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## AP®︎/College Statistics

### Course: AP®︎/College Statistics > Unit 7

Lesson 2: Mutually exclusive events and unions of events# Addition rule for probability

AP.STATS:

VAR‑4 (EU)

, VAR‑4.C (LO)

, VAR‑4.C.2 (EK)

, VAR‑4.E (LO)

, VAR‑4.E.4 (EK)

CCSS.Math: , Venn diagrams and the addition rule for probability. Created by Sal Khan.

## Want to join the conversation?

- Ok, so I have a very basic question, the answer to which seems counter intuitive to me. Ok, so if you flip a coin 4 times only, why do we multiply each possible outcome (2) by each other instead of adding them? So for instance, each flip comes with 2 possible outcomes, heads or tails. If you flip 4 times, then there should be 4X2 or 8 outcomes in my mind. But I know that to be false. There are 16, or 2*2*2*2. It seems as though it should be 2+2+2+2. I seem to conceptualize addition and multiplication does not seem correct to me. Why do we multiply? Ok thanks for any answers.(81 votes)
- Try thinking about the sequence of flips as follows (bear with me, and it should become clear when we get to the third flip!):

After you flip the coin once, you have 2 outcomes:`H`

(you flipped heads)`T`

(you flipped tails)

When you flip the coin a second time, you get another 2 outcomes, which (as you say) seem like they get 'added' to the previous outcomes. So now you have 4 outcomes:`1 2`

(flip number)`H H`

(first flip heads, this flip heads)`H T`

(first flip heads, this flip tails)`T H`

(first flip tails, this flip heads)`T T`

(first flip tails, this flip tails)

So far, it doesn't look like it matters whether you add or multiply, since both 2+2 and 2*2 = 4.

But now consider what happens when you flip the coin a third time. You have to 'add' another 2 outcomes*to each of the previous four outcomes*. So you are adding 2,*four times*. This is what multiplication is - multiple addition! So now there are 8 possible outcomes:`1 2 3`

(flip number)`H H H`

`H T H`

`T H H`

`T T H`

`H H T`

`H T T`

`T H T`

`T T T`

Hopefully you can now see that if you flip a fourth time, you would need to 'add' the two new outcomes to each of the previous 8 possibilities. Adding 2 eight times is the same as 8 x 2, so there are then 16 possible outcomes. I hope this helps! :-)(67 votes)

- Hello everyone! I hope this question is not too hard to answer. I understand why we remove the intersection (5/29), to avoid overestimating the probability. We use 5/29, because this is a given value (we already know that there are 5 yellow cubes). But when we apply the intersection rule [P(yellow)*P(cubes)], we get: 12/29*13/29, which is not equivalent to 5/29. Why is that so? What am I missing here? Thank you!(5 votes)
- I take it that by the "intersection rule" you mean the rule which states:

P( A ∩ B ) = P(A) x P(B)

This rule only applies when the two events are*independent*. This is not always a given. What independence means is that the probability of event B is the same whether or not even A occurred.

In this case, there is (overall) a 12/29 = 0.41 chance of drawing something Yellow. However, if we know that we picked a Cube, the probability that we have something Yellow is no longer 0.41, it's 5/13 = 0.38. Hence, the probability is not constant. So the events are not independent, and we can't just multiply the probabilities to get the intersection.(21 votes)

- why would the probability be zero in case of mutual exclusiveness when you can count the probability that someothing or something else has been taken out of the bag even if they dont overlap? Like if you have green, red and yellow cubes and you ask about the probability of taking out green or red than you can solve that even though they dont overlap. this got me confused.(5 votes)
- this is just a bit of common sense. if two events cannot happen at the same time, how likely is it that they will happen at the same time? the definition of the events gives the answer to the question. it is completely unlikely that they will happen at the same time because they cannot happen at the same time!(2 votes)

- What if you have three or more groups that may or may not overlap, and you want to calculate P(A or B or C ... n)?(6 votes)
- That's a great question. As you might guess, things get very complicated pretty quickly based on how many variables there are and how well the various overlaps behave. Google "Inclusion-Exclusion Principle" to see how deep that rabbit hole goes!(6 votes)

- kk so i need help. Let's say i have 27 blueberry pancakes. How many banana pancakes would i need to add to make the probability of grabbing a banana pancake 10%?(1 vote)
- Let 𝑏 be the number of banana pancakes.

