M ∣ P) = \frac{P (M \cap P)}{P (P)} = \frac{8 / 100}{21 / 100} = \frac{8}{21}$ .

Example 31

In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.

Let event $E$ be that the family has two boys and a girl, and let $F$ be the probability that the family has at least two boys. We want $P open parenthesis E ∣ F close parenthesis$ .

We list the sample space along with the events $E$ and $F$ .

$S = {BBB, BBG, BGB, BGG, GBB, GGB, GGG}$

$E = {BBG, BGB, GBB}$ and $F = {BBB, BBG, BGB, GBB}$

$E \cap F = {BBG, BGB, GBB}$

Therefore, $P open parenthesis F close parenthesis equals 4 / 8$ , and $P open parenthesis E intersection F close parenthesis equals 3 / 8$ .

And

$P (E ∣ F) - \frac{3 / 8}{4 / 8} = \frac{3}{4}$ .

Independent Events

In, we considered conditional probabilities. In some examples, the probability of an event changed when additional information was provided. For instance, the probability of obtaining a king from a deck of cards, changed from $4 / 52$ to $4 / 12$ , when we were given the condition that a face card had already shown. This is not always the case. The additional information may or may not alter the probability of the event. For example consider the following example.

The reader should observe that in the above example,

$P (The card is a king ∣ A red card has shown) = P (The card is a king)$

In other words, the additional information, a red card has shown, did not affect the probability of obtaining a king. Whenever the probability of an event $E$ is not affected by the occurrence of another event $F$ , and vice versa, we say that the two events $E$ and $F$ are independent. This leads to the following definition.

Two Events $E$ and $F$ are independent if and only if at least one of the following two conditions is true.

$P open parenthesis E ∣ F close parenthesis equals P open parenthesis E close parenthesis$ or
$P open parenthesis F ∣ E close parenthesis equals P open parenthesis F close parenthesis$

If the events are not independent, then they are dependent.

Next, we need to develop a test to determine whether two events are independent.

We recall the conditional probability formula.

$P (E ∣ F) = \frac{P (E \cap F)}{P (F)}$

Multiplying both sides by $P (F)$ , we get

$P open parenthesis E intersection F close parenthesis equals P open parenthesis E ∣ F close parenthesis P open parenthesis F close parenthesis$

Now if the two events are independent, then by definition

$P open parenthesis E ∣ F close parenthesis equals P open parenthesis E close parenthesis$

Substituting, $P (E \cap F) = P (E) P (F)$

We state it formally as follows.

Test for Independence Two Events $E$ and $F$ are independent if and only if

$P (E \cap F) = P (E) P (F)$

Example 34

The table below shows the distribution of color-blind people by gender.

**Table 7.3**
	Male(M)	Female(F)	Total
Color-Blind(C)	6	1	7
Not Color-Blind (N)	46	47	93
Total	52	48	100

Where $M$ represents male, $F$ represents female, $C$ represents color-blind, and $N$ not color-blind. Use the independence test to determine whether the events color-blind and male are independent.

According to the test, $C$ and $M$ are independent if and only if $P (C \cap M) = P (C) P (M)$ .

$P (C) = 7 / 100, P (M) = 52 / 100 and P (C \cap M) = 6 / 100$

$P (C) P (M) = (7 / 100) (52 / 100) = . 0364$

and $P open parenthesis C intersection M close parenthesis equals .06$

Clearly $.0364 not equal to .06$

Therefore, the two events are not independent. We may say they are dependent.

Example 35

In a survey of 100 women, 45 wore makeup, and 55 did not. Of the 45 who wore makeup, 9 had a low self-image, and of the 55 who did not, 11 had a low self-image. Are the events "wearing makeup" and "having a low self-image" independent?

Let $M$ be the event that a woman wears makeup, and $L$ the event that a woman has a low self-image. We have

$P open parenthesis M intersection L close parenthesis equals 9 / 100$ , $P open parenthesis M close parenthesis equals 45 / 100$ and $P open parenthesis L close parenthesis equals 20 / 100$

In order for two events to be independent, we must have

$P (M \cap L) = P (M) P (L)$

Since $9 / 100 equals open parenthesis 45 / 100 close parenthesis open parenthesis 20 / 100 close parenthesis$

The two events "wearing makeup" and "having a low self-image" are independent.

Example 36

A coin is tossed three times, and the events $E$ , $F$ and $G$ are defined as follows:

$E$ : The coin shows a head on the first toss.

$F$ : At least two heads appear.

$G$ : Heads appear in two successive tosses.

Determine whether the following events are independent.

$E$ and $F$
$F$ and $G$
$E$ and $G$

To make things easier, we list the sample space, the events, their intersections and the corresponding probabilities.

$S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}$

$E = {HHH, HHT, HTH, HTT}$ , $P open parenthesis E close parenthesis equals 4 / 8$ or $1 / 2$

$F = {HHH, HHT, HTH, THH}$ , $P open parenthesis F close parenthesis equals 4 / 8$ or $1 / 2$

$G = {HHT, THH}$ , $P open parenthesis G close parenthesis equals 2 / 8$ or $1 / 4$

$E \cap F = {HHH, HHT, HTH}$ , $P open parenthesis E intersection F close parenthesis equals 3 / 8$

$E \cap G = {HHT, THH}$ , $P open parenthesis F intersection G close parenthesis equals 2 / 8$ or $1 / 4$

$E \cap G = {HHT}$ $P open parenthesis E intersection G close parenthesis equals 1 / 8$

In order for $E$ and $F$ to be independent, we must have
$P (E \cap F) = P (E) P (F)$ .
But $3 / 8 not equal to 1 / 2 times 1 / 2$
Therefore, $E$ and $F$ are not independent.
$F$ and $G$ will be independent if
$P (F \cap G) = P (F) P (G)$ .
Since $1 / 4 not equal to 1 / 2 times 1 / 4$
$F$ and $G$ are not independent.
We look at
$P (E \cap G) = P (E) P (G)$
$1 / 8 = 1 / 2 \cdot 1 / 4$
Therefore, $E$ and $G$ are independent events.

Example 37

The probability that Jaime will visit his aunt in Baltimore this year is $.30$ , and the probability that he will go river rafting on the Colorado river is $.50$ . If the two events are independent, what is the probability that Jaime will do both?

Let $A$ be the event that Jaime will visit his aunt this year, and $R$ be the event that he will go river rafting.

We are given $P open parenthesis A close parenthesis equals .30$ and $P open parenthesis R close parenthesis equals .50$ , and we want to find $P (A \cap R)$ .

Since we are told that the events $A$ and $R$ are independent,

$P (A \cap R) = P (A) P (R) = (. 30) (. 50) = . 15 .$

Adapted from Applied Finite Mathematics by Rupinder Sekhon (De Anza College), originally published by OpenStax CNX (cnx.org, collection col10613), licensed under CC BY 3.0. Changes were made. License: CC-BY-3.0.