Find the smallest sigma-algebra. 
My attempt at finding the smallest sigma-algebra:
$$
\sigma(A) = \{\varnothing, \Omega, \{1\}, \{2\}, \{3\}, \{2,3,4,5\}, \{1,3,4,5\}, \{1,2,4,5\}, \{4,5\}.
$$
Is this correct? What am I missing? And any idea how to solve $P(\{4,5\})$ or $P(\{4\})$ as given in the question? Thank you.
 A: $\newcommand{\scrM}{\mathscr{M}}$
You can do this systematically by considering the requirements in the definition of a $\sigma$-algebra.

Definition: Given a set $\Omega$ a collection of subsets of $\Omega$ denoted $\scrM$ is a sigma-algebra in $\Omega$ if $\varnothing \in \scrM$, the collection $\scrM$ is closed under countable unions, and $\scrM$ is closed under complements.

So now if you take $S = \{\{1\}, \{2\}, \{3\} \}$, you can ask yourself how can this be enlarged so that it is in fact a $\sigma$-algebra. Of course as you've correctly done we must also have $\varnothing$ in the collection if it is to be a $\sigma$-algebra. Taking unions certainly gives you elements like $\{1\} \cup \{2\} = \{1,2\}$, and so you know immediately that you need to enlarge $S$ to some
$$
S' = \{\varnothing, \{1\},\{2\},\{1,2\},\{3\}\}.
$$
More than this you also need closure under complements, so this tells you that sets like $\{1\}^c = \{2,3,4,5\}$ must also be elements of the $\sigma$-algebra.
But importantly note that once we know $\{1,2\}$ must be an element of the $\sigma$-algebra it follows that $\{1,2\}^c = \{3,4,5\}$ must also be in the $\sigma$-algebra due to closure under complements, so you've missed those sets.
For $p(\{4,5\}$ and $p(\{4\})$ a useful thing to remember is that for probability measures $p(\Omega) = 1$, and if $E$ is any measurable set $E$ and $E^c$ are definitely disjoint so that $$p(E^c)+p(E) =p(E^c \cup E) = p(\Omega) = 1.$$
