ACC100:

Example 1, Part 1 | Example 1, Part 2

Example 1, Part 1

PROFESSOR: So let's look at Example 1 here. We're going to toss a coin three times, and we're going to record all the combinations of heads and tails. So the way I would normally do this is we'd have a first flip. We'd have a second flip, and we'd have a third flip. Each one of those can be heads or tails, and we want all those combinations.

So we could have a heads, a heads, and a heads, and there's a lot of other combinations. How many are there? That's the first question that we might want to answer. Well, each time we flip, we have two options.

And we repeat that three times.

And the way this is going to work is, because there's two options repeated three times. So we have two options, and we take that to the third power. Or 2 the third, which is 8. And you can see that a little lower down on the page in the solution.

Now, if we know that we're going to have 8 rows over here, the easiest way to complete this table without doing a lot of guessing is to take that 8 and divide it in half and get 4. So for this first set of flips, I'm going to have 4 H's in a row. Right? And then I'll do 4 T's. Right?

When we have a total of 8 flips, for that first flip, we should have 4 H's and 4 T's. For the second flip, if I break that in half again and do 2 H's, 2 T's, 2 H's, 2 T's. And for the third flip, break it in half again. H, T, H, T, H, T. One each.

Oops. That should be a T. That's an H. A T. H and T. So we see all of our combinations. This is number one-- H, H, H. This is number two, et cetera. This is the kind of thing that you'll be doing later on in the course when we move to logic also. It's that they'll be T's and F's for trues and falses.

Coin Flips

1st	2nd	3rd
H	H	H
H	H	T
H	T	H
H	T	T
T	H	H
T	H	T
T	T	H
T	T	T

Just as a further example, if we flip four times--

Right? And then we'd have two options, and we'd have that one repeated.

Four times.

OK. And that would be 2 to the fourth, which is 16. So if we are flipping four times, we'd start off with 8 H's and then 8 T's. We'd divide that 16 in 1/2, and then we'd go to 4 H's, 4 T's, 4 H's, 4 T's.

And we'd continue on. There would be, of course, four columns. We'd continue on until we got to 1 H, 1 T, 1 H, 1 T, et cetera. So that's how you can easily fill in all 8 of those options or, in this case, 16 of those options without guessing at any of them. So moving on to the next section.

Example 1, Part 2

PROFESSOR: So in this part of the problem, we're going to look at just the specific situation where we're going to record-- or where we're going to determine how many different ways we can get two heads and one tail. If you look at your sample space that we just drew, it's pretty obvious that THH, HTH, and HHT are the three situations.

But I wanted to look at this in a slightly different way. Let's go back to our combinations. We have three spaces, and we want to record two heads. So if we look at 3 choose 2-- that's the way it would look on your calculator, which is the same thing as the notation we've been using in the lesson, which is C 3 comma 2, This is going to be 6-- whoops. Just checking-- 3. As we have 3 over here.

$\begin{array}{l}{3}^{C_1}=C\left(3,2\right)=3\\ 3^{C_1}=3\end{array}$

Another way to look at that is we could look at that as 3 choose 1. Why can we do that? Well, instead of choosing two heads to put into the three slots, we could be choosing one tail. And that also happens to be 3. So we get the same answer either way.

Now on the other hand, what if we're looking at at least 1 head? At least 1 head. Well, once again, if you look at your sample space, you're going to list a lot of them, and you're going to have HHH, dot dot dot. All those things have at least one head in them, and you can count them up. But if we wanted to use this combination to do this, at least 1 head means 1 head exactly, or 2 heads, or 3 heads.

And we've been talking about this word "or" in our lessons, and or means add. So we have C 3 choose 1, for our 1 head, or meaning add. C 3 choose 2 for our 2 heads, or meeting add. C 3-- that's a C there, not a parentheses-- 3 comma 3, which is going to be 3 plus 3 plus 1, which is 7. OK, so that's one way to do that.

$\begin{array}{l}C\left(3,1\right)+C\left(3,2\right)+C\left(3,3\right)\\ 3+3+1=7\end{array}$

There's one more way to do this using the complement. What if we said, OK, what's not at least 1 head? That would be all tails. If you look at that, of course, there's only one situation with all tails, and if we take 8 minus 1, that is going to be our 7 also. Different way to solve the same problem. Most of these problems can be solved in a number of ways, at least two.

So good luck, try that on your own, and I hope--