- [Voiceover] Let's say that you know
your probability of making a free throw.
You know that the probability...
The probability, and let's say of scoring a free throw
because make and miss, they both start with M
and that can get confusing,
so let's say the probability of scoring...
scoring a free throw, is equal to,
is going to be, say 70%.
If we want to write it as a percent
or we could write it as 0.7
if we write it as a decimal.
Let's say the probability of missing a free throw, then
and this is just going to come
straight out of what we just wrote down,
the probability of missing,
of missing a free throw, is just going to be
100% minus this.
You're either going to make or miss,
you're going to score or miss,
I don't want to use make in this
because they both start with M
so this is going to be a 30% probability,
or if we write it as a decimal, 0.3
One minus this is 0.7.
These are the only two possibilities,
so they have to add up to 100%,
or they have to add up to one.
Now, let's say that you were going to take six attempts.
What we are curious about, is the probability
of exactly two scores
in six attempts.
So let's think about what that is
and I encourage you to get inspired
at any point in this video
you should pause it and you should try
to work through what we're asking right now,
so this is what we want to figure out,
the probability of exactly two scores
in six attempts.
So, let's think about the way,
let's think about the particular ways
of getting two scores in six attempts
and think about the probability for
any one of those particular ways,
and then we can think about how many ways
can we get two scores in six attempts?
So, for example, you could get
you could make the first two free throws,
so it could be score, score and then you miss the next four.
So score, score, and then it's miss, miss, miss, and miss.
So what's the probability of this exact thing happening?
This exact thing?
Well, you have a 0.7 chance of making,
of scoring on the first one,
then you have a 0.7 chance of scoring on the second one,
and then you have a 0.3 chance of missing the next four.
So, the probability of this exact circumstance
is going to be what I am writing down.
Hopefully you don't get the multiplication symbols
confused with the decimals, I'm trying to
write them a little bit higher.
Times 0.3, and what is this going to be equal to?
Well, this is going to be equal to...
This is going to be 0.7 squared
times 0.3 to the one, two, three, fourth--
to the fourth power.
Now, is this the only way to get two scores in six attempts?
No, there's many ways of getting two scores in six attempts.
For example, maybe you miss the first one, the first attempt
and then you make the second attempt, you score,
then you miss the third attempt,
and let's just say you make the fourth attempt,
and then you miss the next two.
You miss, and you miss.
This is another way to get two scores in six attempts,
and what's the probability of this happening?
Well, you'll see, it's going to be exactly this,
it's just we're multiplying in a different order.
This is going to be 0.3 times 0.7
you have a 30% chance of missing the first one,
a 70% chance of making the second one,
and then times 0.3, a 30% chance of missing the third,
times a 70 percent chance of making the fourth,
times a 30% chance
for each of the next two misses
if you wanted the exact circumstance,
this is once again going to be 0.7
and if you just rearrange the order that you're multiplying,
this is going to be 0.7 squared
times 0.3 to the fourth power,
so for any one of these particular ways
to get exactly two scores in six attempts,
the probability is going to be this.
So, the probability of getting exactly two scores
in six attempts, well it's going to be
any one of these probabilities
times the number of ways
you can get two scores in six attempts.
Well, how...
If you have out of six attempts,
you're choosing two of them to have scores,
how many ways are there?
Well, as you can imagine,
this is a combinatorics problem,
so you could write this as
you could write this, let me see how I could...
You're going to take six attempts.
You could write this as six choose,
what we're trying to, you're picking from six things,
six attempts, and you're picking two of them,
or two of them are going to need to be made
if you want to meet these circumstances.
This is going to tell us the number
of different ways you can make
two scores in six attempts.
Of course, we can write this as kind of
a binomial coefficient notation.
We can write this is as six, choose two
and we can just apply the formula
for combinations, and if this looks completely unfamiliar
I encourage you to look up combinations on Khan Academy
and then we go into some detail
on the reasoning behind the formula
that makes a lot of sense.
This is going to be equal to six factorial
over two factorial
times six minus two factorial.
Six minus two factorial,
I'm going to do the factorial in green again,
and what's this going to be equal to?
This is going to be equal to six times five
times four times three times two,
and I'll just throw in the one there
although it doesn't change the value,
over two times one.
And six minus two is four,
so that's going to be four factorial,
so this right over here is four factorial
so times four times three times two times one.
Well, that and that is going to cancel,
and the six divided by two is three, so this is 15.
There's 15 different ways that you could get
two things out of six, I guess, is one way to say it
or there's 15 different ways
that you could get two things out of six.
Another way of thinking about it
is there's 15 different ways
to make two out of six free throws.
Now, the probability for each of those
is this right over here.
The probability of exactly two scores in six attempts,
this is where we deserve a little bit of a drumroll.
This is going to be
six choose two times 0.7 squared.
This is two, you're going to make two
and then it's 0.3
to the fourth power.
These will necessarily add up to six.
So this right over here was a three,
then this right over here would be a three
and then this would be six minus three
or three right over here.
Now, what is this value?
Well, it's going to be equal to
we have our 15, three times five
so we have this business right over here.
It's going to be 15 times
let's see, in yellow, 0.7 times 0.7
is going to be times 0.49
and let's see, three to the fourth power
would be 81.
But, I am multiplying four decimals,
each of them have one space
to the right of the decimal point
so this is, I'm going to have four spaces
to the right of the decimal
so 0.0081
so there you go, whatever this number is,
and actually I might as well get a calculator out
and calculate it, so this is going to be
this is going to be...
Let me...
So, it's 15 times 0.49
times 0.0081
and we get 0.059535
so this is going to be equal to,
let me write it down.
Actually, I wish I had a little bit more real estate
right over here, but I'll write it in a very bold color.
This is going to be,
well, actually, I'm kind of out of bold colors.
I'll write it in a slightly less bold color.
This is going to be equal to 0.05935
if we wanted the exact number, or we could say
this is approximately, if we round to the nearest percentage
this is approximately a six percent chance,
six percent probability of getting exactly two scores
in the six attempts.
I didn't say two or more,
I just said exactly two scores in the six attempts.
Actually, it's a fairly low probability
because I have a pretty high free throw percentage.
If someone has this high of a free throw percentage
it's actually reasonably unlikely
that they're only going to make two scores
in the six attempts.