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Current time:0:00Total duration:6:42

AP Stats: UNC‑3 (EU), UNC‑3.A (LO), UNC‑3.A.2 (EK)

- [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.

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