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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 / 52 size 12{4/"52"} {} to 4 / 12 size 12{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.

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 The card is a king = 4 / 52 = 1 / 13 size 12{P left ("The card is a king " right )=4/"52"=1/"13"} {} .

  2. To find P The card is a king A red card has shown size 12{P left ("The card is a king" \lline " 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 The card is a king A red card has shown = 2 / 26 = 1 / 13 size 12{P left ("The card is a king " \lline " A red card has shown" right )=2/"26"=1/"13"} {} .

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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 size 12{P left ("The card is a king" \lline " A red card has shown" right )=P left ("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 size 12{E} {} is not affected by the occurrence of another event F size 12{F} {} , and vice versa, we say that the two events E size 12{E} {} and F size 12{F} {} are independent . This leads to the following definition.

Two Events E size 12{E} {} and F size 12{F} {} are independent if and only if at least one of the following two conditions is true.

  1. P E F = P E size 12{P left (E \lline F right )=P left (E right )} {} or
  2. P F E = P F size 12{P left (F \lline 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 E F = P E F P F size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } } {}

Multiplying both sides by P F size 12{P left (F right )} {} , we get

P E F = P E F P F size 12{P left (E intersection F right )=P left (E \lline F right )P left (F right )} {}

Now if the two events are independent, then by definition

P E F = P E size 12{P left (E \lline F right )=P left (E right )} {}

Substituting, P E F = P E P F size 12{P left (E intersection F right )=P left (E right )P left (F right )} {}

We state it formally as follows.

Test for independence

Two Events E size 12{E} {} and F size 12{F} {} are independent if and only if
P E F = P E P F size 12{P left (E intersection 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 size 12{M} {} represents male, F size 12{F} {} represents female, C size 12{C} {} represents color-blind, and N size 12{N} {} not color-blind. Use the independence test to determine whether the events color-blind and male are independent.

According to the test, C size 12{C} {} and M size 12{M} {} are independent if and only if P C M = P C P M size 12{P left (C intersection M right )=P left (C right )P left (M right )} {} .

P C = 7 / 100 size 12{P left (C right )=7/"100"} {} ,   P M = 52 / 100 size 12{P left (M right )="52"/"100"} {}   and   P C M = 6 / 100 size 12{P left (C intersection M right )=6/"100"} {}
P C P M = 7 / 100 52 / 100 = . 0364 size 12{P left (C right )P left (M right )= left (7/"100" right ) left ("52"/"100" right )= "." "0364"} {}

and P C M = . 06 size 12{P left (C intersection M right )= "." "06"} {}

Clearly . 0364 . 06 size 12{ "." "0364"<>"." "06"} {}

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

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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 size 12{M} {} be the event that a woman wears makeup, and L size 12{L} {} the event that a woman has a low self-image. We have

P M L = 9 / 100 size 12{P left (M intersection L right )=9/"100"} {} , P M = 45 / 100 size 12{P left (M right )="45"/"100"} {} and P L = 20 / 100 size 12{P left (L right )="20"/"100"} {}

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

P M L = P M P L size 12{P left (M intersection L right )=P left (M right )P left (L right )} {}

Since 9 / 100 = 45 / 100 20 / 100 size 12{9/"100"= left ("45"/"100" right ) left ("20"/"100" right )} {}

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

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A coin is tossed three times, and the events E size 12{E} {} , F size 12{F} {} and G size 12{G} {} are defined as follows:

E size 12{E} {} : The coin shows a head on the first toss.

F size 12{F} {} : At least two heads appear.

G size 12{G} {} : Heads appear in two successive tosses.

Determine whether the following events are independent.

  1. E size 12{E} {} and F size 12{F} {}
  2. F size 12{F} {} and G size 12{G} {}
  3. E size 12{E} {} and G size 12{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 size 12{S= left lbrace ital "HHH", ital "HHT", ital "HTH", ital "HTT", ital "THH", ital "THT", ital "TTH", ital "TTT" right rbrace } {}

E = HHH , HHT , HTH , HTT size 12{E= left lbrace ital "HHH", ital "HHT", ital "HTH", ital "HTT" right rbrace } {} , P E = 4 / 8 size 12{P left (E right )=4/8} {} or 1 / 2 size 12{1/2} {}

F = HHH , HHT , HTH , THH size 12{F= left lbrace ital "HHH", ital "HHT", ital "HTH", ital "THH" right rbrace } {} , P F = 4 / 8 size 12{P left (F right )=4/8} {} or 1 / 2 size 12{1/2} {}

