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Statistics and probability
Course: Statistics and probability > Unit 9
Lesson 1: Discrete random variables Random variables
 Discrete and continuous random variables
 Constructing a probability distribution for random variable
 Constructing probability distributions
 Probability models example: frozen yogurt
 Probability models
 Valid discrete probability distribution examples
 Probability with discrete random variable example
 Probability with discrete random variables
 Mean (expected value) of a discrete random variable
 Expected value
 Mean (expected value) of a discrete random variable
 Expected value (basic)
 Variance and standard deviation of a discrete random variable
 Standard deviation of a discrete random variable
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Constructing a probability distribution for random variable
AP.STATS:
VAR‑5 (EU)
, VAR‑5.A (LO)
, VAR‑5.A.1 (EK)
, VAR‑5.A.2 (EK)
, VAR‑5.A.3 (EK)
CCSS.Math: Sal breaks down how to create the probability distribution of the number of "heads" after 3 flips of a fair coin. Created by Sal Khan.
Want to join the conversation?
 At2:45how can P(X=2) = 3/8. when there are also 2 H in HHH? Why the HHH is not considered for P(X=2)?(12 votes)
 When we say X=2, we mean exactly 2. If we wanted to include HHH, we'd say P( X ≥ 2 ).(46 votes)
 How would you find the probablility when your have P(5)(7 votes)
 I agree, it is impossible to have 5 heads in a coin toss occurring only three times but if you were to have to flip a coin 5 times and finding out the number of times it is heads your answer would be:
p=(X=1/32) because HHHHH is the only answer for 5 heads in a coin toss that occurs five times.
In this situation, Master Salman is doing a coin toss only three times.
So there is no probability distribution for 5 heads because that is impossible.
Thank you!(6 votes)
 Am I seeing potential pattern or connection between pascals triangle and the probability of flipping 1, 2 , or three heads 3 at3:12? Can that be used for anything in probability in the future? Or is this just a coincidence? It seems like it still works for the possibility of 4 coins (1 possibility of all tails, then 4 of 3 tails, six of 2 tails, 4 of 1 tail, 1 of no tails)(15 votes)
 A man has three job interviews. The probability of getting the first interview is .3 the second .4 and third .5 suppose the man stops interviewing after he gets a job offer. Construct a probability distribution for X. I assumed due to the probabilities not adding exactly to one that it can't be done. Any help?(0 votes)
 It may help to draw a tree diagram:
Interview 1


