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Let us now revisit the second
calculation that we carried

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out in the context of
our earlier example.

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In that example, we calculated
the total probability of an

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event that can occur under
different scenarios.

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And it involves the powerful
idea of divide and conquer

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where we break up complex
situations

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into simpler pieces.

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Here is what is involved.

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We have our sample space.

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And our sample space is
partitioned into a number of

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subsets or events.

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In this picture we take that
number to be 3, so we'll have

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it partitioned into three
possible scenarios.

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It is a partition which means
that these events cover the

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entire sample, space
and they're

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disjoint from each other.

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For each one of the scenarios
we're given their

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probabilities.

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If you prefer, you can also
draw this situation

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in terms of a tree.

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There are three different
scenarios that can happen.

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We're interested in a particular
event, B. That

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event B can happen in three
different ways.

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It can happen under scenario
one, under scenario two, or

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under scenario three.

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And this corresponds to these
particular sub-events.

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So for example, this
is the event

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where scenario A1 happens.

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And then event B happens
as well.

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In terms of a tree diagram, the

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picture becomes as follows.

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If scenario A1 materializes,
event B may occur or event B

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might not occur.

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Finally, we are given
conditional probabilities that

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event B will materialize under
each one of the different

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possible scenarios.

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Under those circumstances, can
we calculate the probability

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of event B?

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Of course we can.

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And here's how we do it.

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First we realize that event
B consists of a number of

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disjoint pieces.

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One piece is when event B occurs
together with event A1.

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Another piece is when event
B occurs together with A2.

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Another piece is when event
B occurs together with A3.

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These three sets are disjoint
from each other, as we see in

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this picture.

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And together they form the
event B. Therefore, the

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probability of B is going to be,
by the additivity axiom of

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probabilities, equal to the sum
of the probabilities of

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these sub-events.

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Furthermore, for each one of
these sub-events we can use

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the multiplication rule
and write their

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probabilities as follows.

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The probability that B and A1
both occur is the probability

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that scenario one materializes
times the conditional

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probability that B occurs
given that A1 occurred.

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And then we're going to have
similar terms under the second

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scenario and a similar term
under the third scenario.

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So putting everything together,
we have arrived at a

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formula of this form.

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The total probability of event
B is the sum of the

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probabilities of the different
ways that B may occur, that

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is, B occurring under the
different scenarios.

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And those particular
probabilities are the product

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of the probability of the
scenario times the conditional

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probability of B given
that scenario.

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Now, note that the sum of the
probabilities of the different

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scenarios is of course
equal to 1.

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And this is because the
scenarios form a partition of

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our sample space.

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So if we look at this formula
here, we realize that it is a

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weighted average of the
conditional probabilities of

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event B, weighted average of the
conditional probabilities

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where these probabilities of the
individual scenarios are

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the weights.

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In words, the probability that
an event occurs is a weighted

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average of the probability
that it has under each

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possible scenario, where the
weights are the probabilities

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of the different scenarios.

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One final comment--

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our derivation was for the
case of three events.

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But you can certainly see that
the same derivation would go

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through if we had any finite
number of events.

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But even more, if we had a
partition of our sample space

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into an infinite sequence of
events, the same derivation

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would still go through, except
that in this place in the

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derivation, instead of using the
ordinary additivity axiom

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we would have to use the
countable additivity axiom.

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But other than that, all the
steps would be the same.

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And we would end up with the
same formula, except that now

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this would be an infinite
sum over the

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infinite set of scenarios.