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  1. #1
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    ChessBoard

    Theres a chess board.. regular one with 64 squares.
    If two squares are selected at random Find the probability that they have a common side..
    THnx In advance...!
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  2. #2
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    Quote Originally Posted by umangarora View Post
    Theres a chess board.. regular one with 64 squares.
    If two squares are selected at random Find the probability that they have a common side..
    THnx In advance...!
    Consider three cases: 1. first square choosen is a corner sq, 2. first square on a side but not a corner sq 3. first square an interior square.

    CB
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  3. #3
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    plz can u show working n ans.. im clueless.
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  4. #4
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    Quote Originally Posted by umangarora View Post
    plz can u show working n ans.. im clueless.
    There are 4 corner squares, out of a total of 64, therefore the probability of the first sqyuare being a corner is 1/16. Then there are 8 squares adjacent to a corner square so the probability of choosing one of thes3es as the second square is 8/63. So the probability of choosing two adjacent squares the first of which is a corner is 1/126.

    Now repeat for edges and interior first choices.

    CB
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  5. #5
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    Also, be careful that you do not double-count "pairs of squares" when dealing with the edge squares,
    since the "order" in which squares are chosen does not matter.
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  6. #6
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    Hello, umangarora!

    There's a chessboard ... regular one with 64 squares.

    If two squares are selected at random,
    find the probability that they have a common side.

    Choosing 2 squares from the available 64 squares,
    . . there are: . _{64}C_2 \:=\:{64\choose2} \:=\:2016 choices.


    How many of these are "dominos"? . \square\!\square


    In a row, \square\!\square\!\square\!\square\!\square\!\squa  re\!\square\!\square, there are 7 ways to place a domino.

    . . With 8 rows, there are 56 possible "horizontal" dominos.

    The same is true for the columns: .56 possible "vertical" dominos.


    Hence, there are: . 56 + 56 \:=\:112 possible dominos.


    The probability is: . \dfrac{112}{2016} \:=\:\dfrac{1}{18}

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  7. #7
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    Quote Originally Posted by soroban View Post
    hello, umangarora!


    choosing 2 squares from the available 64 squares,
    . . there are: . _{64}c_2 \:=\:{64\choose2} \:=\:2016 choices.


    How many of these are "dominos"? . \square\!\square


    in a row, \square\!\square\!\square\!\square\!\square\!\squa  re\!\square\!\square, there are 7 ways to place a domino.

    . . with 8 rows, there are 56 possible "horizontal" dominos.

    The same is true for the columns: .56 possible "vertical" dominos.


    Hence, there are: . 56 + 56 \:=\:112 possible dominos.


    The probability is: . \dfrac{112}{2016} \:=\:\dfrac{1}{18}

    awesome!! Ty
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  8. #8
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    Quote Originally Posted by umangarora View Post
    Theres a chess board.. regular one with 64 squares.
    If two squares are selected at random Find the probability that they have a common side..
    THnx In advance...!
    Alternatively,

    If a chosen square is in a corner, then the probabilty that the 2nd chosen square
    is not sharing a side with it is

    \displaystyle\frac{61}{63}

    hence the probability that the squares have one of them in the corner,
    with another not sharing a side is

    \displaystyle\frac{4}{64}\ \frac{61}{63}

    There are 6(4)=24 remaining squares along the edge that are not in a corner.
    The probability of another square not sharing a side with one of these is

    \displaystyle\frac{60}{63}

    hence the probability that the squares include one of the remaining edge squares,
    but the other does not share a side with it is

    \displaystyle\frac{24}{64}\ \frac{60}{63}

    There are 36 squares not on an edge.
    Therefore the probability that the squares include one of these, and the 2nd one is not sharing a side with it is

    \displaystyle\frac{36}{64}\ \frac{59}{63}

    Therefore, the probability that 2 squares are selected and do share a side is

    \displaystyle\ 1-\left(\frac{4(61)+(24)(60)+(36)(59)}{(64)(63)}\rig  ht)=\frac{(64)(63)-3808}{(64)(63)}=\frac{4032-3808}{4032}=\frac{224}{4032}=\frac{1}{18}
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