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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- Sep 1st 2010, 02:43 AMumangaroraChessBoard
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...! - Sep 1st 2010, 02:54 AMCaptainBlack
- Sep 1st 2010, 03:38 AMumangarora
plz can u show working n ans.. im clueless.

- Sep 1st 2010, 04:14 AMCaptainBlack
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 - Sep 1st 2010, 04:28 AMArchie Meade
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. - Sep 1st 2010, 06:10 AMSoroban
Hello, umangarora!

Quote:

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: .$\displaystyle _{64}C_2 \:=\:{64\choose2} \:=\:2016$ choices.

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

In a row, $\displaystyle \square\!\square\!\square\!\square\!\square\!\squa re\!\square\!\square$, 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: .$\displaystyle 56 + 56 \:=\:112$ possible dominos.

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

- Sep 1st 2010, 07:01 AMumangarora
- Sep 1st 2010, 07:55 AMArchie Meade
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

$\displaystyle \displaystyle\frac{61}{63}$

hence the probability that the squares have one of them in the corner,

with another**not**sharing a side is

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

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

$\displaystyle \displaystyle\frac{60}{63}$

hence the probability that the squares include one of the remaining edge squares,

but the other does**not**share a side with it is

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

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

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

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

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