logic
I need to find the Boolean expression for the truth table below where $P$, $Q$, $R$ are inputs, and $S$ is the output. Does anyone have a cool easy way of solving such problems please? Your help will be appreciated.
$$\begin{array}{c|c|c||c}
P & Q & R & S\\
\hline
1 & 1 & 1 & 1\\
\hline
1 & 1 & 0 & 0\\
\hline
1 & 0 & 1 & 1\\
\hline
1 & 0 & 0 & 0\\
\hline
0 & 1 & 1 & 1\\
\hline
0 & 1 & 0 & 1\\
\hline
0 & 0 & 1 & 0\\
\hline
0 & 0 & 0 & 0
\end{array}$$
What you've posted is the truth table for material implication (the conditional) $p \rightarrow q$.
EDIT:
To better understand the material conditional (the connective $\rightarrow$, i.e. implication), see the following posts:
At each of those links, you'll find more linked questions that are also relevant. You are not alone: logical implication (e.g. $p\rightarrow q$) is perhaps the most difficult connective to grasp, in terms of its truth-table and how it is defined, in classical logic, (which is, in part, explained by the fact that in natural language, the term "implies" is used in ways whose meaning is not captured by its narrower meaning, as defined in logic).
If you have any further questions, I'll be happy to try and answer them!
Layout: You want the first line to be true when $H,M$ and $B$ are all true. So your statement $F$ must look like $(H\land M\land B)\lor \text{Something}$.
But you also want it to be true when $H,M$ are true and $B$ is false, equivalently, when $H, M$ and $\neg B$ is true. So, using the information in the paragraph above, $F$ must look like $(H\land M\land B)\lor (H\land M\land \neg B)\lor \text{Something else}$.
Proceed in this fashion to find $F$.
Edit: Firstly note that we need only apply this technique for the true lines, because by exactly pinpointing the true lines, the falsehoods will be determined.
So inspecting the true lines you can find:
$$(H\land M\land B)\lor (H\land M\land \neg B)\lor (H\land \neg M\land B)\lor (\neg H\land M\land B),$$
which has the expected truth table: 
Best Answer
A mechanical way of getting an expression that has a desired truth table is to take the disjunction of the formulas that determine the rows you want with 1s.
For example, for a truth table on P, Q, and R that has $$\begin{array}{c|c|c||c} P & Q & R & \text{Table}\\ \hline 1 & 1 & 1 & 1\\ \hline 1 & 1 & 0 & 0\\ \hline 1 & 0 & 1 & 1\\ \hline 1 & 0 & 0 & 0\\ \hline 0 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1\\ \hline 0 & 0 & 1 & 0\\ \hline 0 & 0 & 0 & 0 \end{array}$$ we note that the rows corresponding to $1$s are: $P\land Q\land R$, $P\land \neg Q\land R$, $\neg P\land Q\land R$, and $\neg P\land Q\land\neg R$. So a formula that has the desired truth table is $$(P\land Q\land R)\lor (P\land\neg Q\land R)\lor (\neg P\land Q\land R) \lor (\neg P\land Q\land \neg R).$$ This is, of course, unlikely to be the simplest formula that works, and may be simplified later.