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    a question about discrete RV




    hello,
    I have a question about discrete RV.
    Five fair coins are tossed. The coins that fall on "Head" are removed and the rest of them are tossed again. Then we removed again the coins that fall on "Head" and so forth until all the coins are removed.
    What is the probability that after the fifth round not all the coins were removed?
    Thanks a lot,
    Avshi

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    Re: a question about discrete RV

    Where are you having troubles with this question?
    Stop cowardice, ban guns!

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    Re: a question about discrete RV

    I tried to solve it but I couldn't, so I'm also interested in the solution to learn something. I started with

    X(1) = number of coins remained after the first run
    X(2) = number of coins remained after the second run
    ....

    P(X(1) > 0) = P(at least one coin tossed tail)
    = 1 - P(all heads)
    = 1 - 1/32
    P(X(2) > 0) = P(X(1) > 0) * P(at least on coin in the second run is tail)

    But I don't know how to get P(at least on coin in the second run is tail). Or my attempt is wrong in the very beginning. Please advise.

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    Re: a question about discrete RV

    Try to think in the following way:

    Let X_i be the number of tosses the i-th coin have before it is removed.

    What is its distribution? (actually it is not a must to know this answer; it just act as a side question to consolidate the concept)

    How does it relate to your original question?

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    Re: a question about discrete RV

    Let me try.... X(i) is a binomial distribution, Bin(5,0.5).

    Let X(i) be the number of heads after the experiment.

    P(after fifth not all coins are removed) = P(first coin all tail) + ... + P(fifth coin all tail)

    = P(X(1)=0) + ... + P(X(5)=0)

    where X(i) = Bin(5,0.5)

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    Re: a question about discrete RV

    Sorry it is incorrect. Actually X_i is the number of tosses to obtain the first head.

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    Re: a question about discrete RV

    X(i) = number of tosses to obtain the first head

    Probability after the fifth round that not all coins are head = P(X(1)>5) + ... + P(X(5)>5) = 5 * P(X(1)>5) because the tosses are independent

    P(X(1)>5) = 1 - P(X(1)<=5) = P(first toss to get a head) + .... + P(fifth toss to get a head)

    = (1/2) + (1/2*1/2) + ... + (1/2*1/2*1/2*1/2*1/2) = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 = 0.9875

    Therefore, the answer would be 5 * (1-0.9875).

    This looks like a Maclaurin series to me.

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    Re: a question about discrete RV

    Closer but not exact. The X_i I defined is corresponding to a specific coin. They are independent as you said, but the events are not mutually exclusive. Therefore, you cannot break the probability of union into sum like that. Instead you should try to work on the intersection which you can use the independent property.

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    Re: a question about discrete RV

    Quote Originally Posted by BGM View Post
    Try to think in the following way:

    Let X_i be the number of tosses the i-th coin have before it is removed.

    What is its distribution? (actually it is not a must to know this answer; it just act as a side question to consolidate the concept)

    How does it relate to your original question?
    Thank you BGM.
    if I understand you correctly each xi has a geometric distribution with p=0.5.
    the probability that after 5 runs all the coins will remove is 1- p(x1<=5 and x2<=5 and ... x5<=5)=1-p(x1<=5)*p(x2<=5)*...*P(x5<=5)=1-[1-(0.5)^5]^5=0.15
    is it correct?

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    Re: a question about discrete RV

    Quote Originally Posted by BGM View Post
    Try to think in the following way:

    Let X_i be the number of tosses the i-th coin have before it is removed.

    What is its distribution? (actually it is not a must to know this answer; it just act as a side question to consolidate the concept)

    How does it relate to your original question?
    Thank you BGM.
    if I understand you correctly each xi has a geometric distribution with p=0.5.
    the probability that after 5 runs not all the coins will remove is 1- p(x1<=5 and x2<=5 and ... x5<=5)=1-p(x1<=5)*p(x2<=5)*...*P(x5<=5)=1-[1-(0.5)^5]^5=0.15
    is it correct?
    Last edited by avsha43; 08-08-2014 at 07:25 AM.

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    Re: a question about discrete RV

    Have not check the numeric answer but the steps look good to me.

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    Re: a question about discrete RV


    Thanks
    your explanation was very useful.
    Avshi

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