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Recognizing binomial variables

AP.STATS:
UNC‑3 (EU)
,
UNC‑3.A (LO)
,
UNC‑3.A.2 (EK)
Some examples and counterexamples of binomial variables.

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  • piceratops ultimate style avatar for user Shashank
    How is each trial being classified as a success or failure?
    (2 votes)
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    • hopper jumping style avatar for user Yuya Fujikawa
      "Success" in this context just means "result wanted". So picking a female's name is a success and picking a male's name is not successful, since X is the random variable that represents # of females chosen. So female = result wanted= success for this very specific problem.

      EDIT:
      I said "Success" = "result wanted" but I should've said, really, the result being sought (I guess I could say), which is one of the possible outcome out of TWO possible outcomes.
      (10 votes)
  • blobby green style avatar for user ju lee
    with the card problem, if taken into consideration of the 10% rule, we can consider the problem as a binomial variable, right?
    (4 votes)
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  • blobby green style avatar for user anushkasingh1412
    I have a doubt in the plant example i.e, I dont understand how the success of each trial is constant. For instance, if we have 100 plants then 60 of them would live, now if out of the 16 plants, 15 of them lived then we would feel less optimistic about the last one living as well right? As most of the healthy plants have already been taken? So how is the success constant here? I am probably missing something, but I am really confused.
    (2 votes)
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    • cacteye blue style avatar for user Jerry Nilsson
      The plants live independently of each other.
      This means that the state of the first 15 plants (alive or not) doesn't affect the probability of the 16th plant being alive.

      Compare it to flipping coins. Just because you just landed 15 Heads in a row doesn't mean that the 16th coin has a greater chance of landing Tails.
      (2 votes)
  • leaf green style avatar for user Amandeep Singh
    I could not understand the second question.
    (2 votes)
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  • blobby green style avatar for user shrivastavashriyam2
    In the dice example isn't it that we also cannot define success or failure until second trail??
    (1 vote)
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    • leaf yellow style avatar for user Timotheos
      Although not explicitly stated in the problem, the wording implies that each trial consists of rolling both dice ("rolling a double"). Also otherwise, success would classify as rolling two consecutive equal values, which is not what the question was about.
      (2 votes)
  • blobby green style avatar for user Mez Cooper
    Plants live independently before or after being moved?
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      We have to assume that the plants live independently both before and after they are moved.

      If they don't live independently before they're moved, then it could be – for example – that a plant with very long roots steals nutrition from neighboring plants, thus making them weaker, and affecting their probability to survive the move.

      If the plants don't live independently after the move, then perhaps vermin can spread from one plant to another, again altering neighboring plants' chance of survival.
      (2 votes)
  • boggle blue style avatar for user Bryan
    In the third scenario at , isn't it also true that you can't really classify X as a success or a failure?
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      𝑋 is the number of rolls until we roll a double, which ranges from 1 to infinity, so 𝑋 can not be classified as success or failure.

      However, the trials that determine 𝑋 can each be classified as success or failure. Either you roll a double (success), or you don't (failure).
      (2 votes)
  • blobby green style avatar for user jazzyval05
    Which of the following choices are binomial random variables?
    Choose all answers that apply:
    Choose all answers that apply:

    (Choice A)
    A
    Sadio randomly chooses a student from his class and asks them if they are left-handed. He stops once he finds the first left-handed person. Let
    L
    =
    L=L, equals the number of students it takes.

    (Choice B)
    B
    Amelia rolls
    5
    55 fair dice. Let
    D
    =
    D=D, equals the number of dice that show a value greater than
    3
    33.

    (Choice C)
    C
    Ms. Glover writes the names of her
    30
    3030 students on individual pieces of paper and randomly selects the names of
    5
    55 students from a hat. Let
    M
    =
    M=M, equals the mean number of letters in the first names of the students.
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      In choice A Sadio keeps sampling students until he finds a student that is left-handed, so the number of trials is not fixed and thus this is not an example of a binomial random variable.

      – – –

      Choice B is an example of a binomial random variable, because each die has the same probability of success (showing a value greater than 3) and there's a fixed number of trials.

      – – –

      In choice C the variable doesn't count the number of times a certain event happens, and so it is not a binomial random variable.
      (1 vote)
  • starky ultimate style avatar for user Michele Franzoni
    Why do we have to check if the probability of success is constant? Isn't that always true when the trials are independent?
    Maybe someone can come up with an example of a binomial variable that is made up of independent trials but with the probability of success not constant in each trial.
    (1 vote)
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    • starky seedling style avatar for user deka
      quite an important point!

