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Thread: probability

  1. #1
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    probability

    Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

    A. 1/18

    B. 64/4032

    C. 63/64

    D. 1/9
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  2. #2
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    Re: probability

    Answer (A)

    Solution:

    The number of ways of choosing the first square is 64. The number of ways of choosing the second square is 63.
    There are a total of 64*63=4032 ways of choosing two squares.
    If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways.
    If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways.
    If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways.
    Hence the desired number of combinations:
    =(4*2)+(24*3)+(36*4)=224

    Therefore, the required probability =224
    4032
    = 1/18
    for practicing more questions go to Probability | Aptitude Test Solved Problems | Interview Question Answers
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  3. #3
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    Re: probability

    Hello, lemini!

    Another approach . . .


    Two squares are chosen at random on a chessboard.
    What is the probability that they have a side in common?

    . . . $\displaystyle (A)\;\tfrac{1}{18}$

    . . . $\displaystyle (B)\;\tfrac{64}{4032}$ . Why isn't this reduced?

    . . . $\displaystyle (C)\;\tfrac{63}{64}$

    . . . $\displaystyle (D)\;\tfrac{1}{9}$

    There are: .$\displaystyle {64\choose2}\,=\,2016$ possible outcomes.


    Consider the top row of the chessboard: .$\displaystyle \square\!\square\!\square\!\square\!\square\! \square\!\square\!\square $

    Consider placing a horizontal domino $\displaystyle \square\!\square$ in the row.
    . . There are 7 possible positions in each row.
    With 8 rows, there are $\displaystyle 7\times 8 \,=\,56$ ways to place a horizontal domino.

    Now consider placing a vertical domino: $\displaystyle \begin{array}{c}\square \\[-2mm] \square\end{array}$ in a column.
    . . There are 7 possible positions in each column.
    with 8 columns, there are $\displaystyle 7\times8 \,=\,56$ ways to place a vertical domino.

    Hence, there are $\displaystyle 56 + 56 \,=\,112$ ways to place a domino.


    The probability is: .$\displaystyle \frac{112}{2016} \:=\:\frac{1}{18}$ . Answer (A)
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