Combining probabilities

Consider two distinct possible outcomes, $X$ and $Y$, of an observation made on the system $S$, with probabilities of occurrence $P(X)$ and $P(Y)$, respectively. Let us determine the probability of obtaining the outcome $X$ or $Y$, which we shall denote $P(X\mid Y)$. From the basic definition of probability

$$P(X\mid Y) = ~_{lt\,{\mit\Omega}({\mit\Sigma})\rightarrow\infty} \frac{ {\mit\Omega}(X \mid Y)}{{\mit\Omega}({\mit\Sigma})},$$ (2)

where ${\mit\Omega}(X \mid Y)$ is the number of systems in the ensemble which exhibit either the outcome $X$ or the outcome $Y$. It is clear that

$${\mit\Omega}(X\mid Y) = {\mit\Omega}(X) + {\mit\Omega}(Y)$$ (3)

if the outcomes $X$ and $Y$ are mutually exclusive (which they must be the case if they are two distinct outcomes). Thus,

$$P(X\mid Y) = P(X) + P(Y).$$ (4)

So, the probability of the outcome $X$ or the outcome $Y$ is just the sum of the individual probabilities of $X$ and $Y$. For instance, with a six sided die the probability of throwing any particular number (one to six) is $1/6$, because all of the possible outcomes are considered to be equally likely. It follows from what has just been said that the probability of throwing either a one or a two is simply $1/6+1/6$, which equals $1/3$.

Let us denote all of the $M$, say, possible outcomes of an observation made on the system $S$ by $X_i$, where $i$ runs from $1$ to $M$. Let us determine the probability of obtaining any of these outcomes. This quantity is clearly unity, from the basic definition of probability, because every one of the systems in the ensemble must exhibit one of the possible outcomes. But, this quantity is also equal to the sum of the probabilities of all the individual outcomes, by (4), so we conclude that this sum is equal to unity. Thus,

$$\sum_{i=1}^{M} P(X_i) =1,$$ (5)

which is called the normalization condition, and must be satisfied by any complete set of probabilities. This condition is equivalent to the self-evident statement that an observation of a system must definitely result in one of its possible outcomes.

There is another way in which we can combine probabilities. Suppose that we make an observation on a state picked at random from the ensemble and then pick a second state completely independently and make another observation. We are assuming here that the first observation does not influence the second observation in any way. The fancy mathematical way of saying this is that the two observations are statistically independent. Let us determine the probability of obtaining the outcome $X$ in the first state and the outcome $Y$ in the second state, which we shall denote $P(X\otimes Y)$. In order to determine this probability, we have to form an ensemble of all of the possible pairs of states which we could choose from the ensemble ${{\mit\Sigma}}$. Let us denote this ensemble ${{\mit\Sigma}}\otimes {{\mit\Sigma}}$. It is obvious that the number of pairs of states in this new ensemble is just the square of the number of states in the original ensemble, so

$${\mit\Omega}({{\mit\Sigma}}\otimes{{\mit\Sigma}}) = {\mit\Omega}({{\mit\Sigma}})\, {\mit\Omega}({{\mit\Sigma}}).$$ (6)

It is also fairly obvious that the number of pairs of states in the ensemble ${{\mit\Sigma}}\otimes {{\mit\Sigma}}$ which exhibit the outcome $X$ in the first state and $Y$ in the second state is just the product of the number of states which exhibit the outcome $X$ and the number of states which exhibit the outcome $Y$ in the original ensemble, so

$${\mit\Omega}(X\otimes Y) = {\mit\Omega}(X) \,{\mit\Omega}(Y).$$ (7)

It follows from the basic definition of probability that

$$P(X\otimes Y) = ~_{lt\,{\mit\Omega}({\mit\Sigma})\rightarrow... ...it\Omega}({{\mit\Sigma}}\otimes {{\mit\Sigma}})}= P(X) \,P(Y).$$ (8)

Thus, the probability of obtaining the outcomes $X$ and $Y$ in two statistically independent observations is just the product of the individual probabilities of $X$ and $Y$. For instance, the probability of throwing a one and then a two on a six sided die is $1/6 \times 1/6$, which equals $1/36$.