      1. let's look at this example
      trials = waking up for 2 days in a row
      success = waking up before 6 a.m.
      say p(success) = around 1/2
      in many cases, we can say the days are independent one another, meaning your chance of waking up earlier than 6 in a day doesn't affect that of the other day

      2. but in some rare cases, this may not be true. say you're in another country after having a 13 hour of flight between 2 days of waking up. what about p(success) now? it must be lower than 1/2 after the flight (because of a jet lag and time difference), even though your waking up on 2 days before and after the flight are independent each other

      3. in other words, the independence of each trial isn't enough to guarantee us there must be one and only probability of success and thus we can play with that in the name of binomial variables. because there could be other (outer) factors affecting them. and that's why we feel somehow weird when we read the sentence like "each plant's living chance after transplant is independent" for the second example in video. it's not realistic. they are depending on one another especially in the same garden. and the survival chance may be changing over time. but for the sake of binomial, it needed to be both "independent" and "constant"
      (1 vote)
  • aqualine ultimate style avatar for user Sammi Z.
    At , how is it that there is no fixed number of trials when the random variable states that the X= the number of rolls in ONE turn. Wouldn't one turn be the fixed number of trials?
    (1 vote)
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Video transcript

- [Instructor] What we're going to do in this video is get some practice classifying whether a random variable is a binomial variable, and we're gonna do it by looking at a few exercises from Khan Academy. So this says a manager oversees 11 female employees and nine male employees. They need to pick three of these employees to go on a business trip, so the manager places all 20 names in a hat and chooses at random. Let X equal the number of female employees chosen. So they're going to do three trials. And on each of those trials, you could say success is if they pick a female employee. And then the random variable X is the number of females out of those three. Is X a binomial variable? Why or why not? So pause this video, and see if you can work through this on your own. All right, now let's go through each choice. Choice A says each trial isn't being classified as a success or failure, so X is not a binomial variable. I disagree with this. Each trial is being classified as a success or failure. It's either going to be female or not. And since we're counting the number of female employees, if in a trial, we pick a female, that would be a success, so each trial is being classified as a success or failure. So this one over here isn't true. There is no fixed number of trials, so X is not a binomial variable. There is a fixed number of trials. They're doing three trials. They're picking three names out of a hat. The trials are not independent, so X is not a binomial variable. So this is interesting. So, for example, trial one, what's the probability of success? Well, there are 20 employees, 20 names in the hat, and 11 of the outcomes would be success, so you have 11/20 probability of success. But in trial two, what's the probability of success given success, success in, I'll say trial one, T one? Well, if you succeeded in trial one, that means that there's now only 10 female names in the hat out of 19. And if you don't have success in trial one, then it'll be 11 out of 19. So your probability does change based on previous outcomes. And so the trials are not independent, and so X is not a binomial variable. So this one is true. The trials are not independent, so that violates that condition for being a binomial variable. In order to be a binomial variable, all your trials have to be independent of each other. And so we'd rule this last one out because this last one says that X has a binomial distribution or is or does meet all the conditions for being a binomial variable. Let's do another example. So here we have different scenarios, and I have the conditions for a binomial variable written right over here. And so once again, pause the video, and look at each of these scenarios for random variables. And look at these conditions, and think about whether these random variables are binomial or not. So let's look at the first one. In a game involving a standard deck of 52 playing cards, an individual randomly draws seven cards without replacement. Let Y be equal to the number of aces drawn. Well, in our introductory video to binomial variables, we talked about, if we're doing without replacement, your probability of getting an ace on a given trial, where a trial is you're taking a card out of the deck, it's going to be dependent on whether you got aces in previous trials. Because if you got an ace in a previous trial, well, that ace, then you're gonna have fewer aces in the deck. So the trials, in this case, are not independent, not independent, not independent trials. Now, on the other hand, if on every trial you looked at whatever card you got and put it back in the deck, then they would be independent trials. The probability of getting an ace on each trial would be the same, but not when you have without replacement. So this is not binomial right over here because you don't have independent trials. The second scenario, 60% of a certain species of tomato live after transplanting from pot to garden. Eli transplants 16 of these tomato plants. Assume that the plants live independently of each other. So whether one plant lives isn't dependent on whether another plant lives. Let T equal the number of tomato plants that live. All right, so let's look at the conditions. The outcome of each trial can be classified as either success or failure. So each trial over here is one of the tomato plants, and we have 16 of those trials. And success is if the tomato plant lives, and failure is if it dies. So we have either success or failure. Each trial is independent of the others. They tell us, the plants live independently of each other. So whether or not a neighboring plants lives or dies doesn't affect whether the plant next to it lives or dies. So each trial is independent of the others. There is a fixed number of trials. Yes, we have 16 right over there. The probability p of success on each trial remains constant. Well, yeah, according to at least the scenario, they're saying that we have a 60% chance for each tomato plant, which is each trial. So it meets all of the conditions right over here, so this one is binomial. Now let's look at this third scenario. In a game of luck, a turn consists of a player continuing to roll a pair of six-sided die until they roll a double, two of the same face values. Let X equal the number of rolls in one turn. So you're gonna keep rolling until they roll a double. Well, the thing that jumps out at me is that you don't have a fixed number of trials, not fixed number of trials. You could say each trial, each roll is a trial. Success is getting a double, which has a fixed probability. Whether or not you get a double on each trial is gonna be independent of the previous roll. So it meets all the other constraints, but it does not meet that there's a fixed number of trials. You're gonna keep, someone, there's some chance you might have to roll 20 times or 200 times or who knows however many times until they roll a double. And so this violates there is a fixed number of trials. So since there is not a fixed number of trials, this variable X, not binomial.