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let A= set of 3x3 determinat having entries +1 or -1, if a determinat A1 is chosen randomly from the set of A,then the probability that the product of the elements of any row or any column of A1 is -1 is _
I am not sure, but here is my best guess.
Sample Space = 2^9
(There are 9 positions, each position has two possiblities (-1 or +1) )
Favourable Outcomes :
Number of -1 in each row or coloumn should be odd.
So we get two cases here.
(i) No. of -1 in each row and coloumn is 3
(ii) No. of -1 in each row and coloumn is 1
Case 1:
This case can occur only when all the elements in the matrix is -1.
No. of favourable cases = 1
Case 2:
Consider the matrix row wise.
In the first row, you can put the -1 in any of the 3 places.
In the second row, you can put the -1 in two places (no below the previous one)
In the third row, you can put the -1 in only one place (remaining one, not below the other two).
So you get 3*2*1 = 6 possibilities in this case.
Total no. of favourable outcome = 6+1 = 7
So required probability is 7/29 ..
Okay I missed a case.
All digits in a single row are -1. Rest of the elements are 1.
So we get 3 more cases.
So Probability = 10/2^9
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