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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

where we get a research paper on Nano chemistry....?
Maira Reply
what are the products of Nano chemistry?
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Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
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please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
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Nasa has use it in the 60's, copper as water purification in the moon travel.
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industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
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How we are making nano material?
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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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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
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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The nanotechnology is as new science, to scale nanometric
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nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
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biomolecules are e building blocks of every organics and inorganic materials.
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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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