In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True? Provided we have this truth table where "$p\implies q$" means "if $p$ then $q$":
$$\begin{array}{|c|c|c|}
\hline
p&q&p\implies q\\ \hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T\\\hline
\end{array}$$
My understanding is that "$p\implies q$" means "when there is $p$, there is q". The second row in the truth table where $p$ is true and $q$ is false would then contradict "$p\implies q$" because there is no $q$ when $p$ is present.
Why then, does the third row of the truth table not contradict "$p\implies q$"? If $q$ is true when $p$ is false, then $p$ is not a condition of $q$.
I have not taken any logic class so please explain it in layman's terms.


Administrative note. You may experience being directed here even though your question was actually about line 4 of the truth table instead. In that case, see the companion question In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? And even if your original worry was about line 3, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.


The statement $(P \land Q) \to P$ should be true, no matter what. So, we should have:
\begin{array}{cc|ccc}
P&Q&(P \land Q) & \to & P\\
\hline
T&T&T&T&T\\
T&F&F&T&T\\
F&T&F&T&F\\
F&F&F&T&F\\
\end{array}
Line 2 shows that we should therefore have that $F \to T = T$
Also note that line 1 forces $T \to T = T$, and that line 4 forces $F \to F=T$, which are another two values of the truth-table for $\to$ that people sometimes wonder about. So, together with the uncontroversial $T \to F = F$, the above give a justification for why we define the $\to$ the way we do.
