Let $A =$ set of all $3$X$3$ determinant having entries $+1$ or $-1$, if a determinant $B$ form set $A$ is chosen randomly, then probability that product of element of any row or any column of $A$ is $-1$.
My approach:
$n(s)=$Total Determinant = $2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2 = 2^9$
$n(E)=$Determinant having product of row or column $(-1)$
$(^3C_3+^3C_1)\cdot2 = 2^3$
so answer should be (2)^3/(2)^9 = 1/64 But given answer is $\frac1{32}$. Please let me know where am i wrong?
It's very difficult to say where you're wrong when you don't explain your reasoning. Here's how you could be right.
Arbitrarily choose the entries that don't lie in the third row or column. There are $2^4$ ways to do this, and this choice determines the remaining entries. Note that the entry in the third row and column is not overdetermined: The product of the top two rows is $1$ and the product of the left two columns is $1$; both products contain the four arbitrarily chosen entries, so the first two entries in the third row and the first two entries in the third column have the same product, allowing you to choose the entry in the third row and column uniquely and consistently.
You correctly counted the total number of possibilities to be $2^9$, so the probability is $2^4/2^9=2^{-5}$.