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P ( A B C ) = P ( A ) P ( A B ) P ( A ) P ( A B C ) P ( A B ) = P ( A ) P ( B | A ) P ( C | A B )

and

P ( A B C D ) = P ( A ) P ( A B ) P ( A ) P ( A B C ) P ( A B ) P ( A B C D ) P ( A B C ) = P ( A ) P ( B | A ) P ( C | A B ) P ( D | A B C )

This pattern may be extended to the intersection of any finite number of events. Also, the events may be taken in any order.

Selection of items from a lot

An electronics store has ten items of a given type in stock. One is defective. Four successive customers purchase one of the items. Each time, the selection is on anequally likely basis from those remaining. What is the probability that all four customes get good items?

SOLUTION

Let E i be the event the i th customer receives a good item. Then the first chooses one of the nine out of ten good ones, the second chooses one of the eight out of ninegoood ones, etc., so that

P ( E 1 E 2 E 3 E 4 ) = P ( E 1 ) P ( E 2 | E 1 ) P ( E 3 | E 1 E 2 ) P ( E 4 | E 1 E 2 E 3 ) = 9 10 8 9 7 8 6 7 = 6 10

Note that this result could be determined by a combinatorial argument: under the assumptions, each combination of four of ten is equally likely; the number of combinationsof four good ones is the number of combinations of four of the nine. Hence

P ( E 1 E 2 E 3 E 4 ) = C ( 9 , 4 ) C ( 10 , 4 ) = 126 210 = 3 / 5
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A selection problem

Three items are to be selected (on an equally likely basis at each step) from ten, two of which are defective. Determine the probability that the first and third selected aregood.

SOLUTION

Let G i , 1 i 3 be the event the i th unit selected is good. Then G 1 G 3 = G 1 G 2 G 3 G 1 G 2 c G 3 . By the product rule

P ( G 1 G 3 ) = P ( G 1 ) P ( G 2 | G 1 ) P ( G 3 | G 1 G 2 ) + P ( G 1 ) P ( G 2 c | G 1 ) P ( G 3 | G 1 G 2 c ) = 8 10 7 9 6 8 + 8 10 2 9 7 8 = 28 45 0 . 62
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(CP2) Law of total probability Suppose the class { A i : 1 i n } of events is mutually exclusive and every outcome in E is in one of these events. Thus, E = A 1 E A 2 E A n E , a disjoint union. Then

P ( E ) = P ( E | A 1 ) P ( A 1 ) + P ( E | A 2 ) P ( A 2 ) + + P ( E | A n ) P ( A n )

A compound experiment.

Five cards are numbered one through five. A two-step selection procedure is carried out as follows.

  1. Three cards are selected without replacement, on an equally likely basis.
    • If card 1 is drawn, the other two are put in a box
    • If card 1 is not drawn, all three are put in a box
  2. One of cards in the box is drawn on an equally likely basis (from either two or three)

Let A i be the event the i th card is drawn on the first selection and let B i be the event the card numbered i is drawn on the second selection (from the box). Determine P ( B 5 ) , P ( A 1 B 5 ) , and P ( A 1 | B 5 ) .

SOLUTION

From [link] , we have P ( A i ) = 6 / 10 and P ( A i A j ) = 3 / 10 . This implies

P ( A i A j c ) = P ( A i ) - P ( A i A j ) = 3 / 10

Now we can draw card five on the second selection only if it is selected on the first drawing, so that B 5 A 5 . Also A 5 = A 1 A 5 A 1 c A 5 . We therefore have B 5 = B 5 A 5 = B 5 A 1 A 5 B 5 A 1 c A 5 . By the law of total probability (CP2) ,

P ( B 5 ) = P ( B 5 | A 1 A 5 ) P ( A 1 A 5 ) + P ( B 5 | A 1 c A 5 ) P ( A 1 c A 5 ) = 1 2 3 10 + 1 3 3 10 = 1 4

Also, since A 1 B 5 = A 1 A 5 B 5 ,

P ( A 1 B 5 ) = P ( A 1 A 5 B 5 ) = P ( A 1 A 5 ) P ( B 5 | A 1 A 5 ) = 3 10 1 2 = 3 20

We thus have

P ( A 1 | B 5 ) = 3 / 20 5 / 20 = 6 10 = P ( A 1 )

Occurrence of event B 1 has no affect on the likelihood of the occurrence of A 1 . This condition is examined more thoroughly in the chapter on "Independence of Events" .

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Often in applications data lead to conditioning with respect to an event but the problem calls for “conditioning in the opposite direction.”

Questions & Answers

Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
Application of nanotechnology in medicine
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
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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
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LITNING Reply
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LITNING Reply
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LITNING
scanning tunneling microscope
Sahil
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Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
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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
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
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 ?
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Kyle
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Adin
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Adin
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Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
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research.net
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Introduction about quantum dots in nanotechnology
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Loga
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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Source:  OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6
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