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