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Probability using combinations

Probability of getting exactly 3 heads in 8 flips of a fair coin. Created by Sal Khan.

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  • leaf green style avatar for user richardcgeddes
    Why are coin tosses referred to as distinct... 2 heads out of 3 tosses {hht, hth, thh}, for example. They all are different arrangements of the same combination, 2 heads and one tail. So I wondered what a permutation of the tosses would be, and realized that the order being referred to, is the sequence in time, and it would be nonsensical to consider different arrangements to the time sequence. Does this make sense?
    (14 votes)
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    • blobby green style avatar for user m.ferreria
      The coin tosses are referred as distinct because otherwise, your total amount of cases would not be equiprobable (that is to say, the chance of getting a special combination is not the same as getting some other)

      Flipping a coin two times, you get the combination {{hh}, {th}, {tt}} (the set of subsets). Yet, the chance of you getting heads-tails and tails-heads is not 1/3, its 2/4. Then, the probabilities of this set becomes {1/4{tt}, 1/2{th}, 1/4{tt}} and it is not an equiprobable set.

      Therefore, you have to arrange it like {(tt), (th), (ht), (hh)} (set of tuples), each of these having probability 1/4. And now you can take simple #P(A)/#P(S) to calculate probabilities.

      Hope this solved your doubts.
      (13 votes)
  • female robot grace style avatar for user Rachel
    Around the 5 minute mark, he starts talking about factorials. What's a factorial?
    (2 votes)
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  • blobby green style avatar for user get it right the 2nd time
    why can't I use this method in the last numerical (previous module: "probability & combination (part 2) where we have to calculate the probability of 3 basket in 5 free throws P(3/5) given that free throw % is 80%

    total outcome= 2^5=32 (since every throw might be basket or a miss, 2 possibility for every throw).

    combination of choosing 3 out of 5= 5!/3!2!= 10

    total probability = 10/32=31.25% but the answer is 20.48%....does it have to do something odds of scoring a basket or missing is not equal.
    (6 votes)
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  • blobby green style avatar for user Shivam Pathak
    I am a bit confused here, Why aren't we considering probability of a coin flip (i.e a single event occurring viz 50% or 1/2) while approaching the problem ?
    I understand that the probability of getting a heads or tails is same but shouldn't it be considered while performing the calculations !
    The solution is directly obtained by putting numbers 8 & 3 where we are not even considering what is heads and what is tails :/ The number 3 can refer to either heads or tails.
    (9 votes)
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    • starky sapling style avatar for user Khushi Viramgami
      I'm pretty sure the "1/2" probability is considered a part of the total outcomes (2^8) which shows that there are only 2 options, and since there are no further calculations, it can be assumed that each option is equally likely. Since the problem is asking for the probability of 3 heads, anyone looking at the problem can consider your answer/work through the context of the question. (However, you are right: the same question asking for the probability of 3/8 tails would also have the same answer.) Hope this helps!
      (1 vote)
  • starky ultimate style avatar for user Brian E.
    Is there such thing as a negative factorial, like "-1!" or "-7!"?
    (3 votes)
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    • leaf yellow style avatar for user Howard Bradley
      No there aren't. The factorials of negative integers are undefined.

      To see why, consider that (n-1)! = n!/n. So 0! = 1!/1 = 1. However (-1)! = 0!/0 = 1/0, and division by zero is undefined. And, since (-1)! is undefined, it follows that factorials of the other negative integers are also undefined.
      (12 votes)
  • leaf green style avatar for user matt.stewart
    Why does he divide all COMBINATIONS of getting 3/8 heads by the total possible PERMUTATIONS? Why doesn't he divide correct permutations by total permutations? (that is, shouldn't he omit the 3! in the denominator at ?)
    (2 votes)
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    • blobby green style avatar for user adibparkar
      He doesn't divide by the total number of permutations, he's dividing by the total number of possible outcomes which is 2^8. To find the number of outcomes out of these 2^8 possible outcomes that have exactly 3 heads we need to figure out how many ways we can choose exactly 3 heads in 8 flips. Since one "heads" is exactly the same as another, the ordering of these 3 heads does not matter, hence the number of ways is 8C3.

