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As promised, we will now start
developing generalizations of

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the different calculations that
we carried out in the

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context of the radar example.

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00:00:09,000 --> 00:00:11,920
The first kind of calculation
that we carried out goes under

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the name of the multiplication
rule.

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And it goes as follows.

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Our starting point is the
definition of conditional

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probabilities.

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The conditional probability of
A given another event, B, is

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the probability that both events
have occurred divided

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by the probability of the
conditioning event.

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We now take the denominator term
and send it to the other

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side of this equality to obtain
this relation, which we

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can interpret as follows.

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The probability that two events
occur is equal to the

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probability that a first event
occurs, event B in this case,

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00:00:47,450 --> 00:00:49,940
times the conditional
probability that the second

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00:00:49,940 --> 00:00:55,360
event, event A, occurs, given
that event B has occurred.

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Now, out of the two events, A
and B, we're of course free to

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choose which one we call the
first event and which one we

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call the second event.

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So the probability of the two
events happening is also equal

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to an expression of this form,
the probability that A occurs

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times the conditional
probability that B occurs,

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given that A has occurred.

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We used this formula in the
context of a tree diagram.

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And we used it to calculate the
probability of a leaf of

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this tree by multiplying the
probability of taking this

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branch, the probability that A
occurs, times the conditional

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probability of taking this
branch, the probability that

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event B also occurs given that
event A has occurred.

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How do we generalize
this calculation?

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Consider a situation in which
the experiment has an

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additional third stage that has
to do with another event,

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C, that may or may not occur.

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For example, if we have arrived
here, A and B have

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both occurred.

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And then C also occurs, then we
reach this particular leaf

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of the tree.

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Or there could be
other scenarios.

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For example, it could be the
case that A did not occur.

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Then event B occurred, and
finally, event C did not

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occur, in which case we end up
at this particular leaf.

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What is the probability of
this scenario happening?

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Let us try to do a calculation
similar to the one that we

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used for the case
of two events.

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However, we need to deal
here with three events.

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What should we do?

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00:02:42,010 --> 00:02:44,400
Well, we look at the
intersection of these three

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events and think of it as the
intersection of a composite

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event, A complement intersection
B, then

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intersected with the
event C complement.

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Clearly, you can form the
intersection of three events

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by first taking the intersection
of two of them

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and then intersecting
with a third.

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00:03:06,900 --> 00:03:09,420
After we group things this way,
we're dealing with the

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probability of two events
happening, this composite

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event and this ordinary event.

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00:03:15,730 --> 00:03:20,230
And the probability of two
events happening is equal to

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the probability that the first
event happens, and then the

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probability that the second
event happens, given that the

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first one has happened.

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Can we simplify this
even further?

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Yes.

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The first term is
the probability

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of two events happening.

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So it can be simplified further
as the probability

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that A complement occurs times
the conditional probability

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that B occurs, given that A
complement has occurred.

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And then we carry over
the last term

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exactly the way it is.

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The conclusion is that we can
calculate the probability of

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this leaf by multiplying the
probability of the first

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00:04:13,190 --> 00:04:17,649
branch times the conditional
probability of the second

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00:04:17,649 --> 00:04:24,070
branch, given that the first
branch was taken, and then

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finally multiply with the
probability of the third

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branch, which is the probability
that C complement

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occurs, given that A
complement and B

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have already occurred.

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00:04:38,250 --> 00:04:41,192
In other words, we can calculate
the probability of a

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leaf by just multiplying the
probabilities of the different

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branches involved and where we
use conditional probabilities

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for the intermediate branches.

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At this point, you can use your
imagination to see that

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such a formula should also be
valid for the case of more

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than three events.

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The probability that a bunch of
events all occur should be

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the probability of the first
event times a number of

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factors, each corresponding
to a branch in a

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tree of this kind.

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00:05:14,500 --> 00:05:25,220
In particular, the probability
that events A1, A2, up to An

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all occur is going to be the
probability that the first

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event occurs times a product of
conditional probabilities

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that the i-th event occurs,
given that all of the previous

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events have already occurred.

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00:05:50,710 --> 00:05:54,980
And we obtain a term of this
kind for every event, Ai,

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after the first one, so this
product ranges from 2 up to n.

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00:06:02,860 --> 00:06:05,730
And this is the most general
version of the multiplication

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00:06:05,730 --> 00:06:08,970
rule and allows you to calculate
the probability of

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00:06:08,970 --> 00:06:13,590
several events happening by
multiplying probabilities and

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conditional probabilities.