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Let us now develop a formula for the conditional probability P E F size 12{P left (E \lline F right )} {} .

Suppose an experiment consists of n size 12{n} {} equally likely events. Further suppose that there are m size 12{m} {} elements in F size 12{F} {} , and c size 12{c} {} elements in E F size 12{E intersection F} {} , as shown in the following Venn diagram.

The figure shows that everything within the square is equal to S. In the Venn diagram m-c is equal to F where c is equal to both E and F.

If the event F size 12{F} {} has occurred, the set of all possible outcomes is no longer the entire sample space, but instead, the subset F size 12{F} {} . Therefore, we only look at the set F size 12{F} {} and at nothing outside of F size 12{F} {} . Since F size 12{F} {} has m size 12{m} {} elements, the denominator in the calculation of P E F size 12{P left (E \lline F right )} {} is m. We may think that the numerator for our conditional probability is the number of elements in E size 12{E} {} . But clearly we cannot consider the elements of E size 12{E} {} that are not in F size 12{F} {} . We can only count the elements of E size 12{E} {} that are in F size 12{F} {} , that is, the elements in E F size 12{E intersection F} {} . Therefore,

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

Dividing both the numerator and the denominator by n size 12{n} {} , we get

P E F = c / n m / n size 12{P left (E \lline F right )= { {c/n} over {m/n} } } {}

But c / n = P E F size 12{c/n=P left (E intersection F right )} {} , and m / n = P F size 12{m/n=P left (F right )} {} .

Substituting, we derive the following formula for P E F size 12{P left (E \lline F right )} {} .

For Two Events E size 12{E} {} and F size 12{F} {} , the Probability of E size 12{E} {} Given F size 12{F} {} is

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 )} } } {}
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A single die is rolled. Use the above formula to find the conditional probability of obtaining an even number given that a number greater than three has shown.

Let E size 12{E} {} be the event that an even number shows, and F size 12{F} {} be the event that a number greater than three shows. We want P E F size 12{P left (E \lline F right )} {} .

E = 2,4,6 size 12{E= left lbrace 2,4,6 right rbrace } {} and F = 4,5,6 size 12{F= left lbrace 4,5,6 right rbrace } {} . Which implies, E F = 4,6 size 12{E intersection F= left lbrace 4,6 right rbrace } {}

Therefore, P F = 3 / 6 size 12{P left (F right )=3/6} {} , and P E F = 2 / 6 size 12{P left (E intersection F right )=2/6} {}

P E F = P E F P F = 2 / 6 3 / 6 = 2 3 size 12{P left (E \lline F right )= { {P left (E intersection F right )} over {P left (F right )} } = { {2/6} over {3/6} } = { {2} over {3} } } {} .

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The following table shows the distribution by gender of students at a community college who take public transportation and the ones who drive to school.

Male(M) Female(F) Total
Public Transportation(P) 8 13 21
Drive(D) 39 40 79
Total 47 53 100

The events M size 12{M} {} , F size 12{F} {} , P size 12{P} {} , and D size 12{D} {} are self explanatory. Find the following probabilities.

  1. P D M size 12{P left (D \lline M right )} {}
  2. P F D size 12{P left (F \lline D right )} {}
  3. P M P size 12{P left (M \lline P right )} {}

We use 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 )} } } {} .

  1. P D M = P D M P M = 39 / 100 47 / 100 = 39 47 size 12{P left (D \lline M right )= { {P left (D intersection M right )} over {P left (M right )} } = { {"39"/"100"} over {"47"/"100"} } = { {"39"} over {"47"} } } {} .
  2. P F D = P F D P D = 40 / 100 79 / 100 = 40 79 size 12{P left (F \lline D right )= { {P left (F intersection D right )} over {P left (D right )} } = { {"40"/"100"} over {"79"/"100"} } = { {"40"} over {"79"} } } {} .
  3. P M P = P M P P P = 8 / 100 21 / 100 = 8 21 size 12{P left (M \lline P right )= { {P left (M intersection P right )} over {P left (P right )} } = { {8/"100"} over {"21"/"100"} } = { {8} over {"21"} } } {} .
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Given P E = . 5 size 12{P left (E right )= "." 5} {} , P F = .7 size 12{P left (F right )=/7} {} , and P E F = .3 size 12{P left (E intersection F right )} {} . Find the following.

