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Q.

Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2 \times 2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :

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a

$\frac{22}{81}$

b

$\frac{23}{81}$

c

$\frac{45}{162}$

d

$\frac{43}{162}$

answer is D.

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Detailed Solution

$(S) = 6 \times 6 \times 6 \times 6 = 6^{4}$

E=matrices have all different entries and are non singular

$\begin{array}{rcl} E & = & [\begin{matrix} a & b \\ c & d \end{matrix}] \Rightarrow | E | = ad - bc \neq 0 \end{array}$

ad=bc is possible if

a	b	c	d
6	2	3	1
6	3	2	1
2	1	6	3
2	6	1	3

6	3	4	2
6	4	3	2
3	6	2	4
3	2	6	4

$\begin{array}{rcl} n (E) & = & ^{6} C_{4} (4!) - 16 = \\ P (E) & = & \frac{344}{6^{4}} = \frac{43}{162} \end{array}$

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Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2×2 matrices. The probability that such formed matrices have all different entries and are non-singular, is :