1it is fiitjee test series 2010 question
they asked same question on 3X3 matrix
my answer came out wrong due to misunderstanding of statement
This one is a question someone asked me a few days back.. I guess it is from a ISI paper... (I may be wrong!)
There is a matrix which is NxN
Each element is filled with either 1 or -1 with an equal probability.
What is the probability that the product of all the elements of each row and of each column in the matrix is -1?
**(As always, for students only ;)
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55 Answers
11for the 3X3 matrix the answer was 24/29
and for 2X2 matrix the answer would be 0
62for a 2 x 2 matrix, it is not zero..
Think of the matrix
1 -1
-1 1
1oops
i considered diagonals+row+column
but in 3X3 matrix the answer is still 24/29
and in 2X2 the answer is 2/24
and also in 1X1 the answer is 20/21
1so now the answer seems to be
\large \frac{2^{{(N-1})^{2}}}{2^{{(N})^{2}}}
here N is order of a square matrix
62That's correct harsh..
Could you explain the logic?
And even in the case of diagonals it would not be zero for a 2x2 matrix:
1 1
1 1
:)
1Nishant Sir i dont have any logic for that
i simply used induction
and also if we consider row+column+diagonals then in 2X2 matrix no matrix is possible such that product of entries of each row,column and diagonal is -1
39harsh have u used induction ? or u simple formulated the answer by the two answers in post #10 ??
62lol..
The answer is not very difficult.. just give it a better shot :)
See the answer and then try to get the method ;)
1i am getting a different answer
\texttt{let the be k (-1)'s and n-k 1's in the first coloumn}\\ \texttt{so the probability that product is -1 is }\\ \frac{1}{2^n}\sum_{k=1}^{n}\binom{n}{k} \ \ \texttt{k takes only odd values}\\ P'(n)=\frac{2^n}{2^{n+1}}\\ \texttt{now all the coloumns must have -1 as the product} \\ P(n)=\left (\frac{2^n}{2^{n+1}} \right )^{2n}
can you please point out wer i made a mistake
edited
39wat do u mean by the sentence product of all the elements of each row and of each column in the matrix is -1? ??
u mean if we take any row and any coloumn and multiply all the terms of that row and multiply all terms of this coloumn and then multiply them both , we should get -1 ?
or we should take all rows and coloumns at a time?
1@harsh
dude u r really cool in formulating results [5]
this is not the first time i am seeing this guy do that
kudos! [4]
62RPF
What happens to the rows?
And again are you sure that there will be -1 in the numerator?
1taking rows in account we just have to put 2n in the power in place of n
62@soumik.. I dont know what you mean to say by
Trust me your answer clearly shows that you dont know when to add and when to multiply in finding probabilities!!!
Where did n.2n-1 come into the picture.. It seems that you are only; trying to reach the anwer by some means or the other!!
@RPF: You are just missing a few points! what happens to the columsn.. you are trying to say that the column's product being -1 is independent of the rows being -1?? (If true.. why?)
1yaar soumik
jaa kar thoda aaram kar , tere sabhi answer strange hote ja rahe hai[4][4]
62You are still missing out the answer to this question:
column's product being -1 is independent of the product of elements in a rows being -1?? (If true.. why?)
I dont see a clear reason..
On the contrary I see a correlation!
Take the case of 1x1 matrix
Probability of the column being -1 is 1/2
row being -1 is 1/2
but column and row being -1 is also 1/2
But 1/2 x 1/2 is not equal to 1/2
Hence these events are not independent
1thanks thinking further
but a doubt
can we generalise this like the answer will be P(row)=P(coloumn)
and not P(row)*P(coloumn)
one may think this is crude
but the symmetry of N*N matrix suggests this
and thats the reason why i put that as answer first
62@soumik
yup .. ab=-1, cd=-1 and ac=-1 and bd=-1
and btw .. you have deleted your post which i was supposed to explain where you went wrong [12][11][2]
@RPF: No we cant generalize that! atleast apparently there is no reason