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## Multiplication rule for independent events

# "At least one" probability with coin flipping

AP.STATS:

VAR‑4 (EU)

, VAR‑4.A (LO)

, VAR‑4.A.4 (EK)

, VAR‑4.E (LO)

, VAR‑4.E.2 (EK)

## Video transcript

Now let's start to do some
more interesting problems. And one of these things that
you'll find in probability is that you can always do
a more interesting problem. So now I'm going to
think about-- I'm going to take a
fair coin, and I'm going to flip it three times. And I want to find the
probability of at least one head out of the three flips. So the easiest way
to think about this is how many equally likely
possibilities there are. In the last video, we saw
if we flip a coin 3 times, there's 8 possibilities. For the first flip,
there's 2 possibilities. Second flip, there's
2 possibilities. And in the third flip,
there are 2 possibilities. So 2 times 2 times 2-- there are
8 equally likely possibilities if I'm flipping a coin 3 times. Now how many of
those possibilities have at least 1 head? Well, we drew all the
possibilities over here. So we just have to
count how many of these have at least 1 head. So that's 1, 2, 3, 4, 5 5, 6, 7. So 7 of these have at
least 1 head in them. And this last one does not. So 7 of the 8 have
at least 1 head. Now you're probably
thinking, OK, Sal. You were able to do
it by writing out all of the possibilities. But that would be
really hard if I said at least one
head out of 20 flips. This had worked well
because I only had 3 flips. Let me make it clear,
this is in 3 flips. This would have
been a lot harder to do or more time consuming
to do if I had 20 flips. Is there some shortcut here? Is there some other
way to think about it? And you couldn't just do
it in some simple way. You can't just say,
oh, the probability of heads times the probability
of heads, because if you got heads the first
time, then now you don't have to get heads anymore. Or you could get heads
again-- you don't have to. So it becomes a little
bit more complicated. But there is an easy way
to think about it where you could use this
methodology right over here. You'll actually see
this on a lot of exams where they make it seem
like a harder problem, but if you just think about in
the right way, all of a sudden it becomes simpler. One way to think about it is
the probability of at least 1 head in 3 flips
is the same thing-- this is the same thing--
as the probability of not getting all tails, right? If we got all tails, then we
don't have at least 1 head. So these two things
are equivalent. The probability of getting
at least 1 head in 3 flips is the same thing
as the probability of not getting all
tails in 3 flips. So what's the probability
of not getting all tails? Well, that's going to be 1
minus the probability of getting all tails. The probability of getting
all tails, since it's 3 flips, it's the probability
of tails, tails, and tails. Because any of the
other situations are going to have at
least 1 head in them. And that's all of the
other possibilities, and then this is the only
other leftover possibility. If you add them together,
you're going to get 1. Let me write it this way. Let me write it a
new color just so you see where this is coming from. The probability of not all
tails plus the probability of all tails-- well, this
is essentially exhaustive. This is all of the
possible circumstances. So your chances of getting
either not all tails or all tails-- and these
are mutually exclusive, so we can add them. The probability of not all
tails or, just to be clear what we're doing, the
probability of not all tails or the probability of all tails
is going to be equal to one. These are mutually exclusive. You're either going to
have not all tails, which means a head shows up. Or you're going
to have all tails. But you can't have both
of these things happening. And since they're
mutually exclusive and you're saying the
probability of this or this happening, you could
add their probabilities. And this is essentially
all of the possible events. So this is essentially,
if you combine these, this is the probability of
any of the events happening. And that's going to
be a 1 or 100% chance. So another way to think
about is the probability of not all tails
is going to be 1 minus the probability
of all tails. So that's what we
did right over here. And the probability of all
tails is pretty straightforward. That's the probability
of it's going to be 1/2, because you have
a 1/2 chance of getting a tails on the
first flip, times-- let me write it here, so we
can have it a little clearer. So this is going to be 1 minus
the probability of getting all tails. You will have a 1/2
chance of getting tails on the first flip,
and then you're going to have to get another
tails on the second flip, and then you're
going to have to get another tails on the third flip. And then 1/2 times
1/2 times 1/2. This is going to be 1/8. And then 1 minus
1/8 or 8/8 minus 1/8 is going to be equal to 7/8. So we can apply that to
a problem that is harder to do than writing
all of the scenarios like we did in
the first problem. Let's say we have 10 flips,
the probability of at least one head in 10 flips-- well,
we use the same idea. This is going to be equal
to the probability of not all tails in 10 flips. So we're just saying
the probability of not getting all of the
flips going to be tail. All of the flips is tails--
not all tails in 10 flips. And this is going to be 1 minus
the probability of flipping tails 10 times. So it's 1 minus
10 tails in a row. And so this is going to be equal
to this part right over here. Let me write this. So this is going to be this one. Let me just rewrite it. This is equal to 1 minus-- and
this part is going to be, well, one tail, another tail. So it's 1/2 times 1/2. And I'm going to
do this 10 times. Let me write this
a little neater. 1/2-- so that's 5,
6, 7, 8, 9, and 10. And so we really just have to--
the numerator is going to be 1. So this is going to be 1. This is going to be equal to 1. Let me do it in that
same color of green. This is going to be equal
to 1 minus-- our numerator, you just have 1 times
itself 10 times. So that's 1. And then on the denominator,
you have 2 times 2 is 4. 4 times 2 is 8, 16, 32, 64, 128,
256, 512, 1,024-- over 1,024. This is the exact same thing
as 1 is 1024 over 1024 minus 1 over 1024, which is equal
to 1,023 over 1,024. We have a common
denominator here. So 1,000-- I'm doing that
same blue-- over 1,024. So if you flip a coin 10
times in a row-- a fair coin-- you're probability of
getting at least 1 heads in that 10 flips is pretty high. It's 1,023 over 1,024. And you can get a calculator
out to figure that out in terms of a percentage. Actually, let me just
do that just for fun. So if we have 1,023 divided
by 1,024 that gives us-- you have a 99.9% chance
that you're going to have at least one heads. So this is if we round. This is equal to 99.9% chance. And I rounded a little bit. It's actually slightly, even
slightly, higher than that. And this is a pretty powerful
tool or a pretty powerful way to think about it because
it would have taken you forever to write all
of the scenarios down. In fact, there would have been
1,024 scenarios to write down. So doing this
exercise for 10 flips would have taken
up all of our time. But when you think about in a
slightly different way, when you just say, look the
probability of getting at least 1 heads in 10 flips
is the same thing as the probably of
not getting all tails. And that's 1 minus
the probability of getting all tails. And this is actually a pretty
easy thing to think about.