<< Chapter < Page Chapter >> Page >

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"} {} .

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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"} {} .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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} {}

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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"} {}

Got questions? Get instant answers now!
Got questions? Get instant answers now!

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 .

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Questions & Answers

anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
Mueller Reply
8. It is known that 80% of the people wear seat belts, and 5% of the people quit smoking last year. If 4% of the people who wear seat belts quit smoking, are the events, wearing a seat belt and quitting smoking, independent?
William Reply
Mr. Shamir employs two part-time typists, Inna and Jim for his typing needs. Inna charges $10 an hour and can type 6 pages an hour, while Jim charges $12 an hour and can type 8 pages per hour. Each typist must be employed at least 8 hours per week to keep them on the payroll. If Mr. Shamir has at least 208 pages to be typed, how many hours per week should he employ each student to minimize his typing costs, and what will be the total cost?
Chine Reply
At De Anza College, 20% of the students take Finite Mathematics, 30% take Statistics and 10% take both. What percentage of the students take Finite Mathematics or Statistics?
Chalton Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied finite mathematics' conversation and receive update notifications?

Ask