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Math Help - Coin toss

  1. #1
    MHF Contributor alexmahone's Avatar
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    Coin toss

    A horizontal flat square board 6 inches width is ruled with a grid of fine lines, 2 inches apart and has a vertical rim like a carrom board, all around the edge. If a coin of diameter 2/3 inches is tossed on to the board. What is the probability that it rests without crossing a line?
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  2. #2
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    Hello, alexmahone!

    A horizontal flat square board 6 inches width is ruled with a grid of fine lines,
    2 inches apart and has a vertical rim like a carrom board, all around the edge.
    If a coin of diameter 2/3 inches is tossed on to the board,
    what is the probability that it rests without crossing a line?

    Each square is 2 inches by 2 inches; there are 9 such squares.

    The coin has a radius of \tfrac{1}{3} inches.
    To avoid all lines, its center must be more than \tfrac{1}{3} inch from any line.

    The center of the coin can be in this shaded region:
    Code:
          : - - - - 2 - - - - :
      -   *---+-----------+---*  -
      :   |   |           |   | 1/3
      :   + - + - - - - - + - *  -
      :   |   |:::::::::::|   |  :
      :   |   |:::::::::::|   |  :
      2   |   |:::::::::::|   | 4/3
      :   |   |:::::::::::|   |  :
      :   |   |:::::::::::|   |  :
      :   + - + - - - - - + - *  -
      :   |   |           |   | 1/3
      -   *---+-----------+---* -
           1/3     4/3     1/3

    The shaded area is: . \left(\frac{4}{3}\right)^2 \:=\:\frac{16}{9} inē.

    In the nine squares, the shaded area is: . 9 \times\frac{16}{9} \:=\:16 inē.

    The total area of the board is: . 6^2 \:=\:36 inē.


    Therefore, the probability is: . \frac{16}{36}\;=\;\frac{4}{9}

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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, alexmahone!


    Each square is 2 inches by 2 inches; there are 9 such squares.

    The coin has a radius of \tfrac{1}{3} inches.
    To avoid all lines, its center must be more than \tfrac{1}{3} inch from any line.

    The center of the coin can be in this shaded region:
    Code:
          : - - - - 2 - - - - :
      -   *---+-----------+---*  -
      :   |   |           |   | 1/3
      :   + - + - - - - - + - *  -
      :   |   |:::::::::::|   |  :
      :   |   |:::::::::::|   |  :
      2   |   |:::::::::::|   | 4/3
      :   |   |:::::::::::|   |  :
      :   |   |:::::::::::|   |  :
      :   + - + - - - - - + - *  -
      :   |   |           |   | 1/3
      -   *---+-----------+---* -
           1/3     4/3     1/3
    The shaded area is: . \left(\frac{4}{3}\right)^2 \:=\:\frac{16}{9} inē.

    In the nine squares, the shaded area is: . 9 \times\frac{16}{9} \:=\:16 inē.

    The total area of the board is: . 6^2 \:=\:36 inē.


    Therefore, the probability is: . \frac{16}{36}\;=\;\frac{4}{9}
    I don't think you're right because my options are:
    A) 0.5625
    B) 0.6525
    C) 0.6255
    D) 0.5562
    E) None of these
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  4. #4
    Junior Member
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    Consider where the centre of the coin can land.

    For each square (2x2) it's centre must be at least \frac {1}{3} from each side, this gives an area of

    (2 - \frac {2}{3})^2 = \frac {16}9 per square.

    There are 9 squares which gives a 'safe' area of 16

    Total area is

    (6 - \frac {2}{3})^2 = 256/9

    16/ \frac {256}{9} = 9/16 = 0.5625



    Edit: Soroban missed that the centre cannot lie in a ring around the edge
    The probability is therefore
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