JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 4)
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 $$\times$$ 2 matrices. The probability that such formed matrix have all different entries and are non-singular, is :
$${{45} \over {162}}$$
$${{21} \over {81}}$$
$${{22} \over {81}}$$
$${{43} \over {162}}$$
Explanation
$$A = \left| {\matrix{
a & b \cr
c & d \cr
} } \right|$$
| A | = ad $$-$$ bc
Total case = 64
For non-singular matrix | A | $$\ne$$ 0 $$\Rightarrow$$ ad $$-$$ bc $$\ne$$ 0
$$\Rightarrow$$ ad $$\ne$$ bc
And a, b, c, d are all different numbers in the set {1, 2, 3, 4, 5, 6}
Now for ad = bc
(i) 6 $$\times$$ 1 = 2 $$\times$$ 3
$$\Rightarrow$$ $$\left. \matrix{ a = 6,b = 2,c = 3,d = 1 \hfill \cr or\,a = 1,b = 2,c = 3,d = 6 \hfill \cr : \hfill \cr : \hfill \cr} \right\}$$ 8 each cases
(ii) 6 $$\times$$ 2 = 3 $$\times$$ 4
$$\Rightarrow$$ $$\left. \matrix{ a = 6,b = 3,c = 4,d = 2 \hfill \cr or\,a = 2,b = 3,c = 4,d = 6 \hfill \cr : \hfill \cr : \hfill \cr} \right\}$$ 8 such cases
favourable cases
= $$^6C_4 \times 4! - 16$$
required probability
$$ = {{{^6C_4 \times 4!} - 16} \over {{6^4}}} = {{43} \over {162}}$$
| A | = ad $$-$$ bc
Total case = 64
For non-singular matrix | A | $$\ne$$ 0 $$\Rightarrow$$ ad $$-$$ bc $$\ne$$ 0
$$\Rightarrow$$ ad $$\ne$$ bc
And a, b, c, d are all different numbers in the set {1, 2, 3, 4, 5, 6}
Now for ad = bc
(i) 6 $$\times$$ 1 = 2 $$\times$$ 3
$$\Rightarrow$$ $$\left. \matrix{ a = 6,b = 2,c = 3,d = 1 \hfill \cr or\,a = 1,b = 2,c = 3,d = 6 \hfill \cr : \hfill \cr : \hfill \cr} \right\}$$ 8 each cases
(ii) 6 $$\times$$ 2 = 3 $$\times$$ 4
$$\Rightarrow$$ $$\left. \matrix{ a = 6,b = 3,c = 4,d = 2 \hfill \cr or\,a = 2,b = 3,c = 4,d = 6 \hfill \cr : \hfill \cr : \hfill \cr} \right\}$$ 8 such cases
favourable cases
= $$^6C_4 \times 4! - 16$$
required probability
$$ = {{{^6C_4 \times 4!} - 16} \over {{6^4}}} = {{43} \over {162}}$$
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