Thereby we have a total of 𝑏 + 27 pancakes.

We want the probability of picking a banana pancake to be 10%:

𝑏∕(𝑏 + 27) = 0.1

Multiplying both sides by 𝑏 + 27, we get

𝑏 = 0.1𝑏 + 2.7

Subtracting 0.1𝑏 from both sides, we get

0.9𝑏 = 2.7

Finally, dividing both sides by 0.9, we get

𝑏 = 2.7∕0.9 = 3

So, we need to add 3 banana pancakes.(10 votes)

- anyone noticed the error @7:00(12+13)/29-5 !=20/29!(2 votes)
- If I know what you mean, I believe it was actually (12 + 13 - 5)/29 = 20/29. :)(8 votes)

- Is the venn diagram necessary?(3 votes)
- Not required, but it explains how to "see" that you are double counting some data, and makes you "see" that you need to subtract that value once to account of the double counting.(5 votes)

- How can all the possibilities be equally likely (2:51), if there are different numbers/colors of cubes/spheres? Am I misunderstanding sth? :((2 votes)
- Sal is talking about the fact that every single object has the same chance of falling out first. So each of the 29 objects could fall out of the bag and every single one of them is equally likely at the start of the experiment. In this case for every single object it would be 1/29. The groups(cubes, yellow,...) on the other hand have, because there are different amounts of them, a different chance of falling out. It's important to remember that even we have grouped these objects by shape and colour, they still are single, and in some sense, unique things with there very own likelihood of falling out. And the likelihood is the same for each of them.(4 votes)

- I am a little confused on the "or" rule for combining probabilities. Here is the problem. I have one die with six sides. If I roll the die 8 times, what is the probability that it will come up a six at least once? Using the "or" rule the formula would be

1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 8/6 or a probability of 133%

Am I applying the "or" rule incorrect? Why?

How does this differ from this example. Say the odds of getting killed in a car accident are 1 death per 100,000 miles driven. Each day I drive 1,000. What is the probability that I will be die in a car accident in 50 days? (50%?) 100 days? (100%?) 200 days (200%)...(2 votes)- The short answer is that there is overlap on the Venn diagram and, therefore, you need to subtract that overlap to avoid double counting.

Roll 1 die: Chance of rolling a 6 is 1/6

Roll 2 dice: Chance of rolling AT LEAST ONE 6 is (chance of rolling 6 on first die) + (chance of rolling 6 on second die) - (chance of rolling 6 on BOTH dice) = 1/6 + 1/6 - 1/36 = 7/36

As the number of dice goes up, subtracting out all the possible overlaps gets really tedious (maybe you roll 6 on 3 of the dice, maybe on 5 of them, etc.)

An easier way in this case is to "flip" the question:

Instead of asking "what's the probability of rolling at least one six?" we ask "What's the probability of never rolling a six?"

P(at least one 6) = 1 - P(no 6's)

P(no 6s) = P(no 6 on first die)P(no 6 on second die)...

We can simply multiply these "AND" conditions because there is no overlap.

Thus, we have,

P(at least one 6) = 1 - P(no 6s)

= 1 - (5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)

for eight dice.