      Let me try and illustrate this through a smaller example, suppose we were dealing with the probability of getting exactly 2 heads in 3 flips. Here the total number of outcomes is 2^3 = 8. Let the flips be numbered 1, 2 and 3. Now if we had to choose 2 flips out of these 3 to have heads, in how many ways could we choose them? Let's call these two flips flipHeadsA and flipHeadsB. We have a total of 3 flips to choose from initially, so for flipHeadsA we have 3 options, and now we have only 2 options left for flipHeadsB. Hence there are 6 permutations for flipHeadsA and flipHeadsB. I'll list them out:

      heads A on flip 1, heads B on flip 2
      heads A on flip 1, heads B on flip 3
      heads A on flip 2, heads B on flip 1
      heads A on flip 2, heads B on flip 3
      heads A on flip 3, heads B on flip 1
      heads A on flip 3, heads B on flip 2

      But as you can see, this means we are differentiating between heads A and heads B, which is unnecessary. If heads A was obtained on flip 1 and heads B was obtained on flip 3, this is equivalent to heads B being obtained on flip 1 and heads A being obtained on flip 3 since they are both heads, so we are overcounting if we use permutations. So we divide by the number of ways the heads can be arranged among themselves (2! in this example) to get the actual number of events that will satisfy the condition exactly 2 heads in 3 flips. Hence, the number of ways to get exactly 2 heads in 3 flips is 3C2 and not 3P2.
      (11 votes)
  • blobby green style avatar for user reshmaak
    If it needs to be exactly 3 heads and not more, don't we need to remove the cases where there may be more than 3 heads in the numerator?
    (5 votes)
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    • starky sapling style avatar for user Khushi Viramgami
      Yes (you could show that), but in turn, you would also have to remove all of the cases where there may be less than 3 heads in the numerator as well. If you do that, you would get the same answer, but you'd end up doing more work. Sal calculated the EXACT number of heads, and since he didn't include anything else in the numerator, we can consider this as the same as removing all of the unwanted outcomes.

      P(3 H) = 0.21
      P(3 H) = 1-P(not 3 H) = 1-0.79 = 0.21

      Hope this helps! (I know you asked this 8 years ago, but for anyone else who has the same question as you!!)
      (5 votes)
  • leafers sapling style avatar for user SHINAZ MOHAMED
    What is the number of outcomes when a coin is tossed and then a die is rolled only in case a head is shown on the coin?
    (1 vote)
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  • mr pink red style avatar for user Daniel Wiczew
    Does n C k = n C (n-k) ? (n C k is other way of writting binomial coffiecient)
    (1 vote)
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    • ohnoes default style avatar for user Cyan Wind
      Yep, that's correct. Let's take some examples:

      16C3 = 560 = 16C13 (n = 16, k = 3)
      22C14 = 319770 = 22C8 (n = 22, k = 14)

      More academic proof: Let's call m = n - k, therefore k = n - m.
      nCk = n! / [k! * (n - k)!]
      nCk = n! / [(n - m)! * m!]
      nCk = nCm
      (7 votes)
  • blobby green style avatar for user TStones01
    I am still struggling to get my head around the intuition on this one; in particular the concept that 3 chosen variables occuring in any order out of 8 unkown variables (outcomes of coin flips) is the same concept as 3 chosen variables arising in any order (e.g.: three people sitting together in any order) out of 8 known variables (e.g.: eight people). Is the unkown nature of the variables irrelevant for the calculation of the combinations? Does this also extend to situations with more unknown variables?

