[ritika_j]

Answer is 1/18

[supervish]

Originally Posted by ritika_j

Answer is 1/18

yayayayayayayayaya !! my answer's right !

Originally Posted by supervish

Sample space = 64C2

Seven unique adjacent square sets in each row and each column.

Thats 7*(8 rows + 8 columns) = 112.

P(ur event) = 112 / 64C2

112/64C2 = 1/18

Me dont have 100% track record with probability, so this helps

Vish

[ritika_j]

May be this explanation helps for ppl like me....



No. of ways of choosing the first square is 64.

No.of ways of choosing the second square is 63.

No. of ways to choose both= 64 x 63 = 4032.


step-1: If the first square is the corner one(4 corners are there), the second square can be chosen in 2 ways.

step-2:If the first square is on the side(24 sides are there), the second square can be chosen in 3 ways.

step-3:If the first square is among the remaining squares(36 are remaining), 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

[TheAkshaT]

Originally Posted by ritika_j

May be this explanation helps for ppl like me....



No. of ways of choosing the first square is 64.

No.of ways of choosing the second square is 63.

No. of ways to choose both= 64 x 63 = 4032.


step-1: If the first square is the corner one(4 corners are there), the second square can be chosen in 2 ways.

step-2:If the first square is on the side(24 sides are there), the second square can be chosen in 3 ways.

step-3:If the first square is among the remaining squares(36 are remaining), 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

Ritika,

ur answer is correct but solution is wrong...

Sample space won't be 64*63
as this will include selection of 1 and 2 as well as 2 and 1
which are apparently same cases!

so it'll be 64*63/2 which is nothing but 64C2

Same mistake in calculating squares with common side...

Always use Combinations when it is a case of selections...

-Akshat

[ritika_j]

Originally Posted by TheAkshaT

ur answer is correct but solution is wrong...

Sample space won't be 64*63
as this will include selection of 1 and 2 as well as 2 and 1
which are apparently same cases!

so it'll be 64*63/2 which is nothing but 64C2

Same mistake in calculating squares with common side...

Always use Combinations when it is a case of selections...

-Akshat

hey akshat
there are 64 sqaure in a chessboard.. rite
so for selecting 2 sqaures one by one..
we'll do 64C1*63C1=64*63
as here the selection of second sqaure is based on the selection of first sqaure...
so i dont think we need to do 64C2...

what say.........

[pendyal]

are we to consider only squares of side 1. : :

[ritika_j]

Originally Posted by pendyal

are we to consider only squares of side 1. : :

we need to consider the sides of which square we choose first only then we can find the probability of sqauresl having common sides...

[pendyal]

Originally Posted by ritika_j

we need to consider the sides of which square we choose first only then we can find the probability of sqauresl having common sides...

yes but if i choose a square of side 2 in the beginning then ur solution wudnt cover that case,right?

[TheAkshaT]

Originally Posted by ritika_j

there are 64 sqaure in a chessboard.. rite
so for selecting 2 sqaures one by one..
we'll do 64C1*63C1=64*63
as here the selection of second sqaure is based on the selection of first sqaure...
so i dont think we need to do 64C2...

what say.........

Selecting 2 things one by one (with or without replacement!) and selecting 2 things at random are two different events!

when selecting one by one the order of selection makes the difference...
Like if u choose square no. 1 and the square no. 2 this will be different than selecting first square no. 2 and then sq. no. 1
When order is important it is permuatation...right...

But in this question order of selection is not important...

Besides! i think pendyal is rite!
The question is :

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

I think we should consider all possible squares on a chessboard...



-Akshat

[gauri_j3]

Originally Posted by pendyal

i am coming up with a different answer.cud u tell me if it is right.

first arrange all the 5 consonants in 5! ways.
now the 3 vowels must go in between the consonantsor at the extremes.so there are 6 possible places for the consonants to occupy.
we can choose 3 places from the six in 6P3 ways i.e.120 ways.

so the total possible no. of ways are 5!*120=14400.

bye..

your soln is perfectly right.