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Free throw binomial probability distribution

AP.STATS:
UNC‑3 (EU)
,
UNC‑3.B (LO)
,
UNC‑3.B.1 (EK)
Sal uses the binomial distribution to calculate the probability of making different number of free throws.

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  • aqualine ultimate style avatar for user 15SAGANmk
    At why is it 0.3 to the 6th power?
    (7 votes)
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    • marcimus pink style avatar for user Tyler
      A simpler way of thinking about it would be this:

      What are the chances of missing a free throw? Well, the probability of making a free throw is 70%, so the probability of missing a free throw must be 30%.

      So what is the probability of missing two free throws in a row? Well, 30% of the times that you go to attempt a free throw, you will miss. And, of those times, 30% of the times when you go to shoot again, you will miss a second time. Thus, the probability of missing twice in a row is 30% * 30%, or 0.3 * 0.3, which equals 0.09 (or 9%).

      The same logic can be applied to 6 misses. The probability of you missing 6 times in a row would therefore be 0.3 * 0.3 * 0.3 * 0.3 * 0.3 * 0.3 = 0.3^6

      I hope this helped.
      (20 votes)
  • blobby green style avatar for user Sydney Stratman
    At when i solved for the equation 0.7^0 multiplied 0.3^6 my answer was 0.000729 not 0.001.
    (7 votes)
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  • blobby green style avatar for user Mandy Smith
    Where are you getting "6 choose 2 equals 15"? Why is it not explained? Am i missing something.
    (6 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      In general, n choose k = n! / [k!(n-k)!], where ! denotes factorial, and the factorial of a whole number is the product of the whole numbers from 1 up to that number.

      So 6 choose 2 = 6! / [2!(6 - 2)!] = 6! / (2! * 4!) = (1*2*3*4*5*6)/(1*2*1*2*3*4) = (5*6)/(1*2) = 15.

      Have a blessed, wonderful day!
      (3 votes)
  • piceratops sapling style avatar for user tanwarkml
    In the first video (no. of heads) we only take combinatorics and not the individual probability (p(H) = 0.5 p(T) =0.5). In this video, we take both P(Score)= 0.7 P(MIss)=0.3 and the combinatorics i.e. the different ways in which you could score and miss. What am I missing?
    (3 votes)
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    • blobby green style avatar for user robshowsides
      Good question. When the two events have equal probability, we can just use combinatorics, because every outcome is equally likely. Say we flip a coin six times. Then p(HHHTTT) = p(HHHHHH) = p(TTHHTT) = ... etc = (0.5)^6 = 1/(2^6) = 1/64
      But when the two possibilities have different probabilities, we need to use those probabilities. So if p(S)= 0.7 and p(M)=0.3, then p(SSSSSS) = (0.7)^6, p(SMSMSM) = (0.7)^3 * (0.3)^3, p(SSMMSS) = (0.7)^4 * (0.3)^2. So there are still 64 different outcomes of 6 shots, like flipping 6 coins, but they have seven different groups with different probabilities that can be collected -- groups with 0,1,2,3,4,5, or 6 scores out of 6 shots. So there are 6C4 different ways he could have 4 scores and 2 misses, all with the same probability. If you DID use the probabilities for H and T, you would still get the correct answer, of course.
      (6 votes)
  • blobby green style avatar for user Dhananjay Reddy
    Suppose that a basketball player has a 50% free throw percentage or that I flip an unbiased coin. I am supposing that both have the same premise.

    Suppose that I flip the coin 6 times and want the probability of 3 heads in 6 flips.

    Using the formula:

    Probability = # of total possible winning outcomes / # of total outcomes

    or

    Probability = "n Choose k" X (f^k) X (1-f)^(n-k) as seen in this video.

    In my example:
    n=6
    k=3
    f=0.5 (heads)

    Solving P(X=0) ... P(X=6), I expect a discrete probability distribution.

    Since the probability of getting heads is exactly 50%, I would expect P(X=3) = 50%, P(X=2) and P(X=4) to be smaller,
    P(X=1) and P(X=5) to be even smaller, and P(X=0) and P(X=6) to be even smaller.

    But that's not how it works out.

