Four cards are placed on a table. Each card has a number on one side and a letter on the other. The cards on display show $3$, $8$, $\text{E}$ and $\text{K}$. Which cards must be turned over to satisfy our requirement?
"If a card has a vowel on one side, then it has an even number on the other side."

This is a classic logic problem (Wason Selection Task). Let's evaluate the conditional rule: **"If Vowel ($\text{P}$), then Even Number ($\text{Q}$)"**.
To test a conditional statement $\text{P}\to \text{Q}$, we must check two cases:

1. The case that confirms $\text{P}$ (to see if it leads to $\text{Q}$).
2. The case that violates $\text{Q}$ (to make sure it doesn't come from $\text{P}$, by Modus Tollens).

Let's verify each visible card:

* Card '3' (Odd Number): This is a violation of an even number ($\text{not-Q}$). If we turn it over and find a vowel, the rule is violated. Therefore, we must turn over card $3$.
* Card '8' (Even Number): This represents $\text{Q}$. The rule states "If vowel $\to$ even", but it does not state that an even number can only have a vowel. An even number can have a consonant on the other side without breaking the rule. So, we do not need to check card $8$.
* Card 'E' (Vowel): This represents $\text{P}$. We must check it to confirm that there is an even number on the other side.
* Card 'K' (Consonant): The rule says nothing about consonants. Turning it over tells us nothing useful.

Thus, we must turn over cards **$3$ and $\text{E}$ **.