Let us now revisit the second calculation that we carried
out in the context of our earlier example.
In that example, we calculated the total probability of an
event that can occur under different scenarios.
And it involves the powerful idea of divide and conquer
where we break up complex situations
into simpler pieces.
Here is what is involved.
We have our sample space.
And our sample space is partitioned into a number of
subsets or events.
In this picture we take that number to be 3, so we'll have
it partitioned into three possible scenarios.
It is a partition which means that these events cover the
entire sample, space and they're
disjoint from each other.
For each one of the scenarios we're given their
probabilities.
If you prefer, you can also draw this situation
in terms of a tree.
There are three different scenarios that can happen.
We're interested in a particular event, B. That
event B can happen in three different ways.
It can happen under scenario one, under scenario two, or
under scenario three.
And this corresponds to these particular sub-events.
So for example, this is the event
where scenario A1 happens.
And then event B happens as well.
In terms of a tree diagram, the
picture becomes as follows.
If scenario A1 materializes, event B may occur or event B
might not occur.
Finally, we are given conditional probabilities that
event B will materialize under each one of the different
possible scenarios.
Under those circumstances, can we calculate the probability
of event B?
Of course we can.
And here's how we do it.
First we realize that event B consists of a number of
disjoint pieces.
One piece is when event B occurs together with event A1.
Another piece is when event B occurs together with A2.
Another piece is when event B occurs together with A3.
These three sets are disjoint from each other, as we see in
this picture.
And together they form the event B. Therefore, the
probability of B is going to be, by the additivity axiom of
probabilities, equal to the sum of the probabilities of
these sub-events.
Furthermore, for each one of these sub-events we can use
the multiplication rule and write their
probabilities as follows.
The probability that B and A1 both occur is the probability
that scenario one materializes times the conditional
probability that B occurs given that A1 occurred.
And then we're going to have similar terms under the second
scenario and a similar term under the third scenario.
So putting everything together, we have arrived at a
formula of this form.
The total probability of event B is the sum of the
probabilities of the different ways that B may occur, that
is, B occurring under the different scenarios.
And those particular probabilities are the product
of the probability of the scenario times the conditional
probability of B given that scenario.
Now, note that the sum of the probabilities of the different
scenarios is of course equal to 1.
And this is because the scenarios form a partition of
our sample space.
So if we look at this formula here, we realize that it is a
weighted average of the conditional probabilities of
event B, weighted average of the conditional probabilities
where these probabilities of the individual scenarios are
the weights.
In words, the probability that an event occurs is a weighted
average of the probability that it has under each
possible scenario, where the weights are the probabilities
of the different scenarios.
One final comment--
our derivation was for the case of three events.
But you can certainly see that the same derivation would go
through if we had any finite number of events.
But even more, if we had a partition of our sample space
into an infinite sequence of events, the same derivation
would still go through, except that in this place in the
derivation, instead of using the ordinary additivity axiom
we would have to use the countable additivity axiom.
But other than that, all the steps would be the same.
And we would end up with the same formula, except that now
this would be an infinite sum over the
infinite set of scenarios.