Four dice are thrown simultaneously and the numbers shown on these...

Let, the outcomes of the four dice are $\left\{a, b, c, d\right\} .$

Then, we have the matrix A = \left[\begin{cases}a & b \\ c & d\end{cases}\right] .

Also, \left|A\right| = \left|$\begin{cases}a & b \\ c & d\end{cases}$\right|$</p><p>$⇒ \left|A\right| = ad - bc$</p><p>Now, for each of the numbers$a, b, c \text{ and } d$there are$6$possibilities, hence total cases$= {6}^{4} .$</p><p>For a non-singular matrix, \left|A\right| \ne 0$

$⇒ ad - bc \ne 0$

$⇒ ad \ne bc$

Also, $a, b, c, d$ are all different numbers from the set ${ 1, 2, 3, 4, 5, 6 } ,$ thus we have total $P4 6 = \dfrac{6 !}{\left(6 - 4\right) !}$

$= \dfrac{6 !}{2 !} = \dfrac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !} = 360 .$

Now, for $ad = bc$

\left(i\right)6 \times 1 = 2 \times 3,$here we have two cases possible.</p><p>When$ad = 6 \times 1 \text{ and } bc = 2 \times 3,$thus, we can choose$a \text{ and } d$in$P2 2 = 2$ways and the same we can do for$b \text{ and } c,$and hence, we have total$2 \times 2 = 4$ways.</p><p>Similarly, we can take$ad = 2 \times 3 \text{ and } bc = 6 \times 1,$and for this also we have total$4$ways.</p><p>Thus, we have total$4 + 4 = 8$ways.</p><p>\left(ii\right)6 \times 2 = 3 \times 4,$ again here also, we have $8$ cases possible.

Thus, the favourable cases $= P4 6 - 16 = 360 - 16 = 344$

And, hence the required probability$= \dfrac{344}{{6}^{4}} = \dfrac{344}{1296} = \dfrac{43}{162} .$