Question

If each element of a $2\times 2$ determinant is either 0 or 1, what is the probability that the value of the determinant is positive?

Answer 1

Hint: We first explain the concept of empirical probability and how the events are considered. We take the given events and find the number of outcomes. Using the probability theorem of $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$, we get the empirical probability of the forming determinant.

Complete answer:
Empirical probability uses the number of occurrences of an outcome within a sample set as a basis for determining the probability of that outcome.
We take two events, one with conditions and other one without conditions. The later one is called the universal event which chooses all possible options.
We find the number of outcomes for both events. We take the conditional event A and the universal event as U and numbers will be denoted as $n\left( A \right)$ and $n\left( U \right)$.
We take the empirical probability of the given problem as $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( U \right)}$.
We first form a $2\times 2$ determinant as $\left| \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right|$ which gives determinant value $ad-bc$. We take this event as U and find its $n\left( U \right)$. For every element we have two choices 0 or 1.
So, $n\left( U \right)={{2}^{4}}=16$.
Now for conditional event A, if $ad-bc$ is positive then we have only one choice $ad-bc=1$.
This is only possible when $ad=1;bc=0$ which gives $a=d=1;b=c=0$.
The particular determinant ${{I}_{2}}=\left| \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right|$ gives $n\left( A \right)=1$.
The empirical probability of the shooting event is $P\left( A \right)=\dfrac{1}{16}$.

Note:
We need to understand the concept of universal events. This will be the main event that is implemented before the conditional event. Empirical probabilities, which are estimates, calculated probabilities involving distinct outcomes from a sample space are exact.