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### Course: Statistics and probability > Unit 9

Lesson 5: Binomial random variables- Binomial variables
- Recognizing binomial variables
- 10% Rule of assuming "independence" between trials
- Identifying binomial variables
- Binomial distribution
- Visualizing a binomial distribution
- Binomial probability example
- Generalizing k scores in n attempts
- Free throw binomial probability distribution
- Graphing basketball binomial distribution
- Binompdf and binomcdf functions
- Binomial probability (basic)
- Binomial probability formula
- Calculating binomial probability

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# Binomial variables

An introduction to a special class of random variables called binomial random variables.

## Want to join the conversation?

- why is the probability of getting heads 0.6 and not 0.5?(6 votes)
- Not all coins are "fair". An example of an unfair coin is one that has two heads; another might be one that is weighted to fall on heads more frequently than tails.

I believe Sal's point to introducing a coin with P(heads)=0.6 is to start viewers out with the idea of trials having a successful or failed outcome that is easy to understand (lots of videos with flipped coins) and is also in keeping with random variables' outcomes having different probabilities.(15 votes)

- Aren't the probability of success is constant, and being independent trials the same ?(6 votes)
- No.

Say you have 2 coins, and you flip them both (one flip = 1 trial), and then the Random Variable X = # heads after flipping each coin once (2 trials).

However, unlike the example in the video, you have 2 different coins, coin 1 has a 0.6 probability of heads, but coin 2 has a 0.4 probability of heads. This WOULD satisfy the requirement of the trials being independent, but not the requirement of the probability of success being the same for each trial.

Hope this helps!(6 votes)

- What does the condition 'finite number of independent trials' mean?(4 votes)
- This means that the number of trials is finite (instead of infinite), and the results of any trials do not affect the success probabilities for other trials.

Have a blessed, wonderful day!(7 votes)

- It is not true when he says
*"If I get a king that looks like that would be a success. If I don't get a king that would be a failure. So it seems to meet that right over there."*because there is chances that both the cards that he picks up are kings. There aren't two possible outcomes there are three possible outcomes:

⚫ 0 kings are picked

⚫ 1 king is picked

⚫ 2 kings are picked(2 votes)- I was thinking the same while watching the video. For me, the way the random variable was defined does not satisfy that condition. But I guess we can always define a success event as either getting a king or two kings, and the other two outputs as failure.(2 votes)

- Should Y be considered a binomial variable (even without replacement) beause of the "10% Rule" for assuming independence between trials? There is a related video where the 10% rule is explained.(3 votes)
- Even with the "10% Rule" for assuming independence between trials, Y cannot be considered a binomial variable without replacement because the probability of success (drawing a king) on each trial is not constant and the trials are not independent. The probability of success on the second trial depends on the outcome of the first trial due to the lack of replacement. Without replacement, the probability of drawing a king changes for each trial, violating the conditions for a binomial variable.(1 vote)

- Doesn't
`Point1`

**imply**`Point4`

? Or am I missing out on some kind of explicit nuance?

For example, in Sal's first coin example, isn't the possibility of someone replacing coins*(which will affect the constancy of the probability of each trial)*a form of dependence?(2 votes)- Same as what I replied to Mohamed, No.

Say you have 2 coins, and you flip them both (one flip = 1 trial), and then the Random Variable X = # heads after flipping each coin once (2 trials).

However, unlike the example in the video, you have 2 different coins, coin 1 has a 0.6 probability of heads, but coin 2 has a 0.4 probability of heads. This WOULD satisfy the requirement of the trials being independent, but not the requirement of the probability of success being the same for each trial.

Hope this helps!(2 votes)

- For a variable to be binomial, does each trial need to have the same probability of success, or is it enough that each trial has an independent constant probability of success? In other words, if I flip two coins in succession, the first one fair and the second one unfair, would that qualify as a binomial variable or not?(2 votes)
- The probability of success must be the same for each trial.(2 votes)

- What's the intuition behind it needing to have a fixed number of trials?(2 votes)
- The requirement of a fixed number of trials ensures that the binomial variable has a well-defined number of events over which the probability distribution can be calculated. This condition allows for the application of specific formulas and statistical methods tailored to binomial variables, providing a clear framework for analysis.(1 vote)

- Why not introduce bernoulli distribution first? At6:34, something seems wrong with the 1st point (made up of indepentant trials) P(K on 1st trial) = 4/52 and P(K on 2nd trial) = 4/52 * 3/51 + 48/52*4/51 => 4/52. It is the same.(1 vote)
- Introducing the Bernoulli distribution before discussing binomial variables could indeed provide a helpful foundation. The Bernoulli distribution deals with the probability of success or failure in a single trial, which is a fundamental concept underlying binomial variables. Understanding the Bernoulli distribution can help clarify the conditions and characteristics of binomial variables. Regarding the calculation for variable Y, the probability of drawing a king on the second trial does depend on the outcome of the first trial, as there is no replacement. Therefore, the probability calculation for Y without replacement differs from the case with replacement.(1 vote)

