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Math Help - probability

  1. #1
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    probability

    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.
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  2. #2
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    Quote Originally Posted by 0123 View Post
    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.
    P_1 = p(2 from 5) = [5!/(3!)(2!)]p^2 (1-p)^3

    P_2 = p(trial 3 and trial 5) = p^2

    so in general P_1 != P_2.

    RonL
    Last edited by CaptainBlack; October 13th 2007 at 01:59 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    P_1 = p(2 from 5) = [5!/(3!)(2!)]p^2 (1-p)^3

    P_2 = p(trial 5 and trial 6) = p^2

    so in general P_1 != P_2.

    RonL
    why p( trial 5 and trial 6) ? the trials are always those 5. Might you explain in words as far as possible? Because I am not understanding.

    Thanks a lot, Captain.
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    Quote Originally Posted by 0123 View Post
    why p( trial 5 and trial 6) ? the trials are always those 5. Might you explain in words as far as possible? Because I am not understanding.

    Thanks a lot, Captain.
    It was a type, it should have read trial 3 and trial 5. But that does not alter
    the probability as it does not matter which two particular trials you specify.

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    It was a type, it should have read trial 3 and trial 5. But that does not alter
    the probability as it does not matter which two particular trials you specify.

    RonL
    And why it does not matter? And what does that " ! " next to the equal sign mean? Is the factorial? And so P_1 and P_2 are different? Might you explain me why without using formulas, I mean, I cannot grasp the intuition... Thank for your patience, Captain.
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    Quote Originally Posted by 0123 View Post
    And why it does not matter?
    Each trial is independent of the others and with equal probabilty of a
    favourable outcome (this is the underlying Bernouli trial assumption that
    is the basis for a binomial distribution).

    RonL
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  7. #7
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    Quote Originally Posted by 0123 View Post
    And what does that " ! " next to the equal sign mean? Is the factorial?
    In ASCII there ar two common methods of typing "not equals" these are "<>"
    and "!=", so:

    a <> b, and a != b

    both mean a is not equal to b.

    RonL
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    Quote Originally Posted by 0123 View Post
    And so P_1 and P_2 are different? Might you explain me why without using formulas, I mean, I cannot grasp the intuition...
    The two expressions are of different forms, and in this case would only be
    equal under special circumstances. To see that they are different just put
    p=0.5 and calculate both.

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    The two expressions are of different forms, and in this case would only be
    equal under special circumstances. To see that they are different just put
    p=0.5 and calculate both.

    RonL
    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.
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  10. #10
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    To avoid it sinks.
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  11. #11
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    Quote Originally Posted by 0123 View Post
    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?
    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.

    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).
    With your teachers interpretation of the question this is not possible this is a
    claim that:

    5! / (3! 2!) = 1

    which is does not.

    Third, you talked about special circumstances. Which are them?
    In the case of my interpretation that would be if:

    5! / (3! 2!) = (1-p)^2

    (which is not possible for real p in [0,1], but does have solutions in theory)

    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.
    No, that the probability of success on the i'th and j'th trials i not equal j
    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.
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    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.
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  13. #13
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    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?
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  14. #14
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    Quote Originally Posted by 0123 View Post
    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?
    I think so

    RonL
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