This "trick" gives the same answer as adding all the P(roll 6)'s then subtracting out all the possible overlaps, it's just a lot less error-prone and tedious.(1 vote)

- At6:32, why does it say on the bottom right, in a type box, 29 is also the denominator of -5?(3 votes)
- It's because the probability of getting a
*cube*or a*yellow*is (12+13-5)/29. Sal mistakenly wrote (12+13)/29 - 5. He forgot to extend the division line (also called**Vinculum**) to include 5.(1 vote)

## Video transcript

Let's say I have a bag. And in that bag--
I'm going to put some green cubes in that bag. And in particular, I'm going
to put eight green cubes. I'm also going to put
some spheres in that bag. Let's say I'm going
to put nine spheres. And these are the green spheres. I'm also going to put some
yellow cubes in that bag. I'm going to put five of those. And I'm also going to put some
yellow spheres in this bag. And let's say I
put seven of those. I'm going to stick
them all in this bag. And then I'm going
to shake that bag. And I'm going to pour it out. And I'm going to look
at the first object that falls out of that bag. And what I want to think
about in this video is what are the
probabilities of getting different types of objects? So for example, what
is the probability of getting a cube of any color? What is the probability
of getting a cube? Well, to think about that
we should think about what-- or this is one way
to think about it-- what are all of the equally
likely possibilities that might pop out of the bag? Well, we have 8 plus 9 is 17. 17 plus 5 is 22. 22 plus 7 is 29. So we have 29 objects. There are 29 objects in the bag. Did I do that right? This is 14, yup 29 objects. So let's draw all of
the possible objects. I'll represent it as this
big area right over here. So these are all the
possible objects. There are 29 possible objects. So there's 29
equal possibilities for the outcome of my experiment
of seeing what pops out of the bag, assuming
that it's equally likely for a cube or a
sphere to pop out first. And how many of them meet our
constraint of being a cube? Well, I have eight green cubes,
and I have five yellow cubes. So there are a
total of 13 cubes. So let me draw
that set of cubes. So there's 13 cubes. We could draw it like
this-- there are 13 cubes. This right here is the
set of cubes, this area. And I'm not drawing it exact. I'm approximating. It represents the
set of all the cubes. So the probability
of getting a cube is the number of events
that meet our criteria. So there's 13
possible cubes that have an equally likely
chance of popping out, over all of the possible equally
likely events, which are 29. That includes the
cubes and the spheres. Now let's ask a
different question. What is the probability of
getting a yellow object, either a cube or a sphere? So once again, how many things
meet our conditions here? Well, we have 5 plus 7. There's 12 yellow
objects in the bag. So we have 29 equally
likely possibilities. I'll do it in that same color. We have 29 equally
likely possibilities. And of those, 12
meet our criteria. So let me draw 12
right over here. I'll do my best attempt. So let's say it looks
something like-- so the set of yellow objects. There are 12 objects
that are yellow. So the 12 that
meet our conditions are 12, over all the
possibilities-- 29. So the probability
of getting a cube-- 13 29ths, probability of
getting a yellow-- 12 29ths. Now let's ask something a
little bit more interesting. What is the probability
of getting a yellow cube? So I'll put it in yellow. So we care about the color, now. So this thing is yellow. What is the probability of-- or
as my son would say, "lello." What is the probability
of getting a yellow cube? Well, there's 29 equally
likely possibilities. And of those 29 equally likely
possibilities, 5 of those are yellow cubes, or
"lello" cubes, five of them. So the probability is 5 29ths. And where would we see
that on this Venn diagram that I've drawn? This Venn diagram is
just a way to visualize the different probabilities. And they become
interesting when you start thinking about
where sets overlap, or even where they
don't overlap. So here we are
thinking about things that are members
of the set yellow. So they're in this set,
and they are cubes. So this area right
over here-- that's the overlap of these two sets. So this area right
over here-- this represents things that
are both yellow and cubes, because they are
inside both circles. So this right over here-- let
me rewrite it right over here. So there's five objects that
are both yellow and cubes. Now let's ask-- and this is
probably the most interesting thing to ask-- what is
the probability of getting something that is yellow or or
a cube, a cube of any color? The probability of getting
something that is yellow or a cube of any
