Main content

## 7th grade

### Unit 7: Lesson 1

Basic probability- Statistics and probability FAQ
- Intro to theoretical probability
- Simple probability: yellow marble
- Simple probability: non-blue marble
- Simple probability
- Experimental probability
- Experimental probability
- Intuitive sense of probabilities
- Comparing probabilities

© 2022 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Simple probability: non-blue marble

In this example we are figuring out the probability of randomly picking a non-blue marble from a bag. Again, we'll have to think about the possible outcomes first. Created by Sal Khan.

## Video transcript

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.