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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 _

3 years ago


Answers : (2)


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 ..

3 years ago

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

3 years ago

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