I need help with this. Consider 5 trials. You can have success or failure. It's binomial distribution. And the events are independent.
The probability of having 2 successes is equal to the probability that the third and fifth trials are successes? Or not? and why? Thank you.
Sorry, I still do not get something.
First, my prof. said that P( 3rd and 5th)= p^2 * (1-p) ^3 and she said the computation does not reflect the order. So we have used a different formula.
Why is that?
Second, she said that the two probability are different but since the events are independent they are equal in this case (indeed we got pretty the same value as a result of the two computations).
Third, you talked about special circumstances. Which are them?
Fourth, I do not get: we said, the events are independent, they all have the same probability to happen. So the probability of success at 3rd trial = probability o success of 5th trial, right? Now, this means the probability of success of 3rd AND 5th is equal to the probability of 4th and 5th, as well as 1st AND 2nd...and so on. This means we are not caring of which couples we are taking. This means probability of success of 2 general events, of a general couple over 5 trials must be equal to the probability of succeding EXACTLY at the 3rd and 5th thrial.
This is my reasoning. And I am afraid I am really not understanding.
I seriuosly beg your pardon for so much confusion.
Please, can you intervene on my reasonment? Thank you so much.
P( 3rd and 5th succeed)= p^2
P( 3rd and 5th succeed and 1st, 2nd and 4th fail) = p^2 * (1-p) ^3
If the latter is what you mean then you or your teacher should be more
careful about what you ask.
With your teachers interpretation of the question this is not possible this is aSecond, she said that the two probability are different but since the events are independent they are equal in this case (indeed we got pretty the same value as a result of the two computations).
claim that:
5! / (3! 2!) = 1
which is does not.
In the case of my interpretation that would be if:Third, you talked about special circumstances. Which are them?
5! / (3! 2!) = (1-p)^2
(which is not possible for real p in [0,1], but does have solutions in theory)
No, that the probability of success on the i'th and j'th trials i not equal jFourth, I do not get: we said, the events are independent, they all have the same probability to happen. So the probability of success at 3rd trial = probability o success of 5th trial, right? Now, this means the probability of success of 3rd AND 5th is equal to the probability of 4th and 5th, as well as 1st AND 2nd...and so on. This means we are not caring of which couples we are taking. This means probability of success of 2 general events, of a general couple over 5 trials must be equal to the probability of succeding EXACTLY at the 3rd and 5th thrial.
This is my reasoning. And I am afraid I am really not understanding.
being independent of the particular i and j does not imply that this is the
probability of exactly two successes, as this latter is the sum of all possible
combinations of i and j.
I seriuosly beg your pardon for so much confusion.
Please, can you intervene on my reasonment? Thank you so much.
Ah, yeah, I think I see: having the 3rd and 5th as success does not specify what is happening to the other trials, right. It is not like saying exactly 2 and only 2 successes. It is diferent.
Provided I got it right (I am so bad with this stuff), it feels way better now. Thank you Captain. You always help me out with probability. Thanks.
How can I do to get into probability Captain? We are doing it in my statistic course. Let me tell you our program over probability.
" Discrete random variable review: probability function, cumulative function,
expected value, variance.
Expected value and variance of linear transformations and standardized
variables.
Discrete random variables distributions: Uniform, Bernoulli and Binomial
Poisson Distribution review.
Continuous random variables: density function, cumulative function,
expected value and variance .
Continuous Uniform distribution.
Expected value and variance of linear transformations and standardized
variables.
Normal Distribution, Standard Normal Distribution and Normal Probability
Plot.
Normal Distribution approximation for Binomial Distribution
Jointly distributed Discrete Random Variables: joint, marginal and
conditional probabilities. Independent Random Variables.
Covariance and correlation..
Sums and differences of discrete random variables (Example: Portfolio
valuation).
Jointly distributed Continuous Random Variables: joint, marginal and
conditional probabilities.
Covariance and correlation..
Sums and differences of continuous random variables "
Can you advice me what can I do? I really need to understand it properly.
I so much wished I could be as good as you with this stuff.Thanks you again, CaptainBlack.
Oh no, no, no! I think I got it now..
You mean those probabilities are different because:
if I tell you
-"tell me the probability that the third and fifth out of 5 trials are successes"
you anwer me " Okay. just the 3rd and fifth"
-"yes"
-"So it is just ONE case, you want that? so you have the probability of one possible cpmbinations(out of all that are possibile) (all combinattions having the same likelyhood, the same probability"
Good, then I tell you- "tell me the prob that out of 5 trials you have 2 successes"
you will tell me- " Well, there are x combinations that will give you 2 successes, each combination is equaly likely. You do not care which you are taking. So clearly let's multiply all the times we are lucky(=all the possible combinations that makes us succees) times each combination probability. And there you go."
Yes, yes, it should be like this. Am I right Captain, am I right?? Have I got it now?