Decomposition of an Event with Respect to a Partition

Given any event $ E \subseteq U $ and a partition $ \{E_1, E_2, \dots, E_n\} $ of the sample space $ U $, the event $ E $ can be represented as the union of simpler events. $$ E = (E \cap E_1) \cup (E \cap E_2) \cup \dots \cup (E \cap E_n) $$ The events $ E \cap E_i $ are pairwise mutually exclusive.

The decomposition of an event does not introduce new probabilistic principles. Rather, it follows directly from elementary set-theoretic arguments.

Nonetheless, it plays a crucial conceptual role in probability theory, as it underpins the law of total probability. By organizing the space of possible outcomes in a systematic way, it makes probability calculations more transparent, more structured, and easier to manage, even in complex scenarios.

Note. The decomposition of an event is a fundamental conceptual step on which Bayes' theorem is based.

Explanation
A practical example

Explanation

Consider the sample space $ U $ associated with a random experiment.

A collection of events $ E_1, E_2, \dots, E_n $ is said to form a partition of $ U $ if the events cover the entire sample space and do not overlap.

$ E_i \ne \varnothing $ for every $ i = 1, 2, \dots, n $
$ E_i \cap E_j = \varnothing $ for $ i \ne j $
$ E_1 \cup E_2 \cup \dots \cup E_n = U $

Equivalently, each event in the partition is nonempty, no two distinct events can occur simultaneously, meaning they are mutually exclusive, and the union of all events coincides with the sample space.

Since $ E_1 \cup E_2 \cup \dots \cup E_n = U $, the intersection of any event $ E $ with the sample space $ U $ is simply the event $ E $ itself.

$$ E = E \cap U $$

Substituting the expression for $ U $ yields:

$$ E = E \cap (E_1 \cup E_2 \cup \dots \cup E_n) $$

Applying the distributive property of intersection over union, we obtain:

$$ E = (E \cap E_1) \cup (E \cap E_2) \cup \dots \cup (E \cap E_n) $$

Because the events in the partition are pairwise mutually exclusive, that is, they do not overlap, if $ i \ne j $ then:

$$ (E \cap E_i) \cap (E \cap E_j) = E \cap (E_i \cap E_j) = \varnothing $$

It follows that the events $ E \cap E_i $ are themselves pairwise mutually exclusive.

Since these events are mutually exclusive, their probabilities can be summed, leading to:

$$ p(E) = p(E \cap E_1) + p(E \cap E_2) + \dots + p(E \cap E_n) $$

This identity is known as the law of total probability.

Accordingly, the probability of an event $ E $ is expressed as a sum of probabilities corresponding to the cases in which $ E $ occurs within the individual events of a partition of the sample space.

A practical example

Consider an urn containing 10 numbered balls in two different colors:

6 white balls, 2 of which have even numbers
4 black balls, 3 of which have even numbers

A single ball is drawn at random from the urn.

In this example, the sample space $ U $ is described solely in terms of the color of the ball drawn, disregarding all other attributes. The possible outcomes are therefore two.

$$ U = \{ white \ , black \} $$

Define two events that form a partition of $ U $:

$ E_1 $: the ball drawn is white
$ E_2 $: the ball drawn is black

The events $ E_1 $ and $ E_2 $ are mutually exclusive, because if a ball is white, it cannot be black, and vice versa.

$$ E_1 \cap E_2 = \emptyset $$

Moreover, $ E_1 $ and $ E_2 $ together exhaust the sample space $ U $.

$$ E_1 \cup E_2 = U $$

Therefore, $ E_1 $ and $ E_2 $ constitute a partition of $ U $.

Now define the event $ E $: the ball drawn has an even number.

We already know that among the 6 white balls, 2 have even numbers, and among the 4 black balls, 3 have even numbers.

At this point, we apply the decomposition of an event with respect to the partition of the sample space $ U $:

$$ E = (E \cap E_1) \cup (E \cap E_2) $$

Here, $ E \cap E_1 $ corresponds to drawing a white ball with an even number, while $ E \cap E_2 $ corresponds to drawing a black ball with an even number. These two events are mutually exclusive.

We now compute the probabilities of the two events:

$$ p(E \cap E_1) = \frac{2}{10} $$

$$ p(E \cap E_2) = \frac{3}{10} $$

Since the two events are mutually exclusive, the probability of the event $ E $ is the sum of their probabilities.

$$ p(E) = p(E \cap E_1) + p(E \cap E_2) $$

$$ p(E) = \frac{2}{10} + \frac{3}{10} = \frac{5}{10} = \frac{1}{2} $$

$$ p(E) = \frac{1}{2} $$

In this way, the probability of the event "even number" is not computed directly, but instead decomposed into the cases "white" and "black".

The decomposition makes explicit that total probability arises as the sum of the contributions associated with the individual scenarios defined by the partition.

Note. The probability of the event "even number" could also be computed without distinguishing the color of the balls. Since there are $ 2+3 = 5 $ balls with even numbers out of $ 10 $, the probability of the event is $$ p(E) = \frac{5}{10} = \frac{1}{2} $$ Clearly, the result is the same.

And so on.