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

Lesson 4: Independent versus dependent events and the multiplication rule- Compound probability of independent events
- Independent events example: test taking
- General multiplication rule example: independent events
- Dependent probability introduction
- General multiplication rule example: dependent events
- The general multiplication rule
- Probability with general multiplication rule
- "At least one" probability with coin flipping
- Probabilities involving "at least one" success
- Probability of "at least one" success

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# "At least one" probability with coin flipping

In this video, we 'll explore the probability of getting at least one heads in multiple flips of a fair coin. Created by Sal Khan.

## Want to join the conversation?

- what if there were would be P(of atleast 2HH in 10 flips). We would probably not use the same method. RIGHT?(132 votes)
- Short Answer: You are right, we would not use the same method.

Long Answer:

You would use a similar method, which involves what we've been doing. However, instead of just subtracting "no tails" from one, you would also subtract "one heads" from it too.

P(at least 2 heads) = 1 - P(No heads) - P(One heads)

Since there are ten repetitions of the experiment, and two possible outcomes per experiment, the number of different outcomes is 2 ^ 10, or 1024.

P(No heads) is simple enough to find, just take the probability of tails to the tenth power.

P(No heads) = (1 / 2) ^ 10 = 1 / 1024

In order to find P(One Heads) you're going to have to think. If you want only one heads out of ten, there are going to be ten different ways to get one head. Heads could be first, second, third, fourth, fifth, sixth, seventh, eighth, ninth, or tenth, so that's ten different ways you would have just one heads. (Scroll to the bottom of comment to see a picture of what I'm talking about) We have the same number of different possibilities as before, so we keep the denominator the same.

P(One Heads) = 10 / 1024

So now we have P(No Heads) and P(One Heads), so we just plug those in to find P(At Least Two Heads):

P(At Least Two Heads) = 1 - (1 / 1024) - (10 / 1024)

P(At Least Two Heads) = (1024 - 1 - 10) / 1024

P(At Least Two Heads) = 1013 / 1024

P(One Heads) Visual

1 2 3 4 5 6 7 8 9 10

H T T T T T T T T T

T H T T T T T T T T

T T H T T T T T T T

T T T H T T T T T T

T T T T H T T T T T

T T T T T H T T T T

T T T T T T H T T T

T T T T T T T H T T

T T T T T T T T H T

T T T T T T T T T H(242 votes)

- Isn't this called complementary counting? Or is it called complimentary counting? What's the difference between the two?(11 votes)
- It is complementary counting. Compliment has a different meaning than complement.(3 votes)

- So the question of P(at least 2 heads in 10 flips) was asked and the answer was

P(at least 2 heads) = 1 - P(No heads) - P(1 heads)

I figure we subtract P(1 heads) because it does not meet our conditions of 2 heads. So I was curious if the rule follows as such:

P(at least 3 heads) = 1 - P(No heads) - P(1 heads) - P(2 heads)

And the generalization being

P(at least n heads) = 1 - P(No heads) + ∑(k=1 to n-1) of -P(k heads)(9 votes)- Well done! Yes, your generalization works (though you could just start the summation from k=0, to avoid separating the P(X=0) each time).

Though be careful about this "rule". Say with ten flips, you wanted the probability of at least 9 heads. With your generalization it would be:

P(X>=9) = 1 - ∑{k=0 to n-1} P(X=k)

But this might have you calculate 9 probabilities (0,...,8), when it might be easier to calculate P(X=9) + P(X=10). It's good to know how to manipulate the probability expressions, and knowing that probability sums to 1 is a very useful 'trick'. Using it, or other little rules, we can flip around probability statements to make what we have to calculate easier. For example, some calculators include functions for P(X <= k), and so it is easiest to express the probability in terms of that (like what you did above) when possible. Or say we wanted:

P( A <= X <= B)

That is, at least A, but no more than B. We could rewrite this as:

P(X <= B) - P(X <= A)

It's just a game of making it easier to calculate for the tools you have available.(10 votes)

- Why do we subtract those from 1?(3 votes)
- "1" represents the total number of possible events, or 100%. If you want to know what the probability is to get at least one Heads, then that is the same as the probability of all the events (100%, or 1) minus the probability of getting all Tails. If you subtract the possibility of having all tails from the probability of anything happening (100%), then you are left with the probability of all the scenarios where there are Heads involved. Because getting all Tails is the only scenario where the "at least one Heads" requirement is not met, all other scenarios are good and have at least one Heads.(5 votes)

- Suppose that a fair coin is tossed 100 times. Find the probability of observing at least 60 heads. How do you solve for this?(3 votes)
- OK cool

But what about at least 2(or other numbers larger than 1) heads in 20 flips?(3 votes)- The general formula for “at least two” would be 1-P(none)-P(one).

Each probability could be calculated as a binomial probability.

For the “at least least two heads in 20 rolls” it would be

1 - 0.5^20 - 20*0.5^20

= 99.998%(4 votes)

- Any kind peeps out there to help me understand this problem in detail please? The problem goes:

"Toss a coin for 50 times. Calculate the probability of getting head for 30 times.Draw Probability Diagram for the case (for 5 times only)"(3 votes)- You will need to use binomial theorem for this question. So the answer is (50 C 30) (0.5^30)(0.5^20) which also equals (50 C 20) (0.5^30)(0.5^20)= (50 C 20)(0.5^50)

The (n C r) means n!/(r!(n-r)!).(2 votes)

- What is the probability that you will get heads four times in a row when flipping a fair coin(2 votes)
- A coin has a 50% chance of landing on heads the each time it is thrown. For the first coin toss, the odds of landing heads is 50%. On the second coin toss, take the 50% from the first toss, and multiply it by another 50%. Repeat this for the third and fourth tosses and it should look something like this:

(1/2)(1/2)(1/2)(1/2) = 1/(2*2*2*2) = 1/(2^4) = 1/16 = 6.25%

The 1/2 is the chance that the desired outcome occurs, the answer remains the same if the question was "What are the chances of landing tails four times in a row?" or "What are the chances of landing heads the first two times and tails the second two times?"(4 votes)

- Why wouldn't you just do .5^10 instead of multiplying all the 1/2s?(5 votes)
- Yes. Normally you would do
`P(at least 1H) = 1- (1/2)^10`

Sal was just writing all of the 1/2's out to illustrate what's going on to people who are new to the concept.(1 vote)

- I am not really sure when you will use this technique. Can you please explain when using 1- P(not event) would be helpful?(2 votes)
- Flip a fair coin 13 times. Find the probability of at least 1 head. We could either do:

P( X ≥ 1 ) = P( X=1 ) + P( X=2 ) + P( X=3 ) + P( X=4 ) + ... + P( X=13 )

or, we could do:

P( X ≥ 1 ) = 1 - P( X = 0 )

Calculating just one probability, P( X=0 ), is much easier than calculating many (in this case, 13) probabilities.(4 votes)

## 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.