    For example if we have a fruit machine with 8 spinning wheels, there are 10 items on each wheel 1 of those 10 is a bell and we say we want to calculate the permutations where there will be 3 of the 8 wheels showing a bell? Would this situation still be 8!/(3!*(8-3)!)?
    (3 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      The coins and the people aren't really that different.
      Each coin has two possible outcomes, Heads or Tails, and each person also has two possible outcomes, sitting or standing.

      The difference is that for the 8 people we are asked the number of ways that we can choose which 3 are sitting, while for the 8 coins we are asked the probability that 3 are Heads and the others are Tails.

      However, since each coin is equally likely to land Heads as to land Tails, then each sequence of 8 coins is also equally likely to occur, and thereby the probability that 3 coins are Heads is equal to the number of sequences that consist of 3 Heads and 5 Tails, divided by the total number of possible sequences.

      So, we want to know in how many ways we can choose which 3 coins are Heads, which of course must be the exact same as the number of ways we can choose which 3 people are sitting.

      – – –

      In the case with the fruit machine, each wheel can be seen as having two possible outcomes, Bell or No Bell.

      However, now the two possibilities have different probabilities:
      P(Bell) = 1∕10
      P(No Bell) = 9∕10

      This means that the probability of a certain sequence depends on the number of Bells that are in the sequence.
      A sequence containing 3 Bells has the probability
      (1∕10)³ ∙ (9∕10)⁵

      But now that we have that probability, all we need to do is multiply it by the number of sequences containing 3 Bells, which again is the same as the number of ways we can choose which 3 wheels are Bells = 8!∕(3! ∙ 5!)
      (3 votes)

