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(B) Let the determinant be $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$.Since each element can be $0$ or $1$, there are $2^4 = 16

Vedclass · Accepted Answer

$\frac{3}{8}$

Vedclass · Answer

(B) Let the determinant be $\Delta = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$.Since each element can be $0$ or $1$, there are $2^4 = 16$ possible determinants.The determinant is zero if $ad = bc$.Case $1$: $ad = 0$ and $bc = 0$.For $ad=0$, the pairs $(a,d)$ can be $(0,0), (0,1), (1,0)$, which are $3$ possibilities.Similarly, for $bc=0$, there are $3$ possibilities.Total cases for $ad=bc=0$ is $3 	imes 3 = 9$.Case $2$: $ad = 1$ and $bc = 1$.This only happens if $a=1, d=1$ and $b=1, c=1$, which is $1$ possibility.Total cases where $\Delta = 0$ is $9 + 1 = 10$.Total cases where $\Delta 
eq 0$ is $16 - 10 = 6$.Probability $= \frac{6}{16} = \frac{3}{8}$.

$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is

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