    P(X=0) = "6 Choose 0" X (0.5)^0 X (0.5)^6 = (6!/0!6!) X (0.5)^6 = 1 X 0.015625 = 0.015625
    P(X=1) = "6 Choose 1" X (0.5)^1 X (0.5)^5 = (6!/1!5!) X (0.5)^6 = 6 X 0.015625 = 0.09375
    P(X=2) = "6 Choose 2" X (0.5)^2 X (0.5)^4 = (6!/2!4!) X (0.5)^6 = 15 X 0.015625 = 0.234375
    P(X=3) = "6 Choose 3" X (0.5)^3 X (0.5)^3 = (6!/3!3!) X (0.5)^6 = 20 X 0.015625 = 0.3125 <-------
    P(X=4) = "6 Choose 4" X (0.5)^4 X (0.5)^2 = (6!/4!2!) X (0.5)^6 = 15 X 0.015625 = 0.234375
    P(X=5) = "6 Choose 5" X (0.5)^5 X (0.5)^1 = (6!/5!1!) X (0.5)^6 = 6 X 0.015625 = 0.09375
    P(X=6) = "6 Choose 6" X (0.5)^6 X (0.5)^0 = (6!/6!0!) X (0.5)^6 = 1 X 0.015625 = 0.015625

    Where's my mistake?

    Intuitively, it makes sense that 3 heads in a row or 3 tails in a row should have a percentage much lower than 50%: (0.5)(0.5)(0.5)=0.125 or 12.5%.

    But I want 3 heads (half of 6) in any order when flipping the coin 6 times. This should mirror the coin's natural tendency to come up heads 50% and tails 50% of the time.
    (3 votes)
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    • leaf blue style avatar for user Dr C
      All of the probabilities you calculated are correct. Your only mistake is:

      > "Since the probability of getting heads is exactly 50%, I would expect P(X=3) = 50%"

      Can you explain why you think that this should be the case?

      Taking a look at the probabilities, we see that P(X=3) is the largest, which aligns with our expectation based on the known probability of 50%. But just because the probability of Heads on a given flip is 50%, does not mean that for 6 flips P(X=3) should be 50% as well.

      This is actually a pretty good example of why it is important to study probability: because we humans often have bad intuition about how probability works.

      Maybe making tree diagrams will help to understand why your intuition was mistaken. Draw a full tree diagram for 1, 2, 3, and 4 flips. For each tree, determine the probability of the most likely value (in the case of 6 flips, this would be 3 heads/3 tails).
      (2 votes)
  • blobby green style avatar for user daddis2UIC.edu
    How are you calculating 6 choose 0 = 1 and 6 choose 6 = 1? This is where I am confused.
    (2 votes)
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  • blobby green style avatar for user Ankit
    But then why do we call this a binomial distribution if the distribution of P(X) is not symmetric. The probability distribution graph wont be symmetric in this case. So why?
    (1 vote)
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  • aqualine ultimate style avatar for user Tyler
    Doesn't each attempt affect the probability of the next? Since the probability is just an average or something like that. I feel like the 70% & 30% should change each attempt. (since this isn't something like a coin toss)
    (2 votes)
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  • aqualine ultimate style avatar for user Cecilie
    But how can I calculate 120c3? It´s easy when it´s only 6c0,1,2,3 and so on, but what do I do when the number is 120c3? My problem is this: 120c3*0,05^3*0.95*117...
    (1 vote)
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    • blobby green style avatar for user robshowsides
      120C3 = (120!)/(117! * 3!), right? Think about what that would look like if you wrote it out:
      (120*119*118*117*116*115*...3*2*1)/((117*116*115...*3*2*1)(3*2*1)), right? Can you see that lots and lots of terms will cancel? Not much will be left after you cancel...
      (3 votes)
  • piceratops sapling style avatar for user Akilash
    A calculator claims it can randomly and independently generate a digit from 0-9. For any four digits generated, the probability of 2 zeros is 0.03. Is the calculators claim correct? Show your working.

    Im no sure how to exactly do this one :/
    (1 vote)
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    • leaf yellow style avatar for user Howard Bradley
      You're in the right section: binomial probability.
      You need to use binomial expansion to work out the probability of two out of four zeroes. Where, if the claim is correct, the probability of a zero is 0.1
      You can use a formula very similar to the one Sal uses in this video:
      P(exactly k zeroes in n digits) = 𝑛C𝑘 · fᵏ · (1-f)ⁿ⁻ᵏ, where f is the 0.1 above.

      Spoiler alert: the calculator's (manufacturer's) claims are not correct.
      (2 votes)