 
Get it Not get it
0.3 0.7

Interview 2


 
Get it Not get it
0.4 0.6

Interview 3


 
Get it Not get it
0.5 0.5
There are four leaves on this tree. Assuming that the jobs are independent, then:
Get job 1: probability = 0.3 = 0.3
Get job 2: probability = 0.7 × 0.4 = 0.28
Get job 3: probability = 0.7 × 0.6 × 0.5 = 0.21
Get no job: probability = 0.7 × 0.6 × 0.5 = 0.21
And these add up to 1.(23 votes)
 what aren't HHT and THH considered the same thing? is it the order that differentiates the two?(4 votes)
 They are considered two different outcomes.
Think of it as of flip of three coins of different value. Each is individual and shows tails or heads so you can tell which coin is tails or heads.
There are 8 different outcomes: heads/tails*head/tails*head/tails.
If HHT, THH and HTH are considered the same thing P(HTT and THH and HTH) = 1/8.
Then P(HTT and TTH and THT) = 1/8. P(HHH)=1/8, P(TTT) = 1/8.
Sum of this possibilities is 4/8. But we know, that the sum of all the possibilities of an event must equal 1. So you see there's flaw in this logic in terms of probability.(7 votes)
 At3:31Sal says 'You can have probability larger than 1", how is that possible?(3 votes)
 If you check the transcript, he is actually saying "You can't have a probability larger than 1". Your intuition was correct, the largest probability comes when every element of the sample space meets the criterion, and that's a probability of 1.(7 votes)
 If for example we have a random variable that contains terms like pi or fraction with non recurring decimal values ,will that variable be counted as discrete or continous ? According my understanding eventhough pi has infinte long decimals , it still represents a single value or fraction 22/7 so if random variables has any of multiples of pi , then it should be discrete(3 votes)
 Correct. Discrete vs continuous only considers the number of possible outcomes (more or less), but not what those outcomes are. The values can be irrational, like pi, but if there are distinct multiples it takes, then it's discrete.(5 votes)
 how can we have probability greater than 1?(2 votes)
 We cannot. A probability equal to 1 means certainty, an event with probability equal to 1 is sure to happen, no questions asked, it's impossible to be more certain, and therefore it's impossible to have a probability greater than 1.
If during a problem you end up with a probability greater than 1, then you have to go back because somewhere before that point you have an error in your calculations.(6 votes)
 I can not understand 'Round answers up to the nearest 0.025.' How can I solve this problem?(0 votes)
 It means, every multiple of 0.025 is what you would be rounding to.
i.e., say you get the number 1.548...
Then you would round this number to the nearest multiple of 0.025, which would be 1.55.(1 vote)
 Is there a possibility to calculate the likelihood of an event without visually displaying the outcome? ie.( for 3 coins flip) what mathematical expression can I use to conclude that P(x =2)=3/8 without relying on visual combinations. I do not have a math background , but I would not think to display the outcomes visually to come to this conclusion.(2 votes)
 ...You probably don't need this anymore, but here (because it'll help me study for a test)
Total outcomes for 3 coins flip:
2^3 = 8
To find P(x=2):
HHT (or any other example) > 3(coins) choose 2(heads, which leaves 1 tail) > 3!/(2!)(1!) or 3C2
> = 3
3/(2^3) = 3/8
Another example:
PPOO possible combinations:
4 letters > 4!
For 2 Ps (and 2 Os) > (2!)(2!)
4!/(2!)(2!) or 4C2(2 votes)
Video transcript
Voiceover:Let's say we define the random variable capital X as the number of heads we get after three flips of a fair coin. So given that definition
of a random variable, what we're going to try
and do in this video is think about the
probability distributions. So what is the probability of the different possible outcomes or the different possible values for this random variable. We'll plot them to see how that distribution is spread out amongst those possible outcomes. So let's think about all
of the different values that you could get when
you flip a fair coin three times. So you could get all heads, heads, heads, heads. You could get heads, heads, tails. You could get heads, tails, heads. You could get heads, tails, tails. You could have tails, heads, heads. You could have tails, head, tails. You could have tails, tails, heads. And then you could have all tails. So there's eight equally, when you do the actual experiment there's eight equally
likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about,
what's the probability, there is a situation
where you have zero heads. So what's the probably
that our random variable X is equal to zero? Well, that's this
situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes
gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. And I think that's all of them. So three out of the eight
equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang
of it at this point. What's the probability
that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome
meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random
variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight equally likely outcomes
that meets that constraint. So it's a 1/8 probability. So now we just have to think about how we plot this, to see
how this is distributed. So let me draw... So over here on the vertical axis this will be the probability. Probability. And it's going to be between zero and one. You can't have a
probability larger than one. So just like this. So let's see, if this
is one right over here, and let's see everything here looks like it's in eighths so let's put everything
in terms of eighths. So that's half. This is a fourth. That's a fourth. That's not quite a fourth. This is a fourth right over here. And then we can do it in terms of eighths. So that's a pretty good approximation. And then over here we
can have the outcomes. Outcomes. And so outcomes, I'll say outcomes for alright let's write this so value for X So X could be zero actually let me do those same colors, X could be zero. X could be one. X could be two. X could be equal to two. X could be equal to three. X could be equal to three. So these are the possible values for X. And now we're just going
to plot the probability. The probability that X has
a value of zero is 1/8. That's, I'll make a little bit of a bar right over here that goes up to 1/8. So let draw it like this. So goes up to, so this
is 1/8 right over here. The probability that X equals one is 3/8. So 2/8, 3/8 gets us right over let me do that in the purple color So probability of one, that's 3/8. That's right over there. That's 3/8. So let me draw that bar, draw that bar. And just like that. The probability that X equals two. The probability that X equals two is also 3/8. So that's going to be on the same level. Just like that. And then, the probability
that X equals three well that's 1/8. So it's going to the same
height as this thing over here. I'm using the wrong color. So it's going to look like this. It's going to look like this. And actually let me just write
this a little bit neater. I can write that three. Cut and paste. Move that three a little closer in so that it looks a little bit neater. And I can actually move that
two in actually as well. So cut and paste. So I can move that two. And there you have it! We have made a probability distribution for the random variable X. And the random variable X can only take on these discrete values. It can't take on the value half or the value pi or anything like that. So this, what we've just done here is constructed a discrete
probability distribution. Let me write that down. So this is a discrete, it only, the random variable only takes on discrete values. It can't take on any values
in between these things. So discrete probability. Probability distribution. Distribution for our random variable X.