Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 …

Question

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 :

Accepted Answer

$$A = \left| {\begin{pmatrix} a & b  \
   c & d  \

\end{pmatrix} } \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

$$\Rightarrow$$ $$\left. \begin{pmatrix} a = 6,b = 2,c = 3,d = 1 \hfill \
  or\,a = 1,b = 2,c = 3,d = 6 \hfill \
  : \hfill \
  : \hfill \ \end{pmatrix}  \right\}$$ 8 each cases

(ii) 6 $$\times$$ 2 = 3 $$\times$$ 4

$$\Rightarrow$$ $$\left. \begin{pmatrix} a = 6,b = 3,c = 4,d = 2 \hfill \
  or\,a = 2,b = 3,c = 4,d = 6 \hfill \
  : \hfill \
  : \hfill \ \end{pmatrix}  \right\}$$ 8 such cases

favourable cases

= $$^6C_4 \times 4!  - 16$$

required probability

$$ = {{{^6C_4 \times 4!} - 16} \over {{6^4}}} = {{43} \over {162}}$$

Solutions · 1