28-04-10: Probability

ok

my approach itself was totally wrong

the fact that both these events are dependent itself says it all

sorry to waste ur time bhaiya

the number of ways of doing such that our comdition is satisfied is same no.of ways of filling (N-1)*(N-1) matrix with 1 and -1

but how is that ?

can u please post the solution ?

for every random arrangment
the last row and coloumn can be filled in a unique way such that the product is -1

so total arrangments is

2^n-12

hence the answer is
\frac{2^{^{n-1}^{2}}}{2^{n^{2}}}

No - see this, I just segregate 1 row and 1 column, and fill the others ramdomly in 2^(n-1)2 ways - and to make the prod negative, I shall likewisely fill the segregated row and column.

thast what i told soumik

Soory - did not see u.

I was actually quite excited having suddenly got the idea and did not have patience to go thru all the stuff.

OK - now got it.

What I did is ok or not?

ok here is my solution

without loss of generality remove the first row and first coloumn and what u have is a (n-1)x(n-1) matrix

fill this (n-1)x(n-1) matrix in any way u want

now what u do is

come to the element a_1-2 that is 1st row 2ound coloumn element

now multiply all the elements of the coloumn below it

if the product of elements in that coloumn is is -1 , let this a_1-2 be 1 and if their product is 1 , let this a_1-2 be -1

similarly for the a_1-3 element , look at the coloumn below it .. if product of all the elements in that coloumn is 1, let this a_1-3 be -1 and if their product is -1 , let this a_1-3 be 1

similarly fill all elements of this first row

use a similar process for first coloumn from a_2-1 onwards

for a_2-1 , if u have product of all the elements in the row beside it as -1 , let this a_2-1 be 1 and of their product is 1, let this a_2-1 be -1

fill all elements of this first coloumn similarly

now we r left with a_1-1

now u see that all remaining elements in the 1st row and first coloumn can be uniquely determined depending on the aarrangement of the (n-1)x(n-1) matrix

so the product of all remaining elements in 1st row and 1st coloumn will be the same (test it out on a matrix if u want to see for it yourself)

then u can accordingly put a_1-1 as 1 or -1

if the product of remaining elements (except a_1-1) in first row is -1 , then product of remaining elements(except a_1-1) in the first coloumn will also be -1 .. then u keep a_1-1 as 1

in the other case u keep a_1-1 as -1

hence u can always make it as prooduct equal to -1

the probability hence comes out to be 2^(n-1)2/2ⁿ²