- It seems to be same as the multiplication rule of probability.(1 vote)
- Yes, the calculation of the probability for variable Y without replacement resembles the multiplication rule of probability. In this case, the probability of drawing a king on the second trial depends on whether a king was drawn on the first trial, leading to a conditional probability calculation.(1 vote)

## Video transcript

- What we're going to do in this video is talk about a special
class of random variables known as binomial variables. And as we will see as we build
up our understanding of them, not only are they interesting
in their own right, but there's a lot of very powerful probability and statistics that we can do based on our understanding of binomial variables. So to make things concrete
as quickly as possible, I'll start with a very tangible example of a binomial variable and then we'll think a
little bit more abstractly abut what makes it binomial. So let's say that I have a coin. This is my coin here. It doesn't even have to be a fair coin. Let me just draw this really fast. So that's my coin. And let's say on a
given flip of that coin, the probability that I get
heads is zero point six and the probability that I'd get tails, well it'd be one minus zero point six or zero point four. And what I'm going to do
is I'm going to define a random variable X as being equal to the number of heads after after ten flips of my coin. Now, what makes this a binomial variable? Well, one of the first conditions that's often given for a binomial variable is that it's made up of a finite number of independent trials. So, it's made up made up of independent independent trials. Now, what do I mean by independent trials? Well, a trial is each flip of my coin. So a flip is equal to a trial in the language of this
statement that I just made. And what do I mean by each flip or each trial being independent? Well the probability of
whether I get heads or tails on each flip are independent of whether
I just got heads or tails on some previous flip. So, in this case, we are made up of independent trials. Now, another condition is each trial can be clearly classified as either a success or failure. Or another way of thinking about it: Each trial clearly has one
of two discreet outcomes. So each trial, and the example I'm giving, the flip is a trial, can be classified classified as either success or failure. So, in the context of
this random variable X, we could define heads as a success because that's what we
are happening to count up. And so you're either going
to have success or failure. You're either going to have heads or tails on each of these trials. Now another condition for
being a binomial variable is that you have a fixed number of trials. Fixed number of trials. So in this case, we're saying that we have ten trials, ten flips of our coin. And then the last condition
is the probability of success, in this context success is a heads, on each trial, each trial, is constant. And we've already talked about it. On each trial on each flip, the probability of heads is going to stay at zero point six. If for some reason that were to change from trial to trial, maybe if you were to swap the coin and each coin had a different probability then this would no longer
be a binomial variable. And so you might say, "Okay, that's reasonable, I get why this is a binomial variable. Can you give me an example of something that is not
a binomial variable?" Well let's say that I were
to define the variable Y and it's equal to the number of kings after taking two cards from a standard deck of cards. Standard deck. Without replacement. Without replacement. So you might immediately say, "Well, this feels like
it could be binomial." We have...each trial can be classified as either a success or failure. Each trial is when I take a card out. If I get a king that looks
like that would be a success. If I don't get a king
that would be a failure. So it seems to meet that right over there. It has a fixed number of trials. I'm taking two cards out of the deck so it seems to meet that. But what about these conditions that it's made up of independent trials or that the probability
of success on each trial is constant? Well, if I get a king the probability of king
on the first trial, probability I say king on first trial would be equal to, well, out of a deck of 52 cards, you're going to have four kings in it. So the probability of a
king on the first trial would be four out of 52. But what about the
probability of getting a king on the second on the second trial? What would this be equal to? Well, it depends on what
happened on the first trial. If the first trial you had a king, well then you would have, so let's see, this would be the situation given first trial, first king, well now there would be three kings left in a deck of 51 cards. But if you did not get a
king on the first trial, now you have four kings
in a deck of 51 cards because, remember, we're
doing it without replacement. You're just taking that
first card, whatever you did, and you're taking it aside. So what's interesting here is this is not made up of independent trials. It does not meet this condition. The probability on your second trial is dependent on what
happens on your first trial. And another way to think about it is because we aren't replacing
each card that we're picking, the probability of success on each trial also is not constant. And so that's why this right over here is not a binomial variable. Now, if Y, if we got rid of without replacement and if we said we did replace
every card after we picked it then things would be different. Then we actually would be
looking at a binomial variable. So instead of without replacement if I just said with replacement, well then your probability
of a king on each trial is going to be four out of 52. You have a finite number of trials. You're probability of success
is going to stay constant and they would be independent. And obviously each trial
could easily be classified as either a success or a failure.