Video transcript

So you might be wondering why I went off into permutations and combinations in the probability playlist, and I think you'll learn in this video. So let's say I want to figure out the probability-- I'm going to flip a coin eight times and it's a fair coin. And I want to figure out the probability of getting exactly 3 out of 8 heads. So I say 3/8 heads, but 3 of my flips are going to be heads and the rest are going to be tails. So how do I think about that? Well, let's go back to one of the early definitions we used for probability. And that says, the probability of anything happening is the probability of the number of equally probable events into which what we're stating is true. So in which the number of events-- I guess trials or situations-- in which we get 3 heads, and exactly 3 heads, we're not saying greater than 3 heads. So 4 heads won't count and 2 heads won't count, 5 heads won't-- only 3 heads. And then, over the total number of equally probable trials-- not trials, total number of equally possible outcomes. I should be using the word outcomes. So just with the word outcomes it should be the total number of outcomes in which what we're saying happens. So we get 3 heads over the total possible outcomes. So let's do the bottom part first. What are the total possible outcomes if I'm flipping a fair coin eight times? Well, the first time I flip it I either get heads or tails, so I get 2 outcomes. And then when I flip it again I get 2 more come for the second one. And then, how many total outcomes? Well, that's 2 times 2 because I could have got 2 in the first, 2 in the second flip. And then essentially we would multiply 2 times the number of flips. So that's 5, 6, 7, 8, and that equals 2 to the eighth. So the number of outcomes is just going to be 2 to the total number of flips. And hopefully that make sense to you. If not, you might want to re-watch some of the earlier videos. But that's the easy part. So there's 2 to the eighth possible outcomes when you flip a fair coin eight times. So how many of those outcomes are going to result in exactly 3 heads? Let's think of it this way. Let's give a name to each of our flips. Let's give a name to them. So let me make a little column, we'll call these the flips. This is my flips column. And I could name them anything. I could name them Larry, Curly, Moe. I could name them-- well, I would need 5 more names for them, but I could name them the 7 dwarfs or the 8 dwarfs really because I have 8 flips. I'll number the flips. Flip 1, 2, 3, 4, 5, 6, 7, 8. And I'm the god of probability. And essentially, I need to just pick 3 of these flips that are going to result in heads. So another way to think about it is, these could be 8 people and I could pick which of these-- how many ways can I pick 3 of these people to put into the car? How many ways can I pick 3 of these people to sit in chairs. And it doesn't matter the order that I pick them in. It doesn't matter if I say the people that are going to get in the car are going to be people 1, 2, 3 3. Or if I say 3, 2, and 1, or if I say 2, 3, and 1. Those are all the same combination. So similarly, if I'm just picking flips and I have to say, OK, 3 of these flips are going to get into the heads car. Heads is like they're sitting, they're people sitting down. I don't want to confuse you too much. But essentially I'm just going to choose 3 things out of the 8. So I'm essentially just saying, how many combinations can I get where I pick 3 out of these 8. And so that should immediately ring a bell that we're essentially saying, out of 8 things we're going to choose 3. How many combinations of 3 can we pick of 8 and that we went over in the last video. And let's do it with the formula first. So let me write the formula up here just so you remember it, but I also want to give you the intuition again, for the formula. So in general, we said, n choose k, that is equal to n factorial over k factorial times n minus k factorial. So in this situation that would equal 8 factorial over 3 factorial times what? 8 minus k-- times 5 factorial. Or another way of writing this, this would be 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 over-- I'll just write 3 factorial for a second. Then times 5 times 4 times 3 times 2 times 1. And of course, that and that cancel out and all you're left with is 8 times 7 time 6 over 3 factorial. And I did this for reason because I want you to re-get the intuition at least for this part of the formula. That's essentially just saying, how many permutations can I-- how many ways can I pick 3 things out of 8? And that's essentially saying, well, before I pick anything I could pick 1 of 8. Then I have 7 left to pick from for the second spot. And then I have 6 left to pick for the third spot. And so that's essentially the number of permutations. But since we don't care what order we picked them in, we need to divide by the number of ways we can rearrange 3 things, and that's where the 3 factorial comes from. And so hopefully I didn't confuse you, but if I did you can go back to this formula for the binomial coefficient. But it's good to have the intuition. And then once we're at this point we can just calculate this. Well, what's this? This is 8 times 7 times 6 over 3 factorial of 3 times 2 times 1. So that's 6. The 6 cancels out, so it's 8 times 7. So there's 8 times 7, or what is that? 56. That's equal to 56. So there's 56 different ways to pick 3 things out of 8. Or if I have 8 people there's 56 ways of picking 3 people to sit in the car or however you want t view it. But if I have 8 flips there's 56 ways of picking 3 of those flips to be heads. So let's go to our original probability problem. What is the probably that I get 3 out of 8 heads? Well, it's the number of ways I can pick 3 out of those 8, so it equals 56, over the total number of outcomes. The total number of outcomes is 2 to the eighth. Another way I could write that-- 56, let me unseparate. That's 8 times 7 over 2 to the eighth. 8 is 2 to the third. Let me erase some of this. Not with that color. Let me erase that. Let me erase all of this just so I space. And I will switch colors for variety. Let me use the small pen. OK, so I'm back. All right, so 8 is the same thing as 2 to the third times 7-- this is all just mathematical simplification, but it's useful-- over 2 to the eighth. And so, if we just divide both sides-- the numerator and the denominator by 2 to the third, this becomes 1. This becomes 2 to the fifth. And so it becomes 7/32. Is that right? So if I were to pick 3 out of 8-- yep, I think that is right. And so what does that turn out to be? Let me get my calculator. [INAUDIBLE] to make careless mistakes. Let's see. My calculator seems to have disappeared. Let me get it back. There it is. OK. 7 divided by 32 is equal to 0.21875. Which is equal to 21.-- you know, if I were to round roughly-- 21.9% chance. So there's a little bit better than 1 in 5 chance that I get exactly 3 out of the 8 flips as heads. Hopefully I didn't confuse you and now you can apply that to pretty much anything. You could say, well, what is the probability of getting-- if I flip a fair coin-- of getting exactly 7 out of 8 heads? Or you could say, what's the probability of getting 2 out of 100 heads? And you could use it the exact same way we did this problem. I'll see you in the next video.