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Math Help - Help with this conditional PMF question

  1. #1
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    Help with this conditional PMF question

    Hi,

    This is one of the past questions from the entrance exam for the graduate school.
    I think I know the answer but I would welcome help with the authoritative answer.

    Question:

    Each time a coin is flipped, it has probability p of a HEAD and q of a TAIL. Let random variable X denote the number of flips until the first HEAD.

    (i) Show that P(X>1) = q.
    (ii) Write an expression for the conditional probability mass function P(X=x|X>1) in terms of p and q.
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  2. #2
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    Quote Originally Posted by itpro View Post
    Hi,

    This is one of the past questions from the entrance exam for the graduate school.
    I think I know the answer but I would welcome help with the authoritative answer.

    Question:

    Each time a coin is flipped, it has probability p of a HEAD and q of a TAIL. Let random variable X denote the number of flips until the first HEAD.

    (i) Show that P(X>1) = q.
    (ii) Write an expression for the conditional probability mass function P(X=x|X>1) in terms of p and q.
    What answers did you get (please show your working as well)?
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  3. #3
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    For part i)
    (and this is why I am seeking help because I have few pieces together but not a complete
    answer - specially not for the part ii)

    Total probability of all possible outcomes is 1. We have two equally possible outcomes p and q
    and therefore each has probability of 0.5 or 1-p and 1-q.

    I recognize P(X>1) as complementary cumulative distribution function ccdf which is equal to 1-cdf
    or 1-P(X<=1) or
    1- [P(0) + P(1)] = 1 - 0.5 = 0.5 which is q.

    Part ii

    Conditional probability P(A|B) = [P(A) ^ P(B)] / P(B) = [P(A) * P(B)] / P(B)

    In this case I have A = P(X=x) and B=P(X>1)

    I have just (attempted to) show that P(X>1) = q

    so P(X=x|X>1) = [P(X=x) * q] / q

    I can cancel out q so P(X=x|X>1) = P(X=x).

    Not sure how much of this is right and how to express P(X=x) as p.
    I know that intuitively given that p and q are independent trials each outcome
    has 0.5 chance of being a head.

    Help please.
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  4. #4
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    Quote Originally Posted by itpro View Post
    For part i)
    (and this is why I am seeking help because I have few pieces together but not a complete
    answer - specially not for the part ii)

    Total probability of all possible outcomes is 1. We have two equally possible outcomes p and q
    and therefore each has probability of 0.5 or 1-p and 1-q.

    I recognize P(X>1) as complementary cumulative distribution function ccdf which is equal to 1-cdf
    or 1-P(X<=1) or
    1- [P(0) + P(1)] = 1 - 0.5 = 0.5 which is q.

    Part ii

    Conditional probability P(A|B) = [P(A) ^ P(B)] / P(B) = [P(A) * P(B)] / P(B)

    In this case I have A = P(X=x) and B=P(X>1)

    I have just (attempted to) show that P(X>1) = q

    so P(X=x|X>1) = [P(X=x) * q] / q

    I can cancel out q so P(X=x|X>1) = P(X=x).

    Not sure how much of this is right and how to express P(X=x) as p.
    I know that intuitively given that p and q are independent trials each outcome
    has 0.5 chance of being a head.

    Help please.
    (a) is OK.

    For (b), do you realise that \Pr(X = x) = q^{x-1}p?
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    (a) is OK.

    For (b), do you realise that \Pr(X = x) = q^{x-1}p?
    I was "suspecting" at geometric distribution but I did not have much practice with it yet,
    so I felt it was as good as guessing.

    Thank you.
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