# 0.12 Probability  (Page 8/8)

 Page 8 / 8

## Independent events

In [link] , 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 $4/\text{52}$ to $4/\text{12}$ , 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.

A card is drawn from a deck. Find the following probabilities.

1. The card is a king.
2. The card is a king given that a red card has shown.
1. Clearly, $P\left(\text{The card is a king}\right)=4/\text{52}=1/\text{13}$ .

2. To find $P\left(\text{The card is a king}\mid \text{A red card has shown}\right)$ , we reason as follows:

Since a red card has shown, there are only twenty six possibilities. Of the 26 red cards, there are two kings. Therefore,

$P\left(\text{The card is a king}\mid \text{A red card has shown}\right)=2/\text{26}=1/\text{13}$ .

The reader should observe that in the above example,

$P\left(\text{The card is a king}\mid \text{A red card has shown}\right)=P\left(\text{The card is a king}\right)$

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.

1. $P\left(E\mid F\right)=P\left(E\right)$ or
2. $P\left(F\mid E\right)=P\left(F\right)$

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.

$P\left(E\mid F\right)=\frac{P\left(E\cap F\right)}{P\left(F\right)}$

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

$P\left(E\cap F\right)=P\left(E\mid F\right)P\left(F\right)$

Now if the two events are independent, then by definition

$P\left(E\mid F\right)=P\left(E\right)$

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

We state it formally as follows.

## Test for independence

Two Events $E$ and $F$ are independent if and only if
$P\left(E\cap F\right)=P\left(E\right)P\left(F\right)$

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

 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 $P\left(C\cap M\right)=P\left(C\right)P\left(M\right)$ .

$P\left(C\right)P\left(M\right)=\left(7/\text{100}\right)\left(\text{52}/\text{100}\right)=\text{.}\text{0364}$

and $P\left(C\cap M\right)=\text{.}\text{06}$

Clearly $\text{.}\text{0364}\ne \text{.}\text{06}$

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

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

$P\left(M\cap L\right)=9/\text{100}$ , $P\left(M\right)=\text{45}/\text{100}$ and $P\left(L\right)=\text{20}/\text{100}$

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

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

Since $9/\text{100}=\left(\text{45}/\text{100}\right)\left(\text{20}/\text{100}\right)$

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

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.

1. $E$ and $F$
2. $F$ and $G$
3. $E$ and $G$

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

$S=\left\{\text{HHH},\text{HHT},\text{HTH},\text{HTT},\text{THH},\text{THT},\text{TTH},\text{TTT}\right\}$

$E=\left\{\text{HHH},\text{HHT},\text{HTH},\text{HTT}\right\}$ , $P\left(E\right)=4/8$ or $1/2$

$F=\left\{\text{HHH},\text{HHT},\text{HTH},\text{THH}\right\}$ , $P\left(F\right)=4/8$ or $1/2$

$G=\left\{\text{HHT},\text{THH}\right\}$ , $P\left(G\right)=2/8$ or $1/4$

$E\cap F=\left\{\text{HHH},\text{HHT},\text{HTH}\right\}$ , $P\left(E\cap F\right)=3/8$

$E\cap G=\left\{\text{HHT},\text{THH}\right\}$ , $P\left(F\cap G\right)=2/8$ or $1/4$

$E\cap G=\left\{\text{HHT}\right\}$ $P\left(E\cap G\right)=1/8$

1. In order for $E$ and $F$ to be independent, we must have

$P\left(E\cap F\right)=P\left(E\right)P\left(F\right)$ .

But $3/8\ne 1/2\cdot 1/2$

Therefore, $E$ and $F$ are not independent.

2. $F$ and $G$ will be independent if

$P\left(F\cap G\right)=P\left(F\right)P\left(G\right)$ .

Since $1/4\ne 1/2\cdot 1/4$

$F$ and $G$ are not independent.

3. We look at

$P\left(E\cap G\right)=P\left(E\right)P\left(G\right)$
$1/8=1/2\cdot 1/4$

Therefore, $E$ and $G$ are independent events.

The probability that Jaime will visit his aunt in Baltimore this year is $\text{.}\text{30}$ , and the probability that he will go river rafting on the Colorado river is $\text{.}\text{50}$ . 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 $P\left(A\right)=\text{.}\text{30}$ and $P\left(R\right)=\text{.}\text{50}$ , and we want to find $P\left(A\cap R\right)$ .

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

$P\left(A\cap R\right)=P\left(A\right)P\left(R\right)=\left(\text{.}\text{30}\right)\left(\text{.}\text{50}\right)=\text{.}\text{15}.$

Given $P\left(B\mid A\right)=\text{.}4$ . If $A$ and $B$ are independent, find $P\left(B\right)$ .

If $A$ and $B$ are independent, then by definition $P\left(B\mid A\right)=P\left(B\right)$

Therefore, $P\left(B\right)=\text{.}4$

Given $P\left(A\right)=\text{.}7$ , $P\left(B\mid A\right)=\text{.}5$ . Find $P\left(A\cap B\right)$ .

By definition $P\left(B\mid A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$

Substituting, we have

$\text{.}5=\frac{P\left(A\cap B\right)}{\text{.}7}$

Therefore, $P\left(A\cap B\right)=\text{.}\text{35}$

Given $P\left(A\right)=.5$ , $P\left(A\cup B\right)=.7$ , if $A$ and $B$ are independent, find $P\left(B\right)$ .

The addition rule states that

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)–P\left(A\cap B\right)$

Since $A$ and $B$ are independent, $P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

We substitute for $P\left(A\cap B\right)$ in the addition formula and get

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)–P\left(A\right)P\left(B\right)$

By letting $P\left(B\right)=x$ , and substituting values, we get

$.7=.5+x–.5x$
$.7=.5+.5x$
$.2=.5x$
$.4=x$

Therefore, $P\left(B\right)=.4$ .

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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