<< Chapter < Page Chapter >> Page >

Roll one fair 6-sided die. The sample space is {1, 2, 3, 4, 5, 6} . Let event A = a face is odd. Then A = {1, 3, 5} . Let event B = a face is even. Then B = {2, 4, 6} .

  • Find the complement of A , A' . The complement of A , A' , is B because A and B together make up the sample space. P(A) + P(B) = P(A) + P(A') = 1 . Also, P(A) = 3 6 and P(B) = 3 6
  • Let event C = odd faces larger than 2. Then C = { 3 , 5 } . Let event D = all even faces smaller than 5. Then D = { 2 , 4 } . P(C and D) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events.
  • Let event E = all faces less than 5. E = { 1 , 2 , 3 , 4 } .

    Are C and E mutually exclusive events? (Answer yes or no.) Why or why not?

    No. C = {3, 5} and E = {1, 2, 3, 4} . P(C AND E) = 1 6 . To be mutually exclusive, P(C AND E) must be 0.

  • Find P(C|A) . This is a conditional. Recall that the event C is {3, 5} and event A is {1, 3, 5} . To find P(C|A) , find the probability of C using the sample space A . You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5} . So, P(C|A) = 2 3

Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6 , P(H) = 0.5 , and P(G AND H) = 0.3 . Are G and H independent?

If G and H are independent, then you must show ONE of the following:

  • P(G|H) = P(G)
  • P(H|G) = P(H)
  • P(G AND H) = P(G) P(H)
The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.

Show that P(G|H) = P(G) .

P(G|H) = P(G AND H) P(H) = 0.3 0.5 = 0.6 = P(G)

Show P(G AND H) = P(G) P(H) .

P(G) P(H)  =  0.6 0.5  =  0.3  =  P(G AND H)

Interpretation of results

Since G and H are independent, then, knowing that a person is taking a science class does not change the chance that he/she is taking math. (Note: IF the two events had not beenindependent - that is, IF they were dependent - then knowing that a person is taking a science class would change the chance he/she is taking math. The next example will illustrate two events that are not independent.)

In a particular college class, 60% of the students are female. 50 % of all students in the class have long hair. 45% of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that the student is female. Let L be the event that the student has long hair.Are the events of being female and having long hair independent?

  • The following probabilities are given in this example:
  • P(F ) = 0.60 ; P(L ) = 0.50
  • P(F AND L) = 0.45
  • P(L|F) = 0.75

If F and L are independent, then the following conditions are true. BUT if you show that any of the conditions are not true , then F and L are not independent .

  • P(L|F) = P(L)
  • P(F|L) = P(F)
  • P(F AND L) = P(F) P(L)
The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know P(F|L) yet, so you can not use the second condition.

Solution 1

Check whether P(F and L) = P(F)P(L): We are given that P(F and L) = 0.45 ; but P(F)P(L) = (0.60)(0.50)= 0.30 The events of being female and having long hair are not independent because P(F and L) does not equal P(F)P(L).

Solution 2

Check whether P(L|F) equals P(L): We are given that P(L|F) = 0.75 but P(L) = 0.50; they are not equal. The events of being female and having long hair are not independent.

Interpretation of results

The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.

In a box there are 3 red cards and 5 blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3,4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.

The sample space S  =  R1, R2, R3, B1, B2, B3, B4, B5 . S has 8 outcomes.

  • P(R) = 3 8 . P(B) = 5 8 . P(R AND B) = 0 . (You cannot draw one card that is both red and blue.)
  • P(E) = 3 8 . (There are 3 even-numbered cards, R2 , B2 , and B4 .)
  • P(E|B) = 2 5 . (There are 5 blue cards: B1 , B2 , B3 , B4 , and B5 . Out of the blue cards, there are 2 even cards: B2 and B4 .)
  • P(B|E) = 2 3 . (There are 3 even-numbered cards: R2 , B2 , and B4 . Out of the even-numbered cards, 2 are blue: B2 and B4 .)
  • The events R and B are mutually exclusive because P(R AND B) = 0 .
  • Let G = card with a number greater than 3. G = { B4 , B5 } . P(G) = 2 8 . Let H = blue card numbered between 1 and 4, inclusive. H = { B1 , B2 , B3 , B4 } . P(G|H) = 1 4 . (The only card in H that has a number greater than 3 is B4 .) Since 2 8 = 1 4 , P(G) = P(G|H) which means that G and H are independent.

Questions & Answers

what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
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 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
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 2

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Collaborative statistics: custom version modified by r. bloom. OpenStax CNX. Nov 15, 2010 Download for free at http://legacy.cnx.org/content/col10617/1.4
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Collaborative statistics: custom version modified by r. bloom' conversation and receive update notifications?

Ask