G = HHT , THH size 12{G= left lbrace ital "HHT", ital "THH" right rbrace } {} , P G = 2 / 8 size 12{P left (G right )=2/8} {} or 1 / 4 size 12{1/4} {}

E F = HHH , HHT , HTH size 12{E intersection F= left lbrace ital "HHH", ital "HHT", ital "HTH" right rbrace } {} , P E F = 3 / 8 size 12{P left (E intersection F right )=3/8} {}

E G = HHT , THH size 12{E intersection G= left lbrace ital "HHT", ital "THH" right rbrace } {} , P F G = 2 / 8 size 12{P left (F intersection G right )=2/8} {} or 1 / 4 size 12{1/4} {}

E G = HHT size 12{E intersection G= left lbrace ital "HHT" right rbrace } {} P E G = 1 / 8 size 12{P left (E intersection G right )=1/8} {}

  1. In order for E size 12{E} {} and F size 12{F} {} to be independent, we must have

    P E F = P E P F size 12{P left (E intersection F right )=P left (E right )P left (F right )} {} .

    But 3 / 8 1 / 2 1 / 2 size 12{3/8<>1/2 cdot 1/2} {}

    Therefore, E size 12{E} {} and F size 12{F} {} are not independent.

  2. F size 12{F} {} and G size 12{G} {} will be independent if

    P F G = P F P G size 12{P left (F intersection G right )=P left (F right )P left (G right )} {} .

    Since 1 / 4 1 / 2 1 / 4 size 12{1/4≠1/2 cdot 1/4} {}

    F size 12{F} {} and G size 12{G} {} are not independent.

  3. We look at

    P E G = P E P G size 12{P left (E intersection G right )=P left (E right )P left (G right )} {}
    1 / 8 = 1 / 2 1 / 4 size 12{1/8=1/2 cdot 1/4} {}

    Therefore, E size 12{E} {} and G size 12{G} {} are independent events.

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The probability that Jaime will visit his aunt in Baltimore this year is . 30 size 12{ "." "30"} {} , and the probability that he will go river rafting on the Colorado river is . 50 size 12{ "." "50"} {} . If the two events are independent, what is the probability that Jaime will do both?

Let A size 12{A} {} be the event that Jaime will visit his aunt this year, and R size 12{R} {} be the event that he will go river rafting.

We are given P A = . 30 size 12{P left (A right )= "." "30"} {} and P R = . 50 size 12{P left (R right )= "." "50"} {} , and we want to find P A R size 12{P left (A intersection R right )} {} .

Since we are told that the events A size 12{A} {} and R size 12{R} {} are independent,

P A R = P A P R = . 30 . 50 = . 15 size 12{P left (A intersection R right )=P left (A right )P left (R right )= left ( "." "30" right ) left ( "." "50" right )= "." "15"} {} .
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Given P B A = . 4 size 12{P left (B \lline A right )= "." 4} {} . If A size 12{A} {} and B size 12{B} {} are independent, find P B size 12{P left (B right )} {} .

If A size 12{A} {} and B size 12{B} {} are independent, then by definition P B A = P B size 12{P left (B \lline A right )=P left (B right )} {}

Therefore, P B = . 4 size 12{P left (B right )= "." 4} {}

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Given P A = . 7 size 12{P left (A right )= "." 7} {} , P B A = . 5 size 12{P left (B \lline A right )= "." 5} {} . Find P A B size 12{P left (A intersection B right )} {} .

By definition P B A = P A B P A size 12{P left (B \lline A right )= { {P left (A intersection B right )} over {P left (A right )} } } {}

Substituting, we have

. 5 = P A B . 7 size 12{ "." 5= { {P left (A intersection B right )} over { "." 7} } } {}

Therefore, P A B = . 35 size 12{P left (A intersection B right )= "." "35"} {}

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Given P ( A ) = .5 , P ( A B ) = .7 , if A and B are independent, find P ( B ) .

The addition rule states that

P ( A B ) = P ( A ) + P ( B ) P ( A B )

Since A and B are independent, P ( A B ) = P ( A ) P ( B )

We substitute for P ( A B ) in the addition formula and get

P ( A B ) = P ( A ) + P ( B ) P ( A ) P ( B )

By letting P ( B ) = x , and substituting values, we get

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

Therefore, P ( B ) = .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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