Q.If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$).

Idea: For a $2\times2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$ with each entry independently $0$ or $1$, list the $2^4=16$ equally likely matrices and count those with $ad-bc>0$.

Step 1 — when is it positive? Since every entry is $0$ or $1$, both $ad$ and $bc$ lie in $\{0,1\}$. So $ad-bc>0$ forces $ad=1$ and $bc=0$. …

Setting up the sample space

A second-order determinant is

Each of $a,b,c,d$ is chosen independently, taking the value $0$ or $1$ with probability $\tfrac12$ each. So there are $2^4=16$ equally likely matrices, and we just count the favourable ones.

When is the determinant positive?

Because every entry is $0$ or $1$, each product satisfies $ad\in\{0,1\}$ and $bc\in\{0,1\}$. Therefore $\Delta=ad-bc$ can only be $-1$, $0$, or $1$. For $\Delta>0$ we must have

Counting the favourable matrices

$ad=1$: a product of $0/1$ values is $1$ only when both are $1$, so $a=1$ and $d=1$. That fixes the diagonal in exactly $1$ way.

$bc=0$: we need at least one of $b,c$ to be $0$, i.e. $(b,c)$ is any pair except $(1,1)$:

• $(b,c)=(0,0)\Rightarrow bc=0$
• $(b,c)=(0,1)\Rightarrow bc=0$
• $(b,c)=(1,0)\Rightarrow bc=0$ …

Method: Classical probability by enumerating an independent-entry sample space

Use this when several slots are each filled independently from a small set of values and you want the probability of some algebraic condition — here, a $2\times2$ determinant of $0$/$1$ entries being positive.

Steps

Step 1: Size the sample space from the independent choices.

With $m$ slots each taking one of $v$ equally likely values, there are $v^{m}$ equally likely configurations ($2^4 = 16$ here). Equal likelihood is what lets you count instead of multiplying probabilities.

Step 2: Turn the target condition into an exact algebraic requirement. …

Common Mistakes

Mistake 1: Forgetting the entries are restricted to $0$ and $1$.

Why it's wrong: with $ad,bc\in\{0,1\}$, "$ad-bc>0$" cannot mean "any $ad$ bigger than $bc$" — it forces $ad=1$ and $bc=0$. Correct approach: pin down that exact condition before counting.

Mistake 2: Counting $\det \ge 0$ or $\det \neq 0$ instead of $\det > 0$. …