Video transcript

- [Voiceover] Now that we've spent a couple of videos exploring a scenario where I'm taking multiple free throws and figuring out the probability of making K of the scores and six attempts or in N attempts. Let's actually define a random variable using this scenario and see if we can construct it's probability distribution and we'll actually see that it's a binomial distribution. So, let's define the random variable X. So, let's say that X is equal to the number, the number of made shots, number of made free throws when taking, when taking six free throws. So, it's how many of the six do you make? And we're going to assume what we assumed in the first video in this series of this, these making free throws. So, we're gonna assume the seventy percent free throw probability right over here. So, assuming assumptions, assuming seventy percent free throw, free throw percentage. All right. So, let's figure out the probabilities of the different values that X could actually take on. So, let's see, what is the probability, what is the probability that X is equal to zero? That even though you have a seventy percent free throw percentage that you make none of the shots and actually you could calculate this through probably some common sense without using all of these fancy things but just to make things consistent, I'm gonna write it out. So, this is going to be, this is going to be, it's going to be equal to six choose zero times zero point seven to the zeroth power times zero point three to the sixth power, and this right over here is gonna end up being one. This over here's going to end up being one. And so, you're just gonna be left with zero point three to the sixth power and I calculated it ahead of time. So, if we round to the nearest, if we do, if we round our percentages to the nearest tenth, this is going to give you approximately, approximately, well, if we round the decimal to the nearest, to the nearest thousandth you're gonna get something like that which is approximately equal to zero point one percent chance of you missing all of them. So, roughly I'm speaking, roughly here a one in a thousand, one in a thousand chance of that happening, of missing all six free throws. Now, let's keep going, this is fun. So, what is the probability that our random variable is equal to 1? Well, this is going to be six choose one times zero point seven to the first power times zero point three to the six minus first power. So, it's a fifth power. And I calculated this and this is approximately zero point zero one or we could say one percent. So, it's still a fairly low probability. Ten times more likely than this roughly but still a fairly low probability. Let's keep going. So, the probability that x is equal to two, well, that's what our first video was essentially. So, this is going to be six choose two times zero point seven squared times zero, zero point three to the fourth power, and we saw that this is approximately going to be zero, zero point zero six, or we could say six percent, and obviously you could type these things in a calculator and get a much more precise answer but just for the sake of just getting a sense of what these probabilities all look like, that's why I'm giving these rough estimates. Kind of, I guess you could say, to the closest, maybe, tenth of a, tenth of a percent. And actually and if you round to the closest tenth of a percent you actually actually get to six point oh percent and this is one point, one point oh percent 'cause this we actually went to a tenth of a percent here but let's keep going. We're obviously going to have to do a few more of these. So, let me just make sure I have enough real estate. All right, so, the probability that our random variable is equal to three is going to be six choosethree and I'm sure you could probably fill this out on your own but I'm going to do it. Zero point seven to the third power times zero point three to the six minus three which is a third power which is approximately equal to, well, it's going to be zero point one eight five or eighteen point five, eighteen point five percent. So, yeah, that's definitely within the realm of possibility. I mean, all of these are in the realm of possibility but it's starting to be a non-insignificant probability. So, now, let's do the probability that our random variable is equal to four. Well, it's going to be six choose four times zero point seven to the fourth power times zero point three to the six minus four or second power which is equal to, this is going to get equal to or approximately, 'cause I am, I am taking away a little bit of the precision when I write these things down. Zero point three two four. So, approximately thirty two point four percent chance of making exactly four out of the six free throws. All right, two more to go. Let's see, I have not used purple as yet. So, the probability that a random variable is equal to five it's gonna be six choose five or zero, at times I should say, zero point seven to the fifth power times zero point three to the first power and that is going to be roughly, roughly zero point three zero three which is thirty point three percent. That's interesting, one more left. So, the probability that I make all of them, of all six, is going to be equal to, is equal to six choose six and zero point seven to the sixth power times zero point three to the zeroth power which is, this right over here is going to be one, this is going to be one, so it's really just zero point seven to the sixth, to the sixth power and this is approximately zero point one one eight. I calculated that ahead of time, which is eleven point eight. Eleven point eight percent. And so, there's something interesting that's going on here. The first time we looked binomial distribution we said, "Hey, there's a symmetry as we kind of "got to some type of a peak and went down, "but I don't see that symmetry here." And the reason you're not seeing that symmetry is that you are more likely to make a free throw than not. It was you have a seventy percent free throw probability. This is no longer just flipping a fair coin. Where you will see the symmetry is in these coefficients. If you calculate these coefficients, six choose zero is one. Six choose six is one. You would see that six choose one is six and six choose five is six. You'd see six choose two is fifteen and six choose four is also fifteen, and then six choose three is twenty. So, you definitely see, you definitely see the symmetry in the coefficients but then these things are weighted by the fact that you're more likely to make something than miss something. If these were both point five then you would also see the symmetry right over here. And you can plot this to essentially visualize what the probability distribution looks like for this example and I encourage you to do that. To take these different cases, just like we did in that first example with the fair coin, and plot these. But this essentially does give you the probability distribution for, for the random variable in question. This is, I just wrote it out instead of just visualizing it but it says, "Okay, well, what's the... "These are the different values "that this random variable can take on." It can't, it can't take on negative one or it can't be fifteen point five or pi or one million. These are the only seven values that this random variable can take on and I've just given you the probabilities, or I guess you could say the rough probabilities of the random variable taking on each of those, each of those seven values.