Question

if each element of a second order determinant is either 0 or 1, the probability that the value of the determinant is positive is​

Answer 1

Answer:

Thus, the probability that the determinants is positive = 3/16.

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that a determinant of order 2.

➢ Let assume that the determinant of 2 × 2 order as

$\rm :\longmapsto\:\triangle = \begin{array}{|cc|}\sf a_{11} &\sf a_{12} \\ \sf a_{21} &\sf a_{22} \\\end{array}$

➢ Now, Further it is given that each element is either 0 or 1.

$\rm \implies\:a_{11} \: can \: assume \: the \: value \: either \: 0 \: or \: 1$

$\rm \implies\:a_{12} \: can \: assume \: the \: value \: either \: 0 \: or \: 1$

$\rm \implies\:a_{21} \: can \: assume \: the \: value \: either \: 0 \: or \: 1$

$\rm \implies\:a_{22} \: can \: assume \: the \: value \: either \: 0 \: or \: 1$

➢ So, it means total possible determinants of order 2 having each element either 0 or 1 are

$\rm \: = \:2 \times 2 \times 2 \times 2$

$\rm \: = \:16$

Now, we have to find the probability that determinant value is positive.

$\rm :\longmapsto\:\triangle > 0$

$\rm :\longmapsto\:a_{11}a_{22} - a_{21}a_{12} > 0$

It means,

$\rm :\longmapsto\:a_{11}a_{22} > 0$

$\rm \implies\:a_{11} = 1 \: \: and \: \: a_{22} = 1$

Also,

$\rm :\longmapsto\:a_{11}a_{22} - a_{21}a_{12} > 0$

$\rm \implies\:a_{21}a_{12} < 1$

$\rm \implies\:a_{21}a_{12} = 0$

So, three possibilities are there, which are as follow

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a_{12} & \bf a_{21} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 0 \\ \\ \sf 0 & \sf 1 \end{array}} \\ \end{gathered}$

So, possible determinants whom value is positive are

$\rm :\longmapsto\:\begin{array}{|cc|}\sf 1 &\sf 0 \\ \sf 0 &\sf 1 \\\end{array}, \: \: \begin{array}{|cc|}\sf 1 &\sf 1 \\ \sf 0 &\sf 1 \\\end{array}, \: \: \begin{array}{|cc|}\sf 1 &\sf 0 \\ \sf 1 &\sf 1 \\\end{array}$

So, total number of favourable outcomes are 3.

We know,

$\boxed{ \bf{ \: \bf \:Probability =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \:}}}$

Hence,

$\red{\bf :\longmapsto\:Required \: probability \: = \: \dfrac{3}{16}}$