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