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We now demonstrate the above results with a tree diagram.

Suppose a jar contains 3 red and 4 white marbles. If two marbles are drawn without replacement, find the following probabilities using a tree diagram.

  1. The probability that both marbles are white.
  2. The probability that the first marble is red and the second white.
  3. The probability that one marble is red and the other white.

Let R size 12{R} {} be the event that the marble drawn is red, and let W size 12{W} {} be the event that the marble drawn is white.

We draw the following tree diagram.

The tree diagram illustrates the different probability of drawing a red marble or a white marble.

Although the tree diagrams give us better insight into a problem, they are not practical for problems where more than two or three things are chosen. In such cases, we use the concept of combinations that we learned in [link] . This method is best suited for problems where the order in which the objects are chosen is not important, and the objects are chosen without replacement.

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Suppose a jar contains 3 red, 2 white, and 3 blue marbles. If three marbles are drawn without replacement, find the following probabilities.

  1. P Two red and one white size 12{P left ("Two red and one white" right )} {}
  2. P One of each color size 12{P left ("One of each color" right )} {}
  3. P None blue size 12{P left ("None blue" right )} {}
  4. P At least one blue size 12{P left ("At least one blue" right )} {}

Let us suppose the marbles are labeled as R 1 size 12{R rSub { size 8{1} } } {} , R 2 size 12{R rSub { size 8{2} } } {} , R 3 size 12{R rSub { size 8{3} } } {} , W 1 size 12{W rSub { size 8{1} } } {} , W 2 size 12{W rSub { size 8{2} } } {} , B 1 size 12{B rSub { size 8{1} } } {} , B 2 size 12{B rSub { size 8{2} } } {} , B 3 size 12{B rSub { size 8{3} } } {} .

  1. P Two red and one white size 12{P left ("Two red and one white" right )} {}

    We analyze the problem in the following manner.

    Since we are choosing 3 marbles from a total of 8, there are 8C3 = 56 size 12{8C3="56"} {} possible combinations. Of these 56 combinations, there are 3C2 × 2C1 = 6 size 12{3C2 times 2C1=6} {} combinations consisting of 2 red and one white. Therefore,

    P Two red and one white = 3C2 × 2C1 8C3 = 6 56 size 12{P left ("Two red and one white" right )= { {3C2 times 2C1} over {8C3} } = { {6} over {"56"} } } {} .

  2. P One of each color size 12{P left ("One of each color" right )} {}

    Again, there are 8C3 = 56 size 12{8C3="56"} {} possible combinations. Of these 56 combinations, there are 3C1 × 2C1 × 3C1 = 18 size 12{3C1 times 2C1 times 3C1="18"} {} combinations consisting of one red, one white, and one blue. Therefore,

    P One of each color = 3C1 × 2C1 × 3C1 8C3 = 18 56 size 12{P left ("One of each color" right )= { {3C1 times 2C1 times 3C1} over {8C3} } = { {"18"} over {"56"} } } {} .

  3. P None blue size 12{P left ("None blue" right )} {}

    There are 5 non-blue marbles, therefore

    P None blue = 5C3 8C3 = 10 56 = 5 28 size 12{P left ("None blue" right )= { {5C3} over {8C3} } = { {"10"} over {"56"} } = { {5} over {"28"} } } {} .

  4. P At least one blue size 12{P left ("At least one blue" right )} {}

    By "at least one blue marble," we mean the following: one blue marble and two non-blue marbles, or two blue marbles and one non-blue marble, or all three blue marbles. So we have to find the sum of the probabilities of all three cases.

    P At least one blue = P one blue, two non-blue + P two blue, one non-blue + P three blue size 12{P left ("At least one blue" right )=P left ("one blue, two non-blue" right )+P left ("two blue, one non-blue" right )+P left ("three blue" right )} {}
    P At least one blue = 3C1 × 5C2 8C3 + 3C2 × 5C1 8C3 + 3C3 8C3 size 12{P left ("At least one blue" right )= { {3C1 times 5C2} over {8C3} } + { {3C2 times 5C1} over {8C3} } + { {3C3} over {8C3} } } {}

    P At least one blue = 30 / 56 + 15 / 56 + 1 / 56 = 46 / 56 = 23 / 28 size 12{P left ("At least one blue" right )="30"/"56"+"15"/"56"+1/"56"="46"/"56"="23"/"28"} {} .

    Alternately,

    we use the fact that P E = 1 P E c size 12{P left (E right )=1 - P left (E rSup { size 8{c} } right )} {} .

    If the event E = At least one blue size 12{E="At least one blue"} {} , then E c = None blue size 12{E rSup { size 8{c} } ="None blue"} {} .

    But from part c of this example, we have E c = 5 / 28 size 12{ left (E rSup { size 8{c} } right )=5/"28"} {}

    Therefore, P E = 1 5 / 28 = 23 / 28 size 12{P left (E right )=1 - 5/"28"="23"/"28"} {} .

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Five cards are drawn from a deck. Find the probability of obtaining two pairs, that is, two cards of one value, two of another value, and one other card.

Let us first do an easier problem–the probability of obtaining a pair of kings and queens.

Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is

P A pair of kings and queens = 4C2 × 4C2 × 44 C1 52 C5 size 12{P left ("A pair of kings and queens" right )= { {4C2 times 4C2 times "44"C1} over {"52"C5} } } {}

To find the probability of obtaining two pairs, we have to consider all possible pairs.

Since there are altogether 13 values, that is, aces, deuces, and so on, there are 13 C2 size 12{"13"C2} {} different combinations of pairs.

P Two pairs = 13 C2 4C2 × 4C2 × 44 C1 52 C5 = . 04754 size 12{P left ("Two pairs" right )="13"C2 cdot { {4C2 times 4C2 times "44"C1} over {"52"C5} } = "." "04754"} {}
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Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
what are the products of Nano chemistry?
Maira Reply
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Bhagvanji
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Hafiz Reply
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what about nanotechnology for water purification
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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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What is meant by 'nano scale'?
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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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if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
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analytical skills graphene is prepared to kill any type viruses .
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Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
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what is Nano technology ?
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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 king of growth are you checking .?
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What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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why we need to study biomolecules, molecular biology in nanotechnology?
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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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how did you get the value of 2000N.What calculations are needed to arrive at it
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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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