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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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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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