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Fon Khe following probability distribution function I = X=123 IMenk=...

Question

Fon Khe following probability distribution function I = X=123 IMenk=

Fon Khe following probability distribution function I = X=123 I Menk=



Answers

Determine the cumulative distribution function of a binomial random variable with $n=3$ and $p=1 / 2$.

So the whole theme in this problem is that your expected value? You know, I'll straight X. And P. Is this is the probability that you'll land on zero or you'll get zero points Maybe like you're playing a dark game, uh if you land on a certain spot is 50% of the time, you get zero points And then you're 20% of the time, you get one And then 20% of the time you get to. And then but The remaining 10% music would land on three. So the whole premise of the expected value of getting that is multiplying. So you would expect to get zero points 50% of the time. You expect to get 1.20% of the time And then uh two points 20% of the time and then three points 10% of the time. This is how you approach the problem. And this isn't too bad because zero times anything is just that thing and one times anything is just that thing. So 0.2 plus 0.4 plus 0.3, I would expect to get just ran through it at the board or whatever game it is Point to .4.6 plus .3, That's how you get the answer.

With this expected value, I don't think you need a calculator but calculate would certainly be useful And I just like to think of the outcome with the probability, so the probability of getting a negative five not 0.5 mess up is 0.2. So you have a 20% chance of getting negative five -1 as a 30% chance of .3. A score of zero gives me a point to chance Um and two has a .1 chance And it was next five has a point to chance and I have no idea why there's this number over here because it says that there's no chance of getting this Now your probabilities should add up to be 1.23.5 6 7.8.9 .1 0. So yes it does work. So what you need to do is multiply your probabilities and then add them together. So um you don't have to actually show this math because zero times anything is zero but you are technically multiplying anything and I would say the same thing about that last one that Know anything Times 00. So that's improbable. It's impossible actually. So I'm just double checking my math that makes five times two is negative one, This would be -3. This will be a point to And then five transpoint 1.2 would be one as well. So as you simplify this, I don't know how you would simplify but you don't need a calculator. Um Those negative one and positive one to cancel negative 10.3 plus 0.2 would give me an expected value of negative 0.1 Yeah.

This problem. We have been given a table, and we want to determine whether the distribution is in fact, a discrete probability distributions. If it's not, we need to determine why. As you remember, there are two things that a probability distribution must follow. And the first is that each probability is greater than equal zero, which means that the probabilities have to be non negative. The second is that the some of the probabilities most people want now start with the first condition there. Look at all of your probability. So look at everything that falls underneath The p of X value in that truck have a 0000 and then one. All of those are greater than or equal to zero. That's what means that holds true. Next, you need to check. Do they add up to one? So 2000 Yeah, plus zero plus one. Does that equal one? And yes, it does. And so it satisfies both of these two conditions. And so that tells us Yes, it is a discrete probability distribution function. Yeah,

Yeah, this problem. We would like to determine if the table that we have been given represents a discreet probability distribution function. It's important to remember whenever we're looking at the is that there are two conditions that must be satisfied. The first is that each probability must be greater than or equal to zero. In other words, you can have a negative probability. The second is that the some of the probabilities must equal one. So these two conditions must be satisfied. In order for this to be a probability distribution function, the first is very easy to check. You just check your table here and make sure that none of the values are negative. And since they're all 0.2 and tell that they're none of them are negative, so the first part is satisfied now for part two, we have 0.2 plus 0.2 was zero point five times Those are all of our probabilities. These add to one which is exactly what they were supposed to do. They were supposed to add up to one, and so that tells us that our second condition is also satisfied. Both conditions are satisfied. And so, yes, it is a probability distribution for a discreet and very


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