JEE Main 2021 — Mathematics Probability & Statistics

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

Then, we have the matrix $A=[\begin{matrix}a & b \\ c & d\end{matrix}].$A=[ac​bd​].

Also, $|A|=|\begin{matrix}a & b \\ c & d\end{matrix}|$∣A∣=∣ac​bd​∣

$$\Rightarrow |A|=ad-bc$$⇒∣A∣=ad−bc

Now, for each of the numbers a,b,c&d there are $6$6 possibilities, hence total cases$={6}^{4}.$=64.

For a non-singular matrix, $|A|\neq 0$∣A∣=0

$$\Rightarrow ad-bc\neq 0$$⇒ad−bc=0

$$\Rightarrow ad\neq bc$$⇒ad=bc

Also, $a,b,c,d$a,b,c,d are all different numbers from the set ${1,2,3,4,5,6},$1,2,3,4,5,6, thus we have total $P46=\frac{6!}{(6-4)!}$P46=(6−4)!6!​

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

Now, for $ad=bc$ad=bc

$(i)6\times 1=2\times 3,$(i)6×1=2×3, here we have two cases possible.

When ad=6\times 1&bc=2\times 3, thus, we can choose a&d in $P22=2$P22=2 ways and the same we can do for b&c, and hence, we have total $2\times 2=4$2×2=4 ways.

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

Thus, we have total $4+4=8$4+4=8 ways.

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

Thus, the favourable cases $=P46-16=360-16=344$=P46−16=360−16=344

And, hence the required probability$=\frac{344}{{6}^{4}}=\frac{344}{1296}=\frac{43}{162}.$=64344​=1296344​=16243​.