The coefficients n-choose-k that we calculated in the
previous segment are known as the binomial coefficients.
They are intimately related to certain probabilities
associated with coin tossing models, the so-called binomial
probabilities.
This is going to be our subject.
We consider a coin which we toss n times in a row,
independently.
For each one of the tosses of this coin, we assume that
there is a certain probability, p, that the
result is heads, which of course, implies that the
probability of obtaining tails in any particular toss is
going to be 1 minus p.
The question we want to address is the following.
We want to calculate the probability that in those n
independent coin tosses, we're going to
observe exactly k heads.
Let us start working our way towards the solution to this
problem by looking first at a simple setting
and then move on.
So let us answer this simple question.
What is the probability that we observe
this particular sequence?
Of course here we take n equal to six, and we wish to
calculate this probability.
Now, because we have assumed that the coin tosses are
independent, we can multiply probabilities.
So the probability of this sequence is equal to the
probability that the first toss is heads times the
probability that the second toss is tails, which is 1
minus p, times the probability that the third toss is tails,
which is 1 minus p, times the probability of heads, times
the probability of heads, times the
probability of heads.
And by collecting terms, this is p to the 4th times 1 minus
p to the second power.
More generally, if I give you a particular sequence of heads
and tails, as in this example, and I ask you, what is the
probability that this particular sequence is
observed, then by generalizing from this answer or from the
derivation of this answer, you see that you're going to get p
to the power number of heads.
And the reason is that each time that there's a head
showing up in this sequence, there's a corresponding factor
of p in this numerical answer.
And then there are factors associated with tails.
Each tail contributes a factor of 1 minus p.
And so we're going to have here 1 minus p to a power
equal to the number of tails.
Now, if I ask you about the probability of a particular
sequence and that particular sequence has happened to have
exactly k heads, what is the probability of that sequence?
Well, we already calculated what it is.
It is the previous answer, except we use the symbol k
instead of just writing out explicitly "number of heads."
And the number of tails is the number of tosses minus how
many tosses resulted in heads.
Now, we're ready to consider the actual problem that we
want to solve, which is calculate the
probability of k heads.
The event of obtaining k heads can happen in
many different ways.
Any particular k-head sequence makes that event to occur.
Any particular k-head sequence has a probability equal to
this expression.
The overall probability of k heads is going to be the
probability of any particular k-head sequence, times the
number of k-head sequences that we have.
Now, the reason why we can carry out this argument is the
fact that any k-head sequence has the same probability.
Otherwise, we wouldn't be able to write down an answer which
is just the product of two terms.
But because every k-head sequence has the same
probability, to find the overall probability, we take
the probability of each one of them and multiply it with the
number of how many of these we have.
So to make further progress, now we need to calculate the
number of possible k-head sequences.
How many are there?
Well, specifying a k-head sequence is
the same as the following.
You think of having n time slots.
These time slots corresponds to the different
tosses of your coin.
And to specify a k-head sequence, you need to tell me
which ones of these slots happen to contain a head.
You need to tell me k of those slots.
So in other words, what you're doing is you're specifying a
subset of the set of these n slots, a
subset that has k elements.
You need to choose k of the slots out of the n and tell me
that those k slots have heads.
That's the way of specifying a particular k-head sequence.
So what's the number of k-head sequences?
Well, it's the same as the number of ways that you can
choose k slots out of the n slots, which is our binomial
coefficient, n-choose-k.
Therefore, the answer to our problem is this expression
here, times n-choose-k, which is shown up here.
At this point, we can pause and consider a simple question
to check your understanding of the binomial probabilities.