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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.

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  • blobby green style avatar for user raj.patel111213
    what if there were would be P(of atleast 2HH in 10 flips). We would probably not use the same method. RIGHT?
    (131 votes)
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    • mr pants teal style avatar for user shawn14parker
      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
      (238 votes)
  • blobby green style avatar for user hms99sun
    Isn't this called complementary counting? Or is it called complimentary counting? What's the difference between the two?
    (11 votes)
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  • duskpin ultimate style avatar for user Nitschke, Charles (Monty)
    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)
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    • leaf blue style avatar for user Dr C
      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)
  • ohnoes default style avatar for user Funkadillon
    Why do we subtract those from 1?
    (4 votes)
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    • mr pink red style avatar for user Jean Rambo
      "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)
  • blobby green style avatar for user 누엔리나
    Suppose that a fair coin is tossed 100 times. Find the probability of observing at least 60 heads. How do you solve for this?
    (2 votes)
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  • blobby green style avatar for user Fraidy Levilev
    Why wouldn't you just do .5^10 instead of multiplying all the 1/2s?
    (5 votes)
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  • primosaur seed style avatar for user Zeba S. Saraf
    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)
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  • blobby green style avatar for user velma clarke
    What is the probability that you will get heads four times in a row when flipping a fair coin
    (2 votes)
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    • blobby green style avatar for user Michael Costello
      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)
  • old spice man green style avatar for user tvmccray825
    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)
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    • leaf blue style avatar for user Dr C
      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)
  • aqualine ultimate style avatar for user Saanzan Bari
    What does the "Fair" in fair coin represent; what does it mean?
    (0 votes)
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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.