If each element of a second order determinant | vedclass

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(A) second order determinant is given by $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element $a, b, c, d \in \{0, 1\}$. The total n

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$\frac{3}{16}$

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(A) second order determinant is given by $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$. Each element $a, b, c, d \in \{0, 1\}$. The total number of possible determinants is $2^4 = 16$. The value of the determinant is $ad - bc$. We want $ad - bc > 0$,which means $ad > bc$. Since $a, b, c, d \in \{0, 1\}$,the product of two elements can only be $0$ or $1$. $ad > bc$ is possible only if $ad = 1$ and $bc = 0$. $ad = 1$ implies $a = 1$ and $d = 1$. $bc = 0$ implies that at least one of $b$ or $c$ must be $0$. The possible pairs $(b, c)$ such that $bc = 0$ are $(0, 0), (0, 1), (1, 0)$. Thus,the favorable cases are: $1$. $\begin{vmatrix} 1 & 0 \ 0 & 1 \end{vmatrix} = 1(1) - 0(0) = 1 > 0$ $2$. $\begin{vmatrix} 1 & 0 \ 1 & 1 \end{vmatrix} = 1(1) - 0(1) = 1 > 0$ $3$. $\begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix} = 1(1) - 1(0) = 1 > 0$ There are $3$ favorable outcomes. Therefore,the required probability is $\frac{3}{16}$.

If each element of a second order determinant is either zero or one,what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently,each value being assumed with probability $\frac{1}{2}$).

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