There is a 7 × 7 chessboard. Two squares are chosen at random. What is

Question

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

firas92 · Accepted Answer

6 pairs of squares share a vertical edge in each row and so total number of pairs of squares sharing a vertical edge = 6*7 = 42

Similarly 6 pairs of squares share a horizontal edge in each column and so total number of pairs of squares sharing a horizontal edge = 6*7 = 42

Total number of pairs of squares that share an edge = 42+42=84

Total number of pairs that can be chosen from 7x7 squares = 49C2

Therefore the probability that the chosen pair will share an edge =  \frac{84}{49C2} = \frac{1}{14}

Answer is  (A)

Hit Kudos if this helped!

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Rakesh1987 · Answer

25 squares have 4 common edges: prob of first square among this is 25/49* prob of second sharing edge 4/48
20 have 3 common edges and: prob of first square among this is 20/49* prob of second sharing edge 3/48
4 have 2 common edges: prob of first among these 4/49* prob of second sharing edge 2/48

Total= 1/49*1/48*(25*4+20*3+4*2)
=1/49*1/48*168
=  1/14. Answer is A.

PreiteeRanjan · Answer

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

nick1816 · Answer

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 · Answer

kindly explain with a drawing.

wizardofoddz · Answer

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.

samarpan.g28 · Answer

Total number of squares 7*7=49. Select any 2 from 49: 49C2=1176.

The number of squares with a common edge = 7 (7 squares on each edge). Select any 2 from 7: 7C2=21.
We have a total of 4 edges therefore 21*4=84.

84/1176=1/14. Option (A) is correct.

Regor60 · Answer

It's not really necessary to calculate a precise figure to arrive at the answer.

The maximum probability of picking squares with a shared edge is if you assume a square surrounded by 4 other squares, yes ?

So given a selected square and 48 other squares to chose among, the probability of selecting a second square that shares an edge is:

4/48 = 1/12

Since this is the maximum probability and we're aware by observation that squares on the periphery will NOT be surrounded by 4 other squares, the actual probability must be less than 1/12.

The only answer that fits is:

1/14

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There is a 7 × 7 chessboard. Two squares are chosen at random. What is

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