Mathematics - Probability Question with Solution | TestHub

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

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

Also, $\left|A\right|=\left|\begin{array}{ll}a& b\\ c& d\end{array}\right|$

$\Rightarrow \left|A\right|=ad-bc$

Now, for each of the numbers $a, b, c \& d$ there are $6$ possibilities, hence total cases$=6^4.$

For a non-singular matrix, $\left|A\right|\ne 0$

$\Rightarrow ad-bc\ne 0$

$\Rightarrow ad\ne bc$

Also, $a, b, c, d$ are all different numbers from the set $\{1, 2, 3, 4, 5, 6\},$ thus we have total ${}^{6}P_4=\frac{6!}{\left(6-4\right)!}$

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

Now, for $ad=bc$

$\left(\mathrm{i}\right) 6\times 1=2\times 3,$ here we have two cases possible.

When $ad=6\times 1 \& bc=2\times 3,$ thus, we can choose $a \& d$ in ${}^{2}P_2=2$ ways and the same we can do for $b \& c,$ and hence, we have total $2\times 2=4$ ways.

Similarly, we can take  $ad=2\times 3 \& bc=6\times 1,$ and for this also we have total $4$ ways.

Thus, we have total $4+4=8$ ways.

$\left(\mathrm{ii}\right) 6\times 2=3\times 4,$ again here also, we have $8$ cases possible.

Thus, the favourable cases $={}^{6}P_4-16=360-16=344$

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