Determining whether a given function can be considered as a probability function Recently I was given the following problem.
The sample space $\Omega = \{ 1, 2, 3, 4 \}$ and the function P is defined as follows: 
$$P(\emptyset) = 0, P(\Omega) = 1, P(\{1,2\}) = \frac12, P(\{3,4\}) = \frac12, P(\{3\}) = \frac13, P(\{4\}) = \frac16 , P(\{1,2,3\}) = \frac56, P(\{1,2,4\}) = \frac23.$$
The question is whether $P$ can be considered a probability function.
My issue with this problem is that I am not sure whether it is definitely possible to answer the question. On the one hand, if P is a probability function, then for each two mutually exclusive events the sum of their probabilities should be $1$. In this case, $\{1\}$ and $\{2,3,4\}$ are mutually exclusive events, but $P(\{1\}) + P(\{2,3,4\})$ is not known. On the other hand, there is not enough information to claim the opposite. It seems to me that the question itself is incorrect, but I am not quite sure because I do not have enough experience in probability theory.
So, is $P$ a probability function?
 A: When I took probability 1, there was a triple $(\Omega, P, \mathcal F)$ consisting of a sample space, a probability measure, and a collection of events, subsets of $\Omega$ to which probabilities are assigned.  The domain of $P$ is  $\mathcal F$.
So I'd give a different answer to this question.  The set of events $\mathcal F$ has not been specified, so we can choose it.  If we take $\mathcal F$ to be all subsets of $\Omega$, then Jeremy's answer is correct: at best $P$ is incompletely specified.  But if we take $\mathcal F$ to be generated by just the subsets $\{1,2\}, \{3\}, \{4\}$, then $P$ is a probability on that $\mathcal F$.  If asked, in this case, what $P(\{1\})$ was,  we could answer: ha-ha, $\{1\}$ is not measurable, is not an event, is not in $\mathcal F$, and it doesn't have (and doesn't need) a probability.
A: Roughly speaking, probability on a sample space $\Omega$ is a function $P:\Sigma\to[0,1]$ defined on a $\color{red}{\sigma\text{-algebra}}$ $\Sigma$ of subsets of $\Omega$. The concept of $\color{red}{\sigma\text{-algebra}}$ means that $\Sigma\subseteq\mathcal{P}(\Omega)$, a subset of the power set of $\Omega$, which has to satisfy several properties:


*

*contains the empty set;

*closed under complements;

*closed under countable unions and intersections.


(In this example, since we have a finite set $\Omega$, the word "countable" can be omitted.) The elements of $\Sigma$ are the measurable sets or the events. Note that the definition does NOT require all subsets to be events. And if you check, you will see that the subsets for which this $P$ is defined do satisfy these properties, and thus they do form a $\sigma$-algebra.
Once we have a $\sigma$-algebra, we can define a probability function $P$ on it. Such a function also has to satisfy certain axioms, such as $P(X^c)=1-P(X)$, and so on. Note how this axiom of probability hinges on the corresponding axiom of $\sigma$-algebras! Again, you can check that all axioms of probability are satisfied in this example.
So the answer is: YES, it is a probability function. The fact that some values, such as $P(\{1\})$ aren't defined only means that they are not admissible events. But in and by itself that doesn't violate the definition.
A: This seems like a trick question. Much information is provided about a possible probability function, $P$. Since the sample space is $\omega = \{1, 2, 3, 4\}$, the probability function must describe the probability of each element of the sample space. Events $1$ and $2$ are defined in three of equalaties together. Since the probability of events $1$ and $2$ aren't defined individually and they are part of the sample space this probability function $P$ is incomplete and not fully defined.
$P$ is not a probability function.
