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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

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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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