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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?
• Short Answer: You are right, we would not use the same method.

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

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
• Isn't this called complementary counting? Or is it called complimentary counting? What's the difference between the two?
• It is complementary counting. Compliment has a different meaning than complement.
• 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)
• 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.
• Why do we subtract those from 1?
• "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.
• Suppose that a fair coin is tossed 100 times. Find the probability of observing at least 60 heads. How do you solve for this?
• Why wouldn't you just do .5^10 instead of multiplying all the 1/2s?
• 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)
• 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)"
• 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)!).
• What is the probability that you will get heads four times in a row when flipping a fair coin
• 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?"
• I am not really sure when you will use this technique. Can you please explain when using 1- P(not event) would be helpful?
• 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.