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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.
Clearly, $P\left(\text{The card is a king}\right)=4/\text{52}=1/\text{13}$ .
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.
Next, we need to develop a test to determine whether two events are independent.
We recall the conditional probability formula.
Multiplying both sides by $P\left(F\right)$ , we get
Now if the two events are independent, then by definition
Substituting, $P\left(E\cap F\right)=P\left(E\right)P\left(F\right)$
We state it formally as follows.
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 $P\left(C\cap M\right)=P\left(C\right)P\left(M\right)$ .
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
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.
To make things easier, we list the sample space, the events, their intersections and the corresponding probabilities.
$E=\left\{\text{HHH},\text{HHT},\text{HTH},\text{HTT}\right\}$ , $P\left(E\right)=4/8$ 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 $1/4$
$E\cap G=\left\{\text{HHT}\right\}$ $P\left(E\cap G\right)=1/8$
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.
$F$ 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.
We look at
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,
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
Therefore, $P\left(A\cap B\right)=\text{.}\text{35}$
Given $P\left(A\right)=.5$ , $P(A\cup B)=.7$ , if $A$ and $B$ are independent, find $P\left(B\right)$ .
The addition rule states that
Since $A$ and $B$ are independent, $P(A\cap B)=P\left(A\right)P\left(B\right)$
We substitute for $P(A\cap B)$ in the addition formula and get
By letting $P\left(B\right)=x$ , and substituting values, we get
Therefore, $P\left(B\right)=.4$ .
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