color-- well, we still know that the denominator
here is going to be 29. These are all of the
equally likely possibilities that might jump out of the bag. But what are the possibilities
that meet our conditions? Well, one way to think about
it is, well, the probability-- there's 12 things that would
meet the yellow condition. So that would be this entire
circle right over here-- 12 things that meet the
yellow condition. So this right over here is 12. This is the number of yellow. That is 12. And then to that, we can't
just add the number of cubes, because if we add
the number of cubes, we've already counted these 5. These 5 are counted
as part of this 12. One way to think
about it is there are 7 yellow objects
that are not cubes. Those are the spheres. There are 5 yellow
objects that are cubes. And then there are 8
cubes that are not yellow. That's one way to think about. So when we counted this
12-- the number of yellow-- we counted all of this. So we can't just add
the number of cubes to it, because then we would
count this middle part again. So then we have to
essentially count cubes, the number of
cubes, which is 13. So the number of
cubes, and we'll have to subtract out this
middle section right over here. Let me do this. So subtract out the middle
section right over here. So minus 5. So this is the number
of yellow cubes. It feels weird to write
the word yellow in green. The number of yellow cubes-- or
another way to think about it-- and you could just do
this math right here. 12 plus 13 minus 5 is 20. Did I do that right? 12 minus, yup, it's 20. So that's one way. You just get this is
equal to 20 over 29. But the more interesting
thing than even the answer of the probability
of getting that, is expressing this in terms
of the other probabilities that we figured out
earlier in the video. So let's think about
this a little bit. We can rewrite this
fraction right over here. We can rewrite this as 12 over
29 plus 13 over 29 minus 5 over 29. And this was the
number of yellow over the total possibilities. So this right over here
was the probability of getting a yellow. This right over
here was the number of cubes over the
total possibilities. So this is plus the
probability of getting a cube. And this right over
here is the number of yellow cubes over
the total possibilities. So this right over here
was minus the probability of yellow, and a cube. I'm not going to
write it that way. Minus the probability
of yellow-- I'll write yellow in yellow--
yellow and a cube. And so what we've
just done here-- and you could play
with the numbers. The numbers I just used
as an example right here to make things a
little bit concrete. But you can see this is
a generalizable thing. If we have the probability
of one condition or another condition-- so let me
rewrite it-- the probability-- and I'll just write it a
little bit more generally here. This gives us an
interesting idea. The probability of getting
one condition of an object being a member of set
a, or a member of set b is equal to the probability
that it is a member of set a, plus the probability that
is a member of set b, minus the probability
that is a member of both. And this is a really
useful result. I think sometimes it's
called the addition rule of probability. But I want to show you that
it's a completely common-sense thing. The reason why you can't just
add these two probabilities is because they might
have some overlap. There's a probability
of getting both. And if you just
added both of these, you would be double
counting that overlap, which we've already seen
earlier in this video. So you have to subtract
one version of the overlap out so you are not
double counting it. I'll throw another
one other idea out. Sometimes, you
have possibilities that have no overlap. So let's say this is the
set of all possibilities. And let's say this is the
set that meets condition a and let me do this
in a different color. And let's say that this is the
set that meets condition b. So in this situation,
there is no overlap. There's no way-- nothing is a
member of both sets, a and b. So in this situation, the
probability of a and b is 0. There is no overlap. And these type of conditions,
or these two events, are called mutually exclusive. So if events are
mutually exclusive, that means that they both
cannot happen at the same time. There's no event that meets
both of these conditions. And if things are
mutually exclusive, then you can say the
probability of a or b is the probability of a plus
b, because this thing is 0. But if things are not
mutually exclusive, you would have to
subtract out the overlap. And probably the
best way to think about it is to
just always realize that you have to
subtract out the overlap. And obviously if something
is mutually exclusive, the probability of getting
a and b is going to be 0.