Main content

## Statistics and probability

### Unit 8: Lesson 4

Combinatorics and probability- Probability using combinations
- Probability & combinations (2 of 2)
- Example: Different ways to pick officers
- Example: Combinatorics and probability
- Getting exactly two heads (combinatorics)
- Exactly three heads in five flips
- Generalizing with binomial coefficients (bit advanced)
- Example: Lottery probability
- Probability with permutations and combinations
- Conditional probability and combinations
- Mega millions jackpot probability
- Birthday probability problem

© 2022 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Probability using combinations

CCSS.Math:

Probability of getting exactly 3 heads in 8 flips of a fair coin. Created by Sal Khan.

## Video transcript

So you might be wondering why I
went off into permutations and combinations in the probability
playlist, and I think you'll learn in this video. So let's say I want to figure
out the probability-- I'm going to flip a coin eight times
and it's a fair coin. And I want to figure out
the probability of getting exactly 3 out of 8 heads. So I say 3/8 heads, but 3 of my
flips are going to be heads and the rest are going to be tails. So how do I think about that? Well, let's go back to one
of the early definitions we used for probability. And that says, the probability
of anything happening is the probability of the number of
equally probable events into which what we're
stating is true. So in which the number of
events-- I guess trials or situations-- in which we
get 3 heads, and exactly 3 heads, we're not saying
greater than 3 heads. So 4 heads won't count and 2
heads won't count, 5 heads won't-- only 3 heads. And then, over the total number
of equally probable trials-- not trials, total number of
equally possible outcomes. I should be using
the word outcomes. So just with the word outcomes
it should be the total number of outcomes in which what
we're saying happens. So we get 3 heads over the
total possible outcomes. So let's do the
bottom part first. What are the total possible
outcomes if I'm flipping a fair coin eight times? Well, the first time I flip
it I either get heads or tails, so I get 2 outcomes. And then when I flip it
again I get 2 more come for the second one. And then, how many
total outcomes? Well, that's 2 times 2 because
I could have got 2 in the first, 2 in the second flip. And then essentially we
would multiply 2 times the number of flips. So that's 5, 6, 7, 8, and
that equals 2 to the eighth. So the number of outcomes is
just going to be 2 to the total number of flips. And hopefully that
make sense to you. If not, you might want
to re-watch some of the earlier videos. But that's the easy part. So there's 2 to the eighth
possible outcomes when you flip a fair coin eight times. So how many of those outcomes
are going to result in exactly 3 heads? Let's think of it this way. Let's give a name to
each of our flips. Let's give a name to them. So let me make a little column,
we'll call these the flips. This is my flips column. And I could name them anything. I could name them
Larry, Curly, Moe. I could name them-- well, I
would need 5 more names for them, but I could name them the
7 dwarfs or the 8 dwarfs really because I have 8 flips. I'll number the flips. Flip 1, 2, 3, 4, 5, 6, 7, 8. And I'm the god of probability. And essentially, I need to just
pick 3 of these flips that are going to result in heads. So another way to think about
it is, these could be 8 people and I could pick which of
these-- how many ways can I pick 3 of these people
to put into the car? How many ways can I pick 3 of
these people to sit in chairs. And it doesn't matter the
order that I pick them in. It doesn't matter if I say the
people that are going to get in the car are going to
be people 1, 2, 3 3. Or if I say 3, 2, and 1,
or if I say 2, 3, and 1. Those are all the
same combination. So similarly, if I'm just
picking flips and I have to say, OK, 3 of these
flips are going to get into the heads car. Heads is like they're sitting,
they're people sitting down. I don't want to
confuse you too much. But essentially I'm
just going to choose 3 things out of the 8. So I'm essentially just saying,
how many combinations can I get where I pick 3 out of these 8. And so that should immediately
ring a bell that we're essentially saying, out of 8
things we're going to choose 3. How many combinations of 3
can we pick of 8 and that we went over in the last video. And let's do it with
the formula first. So let me write the formula up
here just so you remember it, but I also want to give you the
intuition again, for the formula. So in general, we said, n
choose k, that is equal to n factorial over k factorial
times n minus k factorial. So in this situation that
would equal 8 factorial over 3 factorial times what? 8 minus k-- times 5 factorial. Or another way of writing this,
this would be 8 times 7 times 6 times 5 times 4 times 3 times 2
times 1 over-- I'll just write 3 factorial for a second. Then times 5 times 4
times 3 times 2 times 1. And of course, that and that
cancel out and all you're left with is 8 times 7
time 6 over 3 factorial. And I did this for reason
because I want you to re-get the intuition at least for
this part of the formula. That's essentially just saying,
how many permutations can I-- how many ways can I pick
3 things out of 8? And that's essentially saying,
well, before I pick anything I could pick 1 of 8. Then I have 7 left to pick
from for the second spot. And then I have 6 left to
pick for the third spot. And so that's essentially
the number of permutations. But since we don't care what
order we picked them in, we need to divide by the number of
ways we can rearrange 3 things, and that's where the 3
factorial comes from. And so hopefully I didn't
confuse you, but if I did you can go back to this formula
for the binomial coefficient. But it's good to
have the intuition. And then once we're
at this point we can just calculate this. Well, what's this? This is 8 times 7 times
6 over 3 factorial of 3 times 2 times 1. So that's 6. The 6 cancels out,
so it's 8 times 7. So there's 8 times
7, or what is that? 56. That's equal to 56. So there's 56 different ways
to pick 3 things out of 8. Or if I have 8 people there's
56 ways of picking 3 people to sit in the car or however
you want t view it. But if I have 8 flips there's
56 ways of picking 3 of those flips to be heads. So let's go to our original
probability problem. What is the probably that
I get 3 out of 8 heads? Well, it's the number of ways
I can pick 3 out of those 8, so it equals 56, over the
total number of outcomes. The total number of outcomes
is 2 to the eighth. Another way I could write
that-- 56, let me unseparate. That's 8 times 7 over
2 to the eighth. 8 is 2 to the third. Let me erase some of this. Not with that color. Let me erase that. Let me erase all of
this just so I space. And I will switch
colors for variety. Let me use the small pen. OK, so I'm back. All right, so 8 is the same
thing as 2 to the third times 7-- this is all just
mathematical simplification, but it's useful-- over
2 to the eighth. And so, if we just divide both
sides-- the numerator and the denominator by 2 to the
third, this becomes 1. This becomes 2 to the fifth. And so it becomes 7/32. Is that right? So if I were to pick 3 out of
8-- yep, I think that is right. And so what does that
turn out to be? Let me get my calculator. [INAUDIBLE] to make careless mistakes. Let's see. My calculator seems
to have disappeared. Let me get it back. There it is. OK. 7 divided by 32 is
equal to 0.21875. Which is equal to 21.-- you
know, if I were to round roughly-- 21.9% chance. So there's a little bit better
than 1 in 5 chance that I get exactly 3 out of the
8 flips as heads. Hopefully I didn't confuse you
and now you can apply that to pretty much anything. You could say, well, what is
the probability of getting-- if I flip a fair coin-- of getting
exactly 7 out of 8 heads? Or you could say, what's
the probability of getting 2 out of 100 heads? And you could use it the exact
same way we did this problem. I'll see you in the next video.