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Hi.
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Today, we're going to do a
really fun problem called
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geniuses and chocolates.
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And what this problem is
exercising is your knowledge
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of properties of probability
laws.
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So let me just clarify
what I mean by that.
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Hopefully, by this point, you
have already learned what the
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axioms of probability are.
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And properties of probability
laws are essentially any rules
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that you can derive
from those axioms.
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So take for example the fact
that the probability of A
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union B is equal to the
probability of A plus the
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probability of B minus the
probability of the
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intersection.
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That's an example of a property
of a probability law.
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So enough with the preamble.
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Let's see what the problem
is asking us.
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In this problem, we have
a class of students.
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And we're told that 60% of the
students are geniuses.
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70% of the students
love chocolate.
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So I would be in
that category.
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And 40% fall into
both categories.
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And our job is to determine
the probability that a
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randomly selected student is
neither a genius nor a
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chocolate lover.
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So first I just want to write
down the information that
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we're given in the problem
statement.
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So if you let G denote the event
that a randomly selected
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student is a genius then the
problem statement tells us
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that the probability of
G is equal to 0.6.
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Similarly, if we let C denote
the event that a randomly
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selected student is a chocolate
lover, then we have
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that the probability of
C is equal to 0.7.
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Lastly, we are told that the
probability a randomly
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selected student falls into
both categories is 0.4.
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And the way we can express
that using the notation
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already on the board is
probability of G intersect C
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is equal to 0.4.
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OK, now one way of approaching
this problem is to essentially
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use this information and sort of
massage it using properties
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of probability laws to
get to our answer.
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Instead, I'm going to take a
different approach, which I
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think will be helpful.
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So namely, we're going
to use something
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called a Venn diagram.
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Now a Venn diagram is just a
tool that's really useful for
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telling you how different sets
relate to each other and how
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their corresponding
probabilities
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relate to each other.
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So the way you usually draw this
is you draw a rectangle,
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which denotes your sample space,
which of course, we
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call omega.
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And then you draw two
intersecting circles.
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So one to represent our geniuses
and one to represent
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our chocolate lovers.
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And the reason why I drew them
intersecting is because we
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know that there are 40% of the
students in our class are both
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geniuses and chocolate lovers.
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OK, and the way you sort of
interpret this diagram is the
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space outside these two circles
correspond to students
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who are neither geniuses
nor chocolate lovers.
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And so just keep in mind that
the probability corresponding
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to these students on the
outside, that's actually what
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we're looking for.
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Similarly, students in this
little shape, this tear drop
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in the middle, those would
correspond to geniuses and
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chocolate lovers.
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You probably get the idea.
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So this is our Venn diagram.
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Now I'm going to give you guys
a second trick if you will.
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And that is to work
with partitions.
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So I believe you've seen
partitions in lecture by now.
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And a partition is essentially a
way of cutting up the sample
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space into pieces.
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But you need two properties
to be true.
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So the pieces that you cut up
your sample space into, they
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need to be disjoint, so
they can't overlap.
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So for instance, G and C are
not disjoint because they
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overlap in this tear
drop region.
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Now the second thing that a
partition has to satisfy is
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that if you put all the pieces
together, they have to
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comprise the entire
sample space.
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So I'm just going to put these
labels down on my graph.
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X, Y, Z, and W. So X is
everything outside the two
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circles but inside
the rectangle.
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And just note, again, that what
we're actually trying to
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solve in this problem is the
probability of X, the
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probability that you're neither
genius, because you're
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not in this circle, and you're
not a chocolate lover, because
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you're not in this circle.
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So Y I'm using to refer
to this sort of
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crescent moon shape.
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Z, I'm using to refer
to this tear drop.
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And W, I'm using to refer
to this shape.
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So, hopefully, you agree that
X, Y, Z, and W form a
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partition because they
don't overlap.
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So they are disjoint.
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And together they form omega.
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So now we're ready to
do some computation.
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The first step is to sort of
get the information we have
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written down here in terms
of these new labels.
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So hopefully, you guys buy that
G is just the union of Y
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and Z. And because Y and Z are
disjoint, we get that the
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probability of the union is the
sum of the probabilities.
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And, of course, we have from
before that this is 0.6.
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Similarly, we have that the
probability of C is equal to
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the probability of Z union W.
And, again, using the fact
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that these two guys are
disjoint, you get this
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expression.
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And that is equal to 0.7.
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OK, and the last piece of
information, G intersects C
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corresponds to Z, or our tear
drop, and so we have that the
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probability of Z is
equal to 0.4.
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And now, if you notice,
probability of Z shows up in
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these two equations.
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So we can just plug it in.
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So plug in 0.4 into
this equation.
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We get P of Y plus 0.4 is 0.6.
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So that implies that
P of Y is 0.2.
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That's just algebra.
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And similarly we have point.
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0.4 plus P of W is
equal to 0.7.
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So that implies that
P of W is 0.3.
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Again, that's just algebra.
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So now we're doing really well
because we have a lot of
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information.
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We know the probability of Y,
the probability of Z, the
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probability of W. But remember
we're going for, we're trying
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to find the probability of X. So
the way we finally put all
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this information together to
solve for X is we use the
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axiom that tells us that 1 is
equal to the probability of
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the sample space.
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And then, again, we're going
to use sort of this really
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helpful fact that X, Y, Z, and W
form a partition of omega to
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go ahead and write this as
probability of X plus
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probability of Y plus
probability,
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oops, I made a mistake.
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Hopefully, you guys
caught that.
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It's really, oh, no.
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I'm right.
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Never mind.
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Probability of X plus
probability of Y plus
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probability of Z plus
probability of W. And now we
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can go ahead and plug-in the
values that we solved for
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previously.
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So we get probability of X plus
0.2 plus 0.4 plus 0.3.
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These guys sum to 0.9.
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So, again, just simple
arithmetic, we get that the
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probability of X is
equal to 0.1.
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So we're done because we've
successfully found that the
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probability that a randomly
selected student is neither a
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genius nor a chocolate
lover is 0.1.
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So this was a fairly
straightforward problem.
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But there are some important
takeaways.
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The first one is that
Venn diagrams are
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a really nice tool.
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Whenever the problem is asking
you how different sets relate
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to each other or how different
probabilities relate to each
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other, you should probably draw
Venn diagram because it
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will help you.
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And the second takeaway is that
it's frequently useful to
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divide your sample space into
a partition mainly because
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sort of the pieces
that compose a
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partition are disjoint.
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So we will be back soon to
solve more problems.
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