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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

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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J, combine like terms 7x-4y
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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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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