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Let $(a, b, c, d)$ be integers where they are not necessarily different. If each one of them is in the set ($0$, $1$, $2$, $3$), what is the probability that $a \cdot d-b \cdot c$ is odd? (For example, ($0$, $3$, $1$, $1$) meet the condition, because $0 \cdot 1-3 \cdot 1=-3$ is odd.) 
Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
In order for $a \cdot d-b \cdot c$ to be odd, we need to consider the parity. We must have (even)-(odd) or (odd)-(even). There are $2\times2+2\times4=12$ ways to pick numbers to obtain an even product. There are $2 \cdot 2=4$ ways to obtain an odd product. Therefore, the total amount of ways to make $a \cdot d-b \cdot c$ odd is $2 \cdot(12 \cdot 4)=96.$ Thus, the answer is $\frac{96}{256}=\frac38$ 