ChessBoard

**umangarora** · September 1st 2010, 02:43 AM

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...!

**CaptainBlack** · September 1st 2010, 02:54 AM

Originally Posted by umangarora

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

**umangarora** · September 1st 2010, 03:38 AM

plz can u show working n ans.. im clueless.

**CaptainBlack** · September 1st 2010, 04:14 AM

Originally Posted by umangarora

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

**Archie Meade** · September 1st 2010, 04:28 AM

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.

**Soroban** · September 1st 2010, 06:10 AM

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: . $\dfrac{112}{2016} \:=\:\dfrac{1}{18}$

**umangarora** · September 1st 2010, 07:01 AM

Originally Posted by soroban

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

**Archie Meade** · September 1st 2010, 07:55 AM

Originally Posted by umangarora

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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