Question

The entries in a $2X2$ determinant $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$ are integers chosen randomly and independently, and for each entry, the probability that the entry is odd is $p$. If the probability that value of determinant is even is$\frac{1}{2}$, then find the value of $p$

• A. $\frac{1}{4\sqrt{2}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}$

$|D|=ad-bc$ is $|D|$ is even than
$\left(i\right)ad$ and $bc$ are even : $a$ or $d$ is even and $b$or$c$ is even.
$\left(ii\right)ad$and $bc$ are odd: $a,d$ are odd and $b,c$ are odd
$\therefore P\left(|D|iseven\right)=p^4+(1-p^2)(1-p^2)=\frac{1}{2}$
$p^4+1+p^4-2p^2=\frac{1}{2}$
$4p^4-4p^2+1=0$
$(2p^2-1{)}^2=0$
$\Rightarrow 2p^2-1=0\Rightarrow p^2=\frac{1}{2}$
$p=\frac{1}{\sqrt{2}}\because 0\le p\le 1$