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

We use 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 )} } } {} .

  1. P E F = . 3 . 7 = 3 7 size 12{P left (E \lline F right )= { { "." 3} over { "." 7} } = { {3} over {7} } } {} .
  2. P F E = . 3 / . 5 = 3 / 5 size 12{P left (F \lline E right )= "." 3/ "." 5=3/5} {} .
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Given two mutually exclusive events E size 12{E} {} and F size 12{F} {} such that P E = . 4 size 12{P left (E right )= "." 4} {} , P F = . 9 size 12{P left (F right )= "." 9} {} . Find P E F size 12{P left (E \lline F right )} {} .

Since E size 12{E} {} and F size 12{F} {} are mutually exclusive, P E F = 0 size 12{P left (E intersection F right )=0} {} . Therefore,

P E | F = 0 .9 = 0 size 12{P left (E intersection F right )=0} {} .

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Given P F E = . 5 size 12{P left (F \lline E right )= "." 5} {} , and P E F = . 3 size 12{P left (E intersection F right )= "." 3} {} . Find P E size 12{P left (E right )} {} .

Using 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 )} } } {} , we get

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

Substituting,

. 5 = . 3 P E size 12{ "." 5= { { "." 3} over {P left (E right )} } } {} or P E = 3 / 5 size 12{P left (E right )=3/5} {}

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In a family of three children, find the conditional probability of having two boys and a girl, given that the family has at least two boys.

Let event E size 12{E} {} be that the family has two boys and a girl, and let F size 12{F} {} be the probability that the family has at least two boys. We want P E F size 12{P left (E \lline F right )} {} .

We list the sample space along with the events E size 12{E} {} and F size 12{F} {} .

S = BBB , BBG , BGB , BGG , GBB , GGB , GGG size 12{S= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "BGG", ital "GBB", ital "GGB", ital "GGG" right rbrace } {}

E = BBG , BGB , GBB size 12{E= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {} and F = BBB , BBG , BGB , GBB size 12{F= left lbrace ital "BBB", ital "BBG", ital "BGB", ital "GBB" right rbrace } {}

E F = BBG , BGB , GBB size 12{E intersection F= left lbrace ital "BBG", ital "BGB", ital "GBB" right rbrace } {}

Therefore, P F = 4 / 8 size 12{P left (F right )=4/8} {} , and P E F = 3 / 8 size 12{P left (E intersection F right )=3/8} {} .

And

P E F 3 / 8 4 / 8 = 3 4 size 12{P left (E \lline F right ) - { {3/8} over {4/8} } = { {3} over {4} } } {} .

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At a community college 65% of the students use IBM computers, 50% use Macintosh computers, and 20% use both. If a student is chosen at random, find the following probabilities.

  1. The student uses an IBM given that he uses a Macintosh.
  2. The student uses a Macintosh knowing that he uses an IBM.

Let event I size 12{I} {} be that the student uses an IBM computer, and M size 12{M} {} the probability that he uses a Macintosh.

  1. P I M = . 20 . 50 = 2 5 size 12{P left (I \lline M right )= { { "." "20"} over { "." "50"} } = { {2} over {5} } } {}
  2. P M I = . 20 . 65 = 4 13 size 12{P left (M \lline I right )= { { "." "20"} over { "." "65"} } = { {4} over {"13"} } } {} .
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Questions & Answers

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
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What is meant by 'nano scale'?
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LITNING
scanning tunneling microscope
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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
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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
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Adin Reply
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
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Adin
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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
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Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
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absolutely yes
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there is no specific books for beginners but there is book called principle of nanotechnology
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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
If March sales will be up from February by 10%, 15%, and 20% at Place I, Place II, and Place III, respectively, find the expected number of hot dogs, and corn dogs to be sold
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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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