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# Recognizing binomial variables

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

## Video transcript

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 going to do it by looking at a few exercises from Khan Academy so this has a manager over she's 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 of 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 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 hats 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 1 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 eleven twentieths probability of success but in trial two what's the probability of success given success success in I'll say trial 1 t1 well if you succeeded in trial 1 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 1 then you have then it will 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 it 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 binomial variable lit 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 and a game involving a standard deck of 52 playing cards an individual randomly draws seven cards without a 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 us on a given trial when your wear 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 decks so the the trials in this case are not independent not in the pendent 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 in two independent trials the Trott the probability of getting an ace on each trial would be the same but not when you have without replacement so this is not not binomial right over here because you don't have independent trials the second scenario 60% of a certain species of tomato live 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 plant lives or dies doesn't affect whether the plant next to it lives or dies so each trial is independent of the others there's 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 as a trial success is getting a double which has a fixed probability whether or not you get a double on each trial is going to 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 you are 200 times or or or who knows however many times until they roll a double and so this violates there's a fixed number of trials so since there is not a fixed number of trials this variable X naught by no Meah
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