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Mr. Washington is seeking a mathematics instructor's position at his favorite community college in Cupertino. His employment depends on two conditions: whether the board approves the position, and whether the hiring committee selects him. There is a 80% chance that the board will approve the position, and there is a 70% chance that the hiring committee will select him. If there is a 90% chance that at least one of the two conditions, the board approval or his selection, will be met, what is the probability that Mr. Washington will be hired?

Let A size 12{A} {} be the event that the board approves the position, and S be the event that Mr. Washington gets selected. We have,

P A = . 80 size 12{P left (A right )= "." "80"} {} , P S = . 70 size 12{P left (S right )= "." "70"} {} , and P A S = . 90 size 12{P left (A union S right )= "." "90"} {} .

We need to find, P A S size 12{P left (A intersection S right )} {} .

The addition formula states that,

P A S = P A + P S P A S size 12{P left (A union S right )=P left (A right )+P left (S right ) - P left (A intersection S right )} {}

Substituting the known values, we get

. 90 = . 80 + . 70 P A S size 12{ "." "90"= "." "80"+ "." "70" - P left (A intersection S right )} {}

Therefore, P A S = . 60 size 12{P left (A intersection S right )= "." "60"} {} .

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The probability that this weekend will be cold is . 6 size 12{ "." 6} {} , the probability that it will be rainy is . 7 size 12{ "." 7} {} , and probability that it will be both cold and rainy is . 5 size 12{ "." 5} {} . What is the probability that it will be neither cold nor rainy?

Let C size 12{C} {} be the event that the weekend will be cold, and R size 12{R} {} be event that it will be rainy. We are given that

P C = . 6 size 12{P left (C right )= "." 6} {} , P R = . 7 size 12{P left (R right )= "." 7} {} , P C R = . 5 size 12{P left (C intersection R right )= "." 5} {}

P C R = P C + P R P C R = . 6 + . 7 . 5 = . 8 size 12{P left (C union R right )=P left (C right )+P left (R right ) - P left (C intersection R right )= "." 6+ "." 7 - "." 5= "." 8} {}

We want to find P C R c size 12{P left ( left (C union R right ) rSup { size 8{c} } right )} {} .

P C R c = 1 P C R = 1 . 8 = . 2 size 12{P left ( left (C union R right ) rSup { size 8{c} } right )=1 - P left (C union R right )=1 - "." 8= "." 2} {}
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We summarize this section by listing the important rules.

The Addition Rule

For Two Events E size 12{E} {} and F size 12{F} {} , P E F = P E + P F P E F size 12{P left (E union F right )=P left (E right )+P left (F right ) - P left (E intersection F right )} {}

The Addition Rule for Mutually Exclusive Events

If Two Events E size 12{E} {} and F size 12{F} {} are Mutually Exclusive, then P E F = P E + P F size 12{P left (E union F right )=P left (E right )+P left (F right )} {}

The Complement Rule

If E c size 12{E rSup { size 8{c} } } {} is the Complement of Event E size 12{E} {} , then P E c = 1 P E size 12{P left (E rSup { size 8{c} } right )=1 - P left (E right )} {}

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Probability using tree diagrams and combinations

In this section, we will apply previously learnt counting techniques in calculating probabilities, and use tree diagrams to help us gain a better understanding of what is involved.

We begin with an example.

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn with replacement, what is the probability that both marbles are red?

Let E size 12{E} {} be the event that the first marble drawn is red, and let F size 12{F} {} be the event that the second marble drawn is red.

We need to find P E F size 12{P left (E intersection F right )} {} .

By the statement, "two marbles are drawn with replacement," we mean that the first marble is replaced before the second marble is drawn.

There are 7 choices for the first draw. And since the first marble is replaced before the second is drawn, there are, again, seven choices for the second draw. Using the multiplication axiom, we conclude that the sample space S size 12{S} {} consists of 49 ordered pairs. Of the 49 ordered pairs, there are 3 × 3 = 9 size 12{3 times 3=9} {} ordered pairs that show red on the first draw and, also, red on the second draw. Therefore,

P E F = 9 49 = 3 7 3 7 size 12{P left (E intersection F right )= { {9} over {"49"} } = { {3} over {7} } cdot { {3} over {7} } } {}

Further note that in this particular case

P E F = P E P F size 12{P left (E intersection F right )=P left (E right ) cdot P left (F right )} {}
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If in the [link] , the two marbles are drawn without replacement, then what is the probability that both marbles are red?

By the statement, "two marbles are drawn without replacement," we mean that the first marble is not replaced before the second marble is drawn.

Again, we need to find P E F size 12{P left (E intersection F right )} {} .

There are, again, 7 choices for the first draw. And since the first marble is not replaced before the second is drawn, there are only six choices for the second draw. Using the multiplication axiom, we conclude that the sample space S size 12{S} {} consists of 42 ordered pairs. Of the 42 ordered pairs, there are 3 × 2 = 6 size 12{3 times 2=6} {} ordered pairs that show red on the first draw and red on the second draw. Therefore,

P E F = 6 42 = 3 7 2 6 size 12{P left (E intersection F right )= { {6} over {"42"} } = { {3} over {7} } cdot { {2} over {6} } } {}

Here 3 / 7 size 12{3/7} {} represents P E size 12{P left (E right )} {} , and 2 / 6 size 12{2/6} {} represents the probability of drawing a red on the second draw, given that the first draw resulted in a red. We write the latter as P Red on the second red on first size 12{P left ("Red on the second" \lline "red on first" right )} {} or P F E size 12{P left (F \lline E right )} {} . The "|" represents the word "given." Therefore,

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

The above result is an important one and will appear again in later sections.

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Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
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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
?
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
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sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
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nano basically means 10^(-9). nanometer is a unit to measure length.
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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
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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
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
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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.
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Do you know which machine is used to that process?
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for screen printed electrodes ?
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of graphene you mean?
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or in general
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in general
s.
Graphene has a hexagonal structure
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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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