Test Index

Probability Part 1

Section: Mathematics

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Question : 6 of 100

Marks: +1, -0

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:

[22 Jul 2021 Shift 2]

$\frac{45}{162}$
$\frac{23}{81}$
$\frac{22}{81}$
$\frac{43}{162}$

Solution:

A=\begin{vmatrix} a & b \\ c & d \end{vmatrix} \;\;\;\; |A|=ad-bc

Total case

=6^{4}

For non-singular matrix

|A|\neq 0 \Rightarrow ad-bc \neq 0

\Rightarrow ad \neq 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 \begin{array}{l} a=6, b=2, c=3, d=1 \\ \text{or } a=1, b=2, c=3, d=6 \end{array}

8 such cases

(ii)

6 \times 2 = 3 \times 4

\Rightarrow \begin{array}{l} a=6, b=3, c=4, d=2 \\ \text{or } a=2, b=3, c=4, d=6 \end{array}

8 such cases

favourable cases

={}^{6}C_{4!}-16

required probability

=\frac{{}^6C_4 \cdot 4! - 16}{6^4} = \frac{43}{162}

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