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## Binomial random variables

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# Binomial probability example

AP Stats: UNC‑3 (EU), UNC‑3.B (LO), UNC‑3.B.1 (EK)

## Video transcript

- [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.

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