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Re: There is a 7 × 7 chessboard. Two squares are chosen at random. What is [#permalink]
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nick1816 wrote:
There is a 7 × 7 chessboard. Two squares are chosen at random. What is the probability that the 2 squares have a common edge?

A. 1/14
B. 1/7
C. 2/7
D. 1/3
E. 1/2


B IS OA as per me.Total sample space will be 49c2
numerator will be (4*2)*(20*3)*(25*4)
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Re: There is a 7 × 7 chessboard. Two squares are chosen at random. What is [#permalink]
Think about the case of 2*2 square. It shares boundary with 4 squares (1*2, 2*1, 2*3, 3*2). But you consider those pairs again when you you will consider the case of 1*2, 2*1, 2*3 and 3*2. If you don't get it, i'll explain you with diagram.


PreiteeRanjan wrote:
nick1816 wrote:
There is a 7 × 7 chessboard. Two squares are chosen at random. What is the probability that the 2 squares have a common edge?

A. 1/14
B. 1/7
C. 2/7
D. 1/3
E. 1/2


B IS OA as per me.Total sample space will be 49c2
numerator will be (4*2)*(20*3)*(25*4)
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GMAT 1: 710 Q49 V37
GPA: 4
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Re: There is a 7 × 7 chessboard. Two squares are chosen at random. What is [#permalink]
nick1816 wrote:
Think about the case of 2*2 square. It shares boundary with 4 squares (1*2, 2*1, 2*3, 3*2). But you consider those pairs again when you you will consider the case of 1*2, 2*1, 2*3 and 3*2. If you don't get it, i'll explain you with diagram.


PreiteeRanjan wrote:
nick1816 wrote:
There is a 7 × 7 chessboard. Two squares are chosen at random. What is the probability that the 2 squares have a common edge?

A. 1/14
B. 1/7
C. 2/7
D. 1/3
E. 1/2


B IS OA as per me.Total sample space will be 49c2
numerator will be (4*2)*(20*3)*(25*4)


kindly explain with a drawing.
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There is a 7 × 7 chessboard. Two squares are chosen at random. What is [#permalink]
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Two squares are chosen at random. What is the probability that the 2 squares have a common edge?

Notice here that the act of choosing the two squares has already taken place. But we don't know which two have been chosen, yet.
Now, the question is : What is the probability that the 2 squares that have already been chosen, happen to have a common edge?

Since two squares have already been chosen, let's narrow our choices down by finding out the total number of pairs of squares that can be chosen from this 7x7 chess board.

Since total number of squares in the chess board is 49, number of ways in which unique pairs of squares can be chosen from among them is 49C2 = 49*24.

Now, let's see how many unique pairs of squares can have edges in common.

Fact 1: A chess board has squares that have 4 adjacent squares (towards the middle) or 3 adjacent squares (along the edges of the board) or 2 adjacent squares (the squares in the four corners).

Fact 2: If square A touches the left edge of square B, it can also be said that the square B touches the right edge of A. However, A & B form a single pair. Therefore to eliminate counting each pair twice, we will start at the Bottom-Right hand corner of the board and count left-ward and then up-ward (eliminating right-ward and down-ward counting).

Number of squares with adjacent squares to it's left in the bottom row = 6.
Number of rows in the entire board = 7.
Number of pairs of squares found thus = 6*7 = 42

Similarly, the number of squares having an adjacent square above them, on the right most column = 6
Number of columns in the entire board = 7.
Number of pairs of squares found thus = 6*7 = 42

Total number of unique pairs of squares sharing an edge with each other = 42 + 42 = 84.


The probability that a pair of squares have a common edge out of random chosen pairs of squares chosen from a 7x7 chess board = 84/(49*24) = 1/14.

Therefore A is the answer.
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There is a 7 × 7 chessboard. Two squares are chosen at random. What is [#permalink]
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