Let's do a couple of exercises from our probability one
module.
So we have a bag with 9 red marbles, 2 blue marbles,
and 3 green marbles in it.
What is the probability of randomly selecting
a non-blue marble from of the bag?
So let's draw this bag here.
So that's my bag, and we're going
to assume that it's a transparent bag, so it looks
like a vase.
But we have 9 red marbles, so let me draw 9 red marbles.
1, 2, 3, 4, 5, 6, 7, 8, 9 red marbles.
They're kind of orange-ish, but it does the job.
2 blue marbles, so we have 1 blue marble, 2 blue marbles.
And then we have 3 green marbles, let me draw those
3, so 1, 2, 3.
What is the probability of randomly selecting
a non-blue marble from the bag?
So maybe we mix them all up, and we have an equal probability
of selecting any one of these.
And the way you just think about it
is what fraction of all of the possible events
meet our constraint?
So let's just think about all of the possible events first.
How many different possible marbles can we take out?
Well that's just the total number of marbles there are.
So are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14 possible marbles.
So this is the number of possibilities.
And then we just have to think what fraction
of those possibilities meet our constraints.
And the other way you could have gotten 14
is just taking 9 plus 2 plus 3.
So what number of those possibilities
meet our constraints?
And remember, our constraint is selecting a non-blue marble
from the bag.
Another way to think about it is a red or green marble,
because the only other two colors
we have are red and green.
So how many non-blue marbles are there?
Well, there's a couple ways to think about it.
You could say there's 14 total marbles.
2 are blue.
So there are going to be 14 minus 2,
which is 12 non-blue marbles.
Or you could just count them.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
So there are 12 non-blue marbles.
So these are the possibilities that meet our constraints
over all of the possibilities.
And then if we want to-- this isn't in simplified form
right here, since both 12 and 14 are divisible by 2.
So let's divide both the numerator and the denominator
by 2, and you get 6 over 7.
So we have a 6/7 chance of selecting a non-blue marble
from the bag.
Let's do another one.
If a number is randomly chosen from the following list,
what is the probability that the number is a multiple of 5?
So once again, we want to find the fraction
of the total possibilities that meet our constraint,
and our constraint is being a multiple of 5.
So how many total possibilities are there?
Let's think about that.
How many do we have?
Well that's just the total number
of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12.
So there are 12 possibilities.
We have an equal chance of picking any one of these 12.
Now which of these 12 are a multiple of 5?
So let's do this in a different color.
So let me pick out the multiples of 5.
32 is not a multiple of 5, 49 is not a multiple of 5.
55 is a multiple of 5.
Really, we're just looking for the numbers
that in the ones place you either have a 5 or a 0.
55 is a multiple of 5, 30 is a multiple
of 5, that's 6 times 5.
55 is 11 times 5.
Not 56, not 28.
This is clearly 5 times 10, this is 8 times 5,
this is the same number again, also 8 times 5.
So all of these are multiples of 5.
45, that's 9 times 5.
3 is not a multiple of 5.
25, clearly 5 times 5.
So I've circled all the multiples of 5.
So of all the possibilities, the ones that
meet our constraint of being a multiple of 5,
there are 1, 2, 3, 4, 5, 6, 7 possibilities.
So 7 meet our constraint.
So in this example, the probability
of a selecting a number that is a multiple of 5 is 7/12.
Let's do another one.
the circumference of a circle is 36 pi.
Let's draw this circle.
The circumference of a circle is 36 pi,
so let's say the circle looks-- I
can draw a neater circle than that.
So let's say the circle looks something like that.
And its circumference-- we have to be careful
here, they're giving us interesting--
the circumference is 36 pi.
Then they tell us that contained in that circle
is a smaller circle with area 16 pi.
So inside the bigger circle, we have a smaller circle
that has an area of 16 pi.
A point is selected at random from inside the larger circle,
so we're going to randomly select
some point in this larger circle.
What is the probability that the point also
lies in this smaller circle?
So here's a little bit interesting,
because you actually have an infinite number of points
in both of these circles, because it's not
kind of a separate balls or marbles,
like we saw in the first example, or separate numbers.
There's actually an infinite number
of points you could pick here.
And so, when we talk about the probability
that the point also lies in the smaller circle,
we're really thinking about the percentage of the points
in the larger circle that are also in the smaller circle.
Or another way to think about it is the probability
that if we pick a point from this larger circle,
the probably that it's also in the smaller circle
is really just going to be the percentage of the larger
circle that is the smaller circle.
I know that might sound confusing,
but we're really just have to figure out
the areas for both of them, and it's really just going
to be the ratios so let's think about that.
So there's a temptation to just use this 36 pi up here,
but we have to remember, this was the circumference,
and we need to figure out the area of both of these circles.
And so for area, we need to know the radius,
because area is pi r squared.
So we can figure out the radius from
the circumference by saying, well, circumference is equal
to 2 times pi times the radius of the circle.
Or if you say 36 pi, which we were told is the circumference,
is equal to 2 times pi times the radius,
we can divide both sides by 2 pi, and on the left hand side,
36 divided by 2 is 18 the pi's cancel out,
we get our radius as being equal to 18 for this larger circle.
So if we want to know its area, its area
is going to be pi r squared, which
is equal to pi times 18 squared.
And let's figure out what 18 squared is.
18 times 18, 8 times 8 is 64, eight times 1
is 8 plus 6 is 14, and then we put
that 0 there because we're now in the tens place, 1 times 8
is 8, 1 times 1 is 1.
And really, this is a 10 times the 10,
and that's why it gives us 100.
Anyway, 4 plus 0 is a 4, 4 plus 8 is a 12,
and then 1 plus 1 plus 1 is a 3, so it's 324.
So the area here is equal to pi times 324,
or we could say 324 pi.
So the area of the entire larger circle, the part
that I shaded in yellow, including
what's kind of under this orange circle,
if you want to view it that way, this area right over here
is equal to 324 pi.
So the probability that a point that we select from this larger
circle is also in the smaller circle
is really just a percentage of the larger
circle that is the smaller circle.
So our probability-- I'll just write it
like this-- the probability that the point also
lies in the smaller circle-- so all of that stuff
I'll put in it.
The probability of that is going to be
equal to the percentage of this larger circle that
is this smaller one, and that's going to be--
or we could say the fraction of the larger circle's area
that is the smaller circle's area.
So it's going to be 16 pi over 324 pi.
And the pi's cancel out, and it looks like both of them
are divisible by 4.
If we divide the numerator by 4, we get 4,
if we divide the denominator by 4, what do we get?
4 goes into 320 80 times, it goes into 4 once,
so we get an 81.
So a probability-- I didn't even draw
this to scale, this area is actually much smaller when you
do it to scale-- the probability that if you were to randomly
select a point from the larger circle,
that it also lies in the smaller one
is the ratio of their areas, the ratio of the smaller
circle to the larger one.
And that is 4/81, I guess is the best way to say it.