WEBVTT
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In this problem, we're given an
urn with n balls in it, out
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of which m balls
are red balls.
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To visualize it, we can draw a
box that represents the set of
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all n balls.
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Somewhere in the middle or
somewhere else we have a cut,
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such that to the left we have
all the red balls (there are
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m), and non-red balls.
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Let's for now call
it black balls.
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That is n minus m.
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Now, from this box, we are to
draw k balls, and we'd like to
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know the probability that
i out of those k
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balls are red balls.
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For the rest of the problem,
we'll refer to this
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probability as p-r, where r
stands for the red balls.
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So from this picture, we know
that we're going to draw a
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subset of the balls, such that
i of them are red, and the
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remaining k minus i are black.
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And we'll like to know what is
the probability that this
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event would occur.
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To start, we define our sample
space, omega, as the set of
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all ways to draw k balls
out of n balls.
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We found a simple counting
argument -- we know that size
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of our sample space has
n-choose-k, which is the total
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number of ways to draw k
balls out of n balls.
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Next, we'd like to know how
many of those samples
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correspond to the event that
we're interested in.
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In particular, we would like to
know c, which is equal to
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the number of ways to get
i red balls after
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we draw the k balls.
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To do so, we'll break c into
a product of two numbers --
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let's call it a times b --
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where a is the total number of
ways to select i red balls out
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of m red balls.
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So the number of ways to get
i out of m red balls.
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Going back to the picture, this
corresponds to the total
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number of ways to
get these balls.
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And similarly, we define b as
the total number of ways to
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get the remaining k minus i
balls out of the set n minus m
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black balls.
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This corresponds to the total
number of ways to select the
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subset right here in the
right side of the box.
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Now as you can see, once we have
a and b, we multiply them
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together, and this yields
the total number of ways
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to get i red balls.
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To compute what these numbers
are, we see that a is equal to
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m-choose-i number of ways to
get i red balls, and b is n
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minus m, the total number of
black balls, choose k minus i,
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the balls that are not red
within those k balls.
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Now putting everything back, we
have p-r, the probability
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we set out to compute, is equal
to c, the size of the
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event, divided by the size of
the entire sample space.
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From the previous calculations,
we know that c
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is equal to a times b, which
is then equal to m-choose-i
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times (n minus m)-choose-(k
minus i).
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And on the denominator, we have
the entire sample space
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is a size n-choose-k.
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And that completes
our derivation.
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Now let's look at a numerical
example of this problem.
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Here, let's say we have
a deck of 52 cards.
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And we draw a box with n equals
52, out of which we
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know that there are 4 aces.
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So we'll call these the left
side of the box, which is we
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have m equals 4 aces.
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Now if we were to draw
seven cards--
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call it k equal to 7--
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and we'd like to know what is
the probability that out of
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the 7 cards, we have 3 aces.
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Using the notation we did
earlier, if we were to draw a
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circle representing the seven
cards, we want to know what is
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the probability that we have 3
aces in the left side of the
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box and 4 non-aces for the
remainder of the deck.
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In particular, we'll
call i equal to 3.
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So by this point, we've cast the
problem of drawing cards
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from the deck in the same way
as we did earlier of drawing
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balls from an urn.
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And from the expression right
here, which we computed
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earlier, we can readily compute
the probability of
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having 3 aces.
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In particular, we just have to
substitute into the expression
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right here the value of m equal
to 4, n equal to 52, k
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equal to 7, finally,
i equal to 3.
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So we have 4-choose-3 times n
minus m, in this case would be
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48, choose k minus i, will be 4,
and on the denominator, we
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have 52 total number of cards,
choosing 7 cards.
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That gives us [the]
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numerical answer [for]
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the probability of getting 3
aces when we draw 7 cards.