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Thread: [SOLVED] Binomial Distribution Problem

  1. #1
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    [SOLVED] Binomial Distribution Problem

    Hello, I am lost with the following problem.

    Consider two players, player A and player B, in a game that involves tossing a (not necessarily fair) coin with $\displaystyle p = P(H)$ such that A gains one dollar when a head $\displaystyle (H)$ occurs and B gains one dollar if a $\displaystyle T (tail)$ comes up. Assume that player A needs $\displaystyle n$ while player B needs $\displaystyle m$ more dollars to win the game (both $\displaystyle n$ and $\displaystyle m$ are non-negative integers)

    Let $\displaystyle A$ be the event that $\displaystyle n$ heads occur before $\displaystyle m$ tails. $\displaystyle A$ occurs if and only if there are at least $\displaystyle n$ heads in the fi…rst $\displaystyle n +m- 1$ trials. Explain why this reasoning is correct. Then use the binomial distribution to write an expression for $\displaystyle P(A) = P_{n,m}.$


    I know it is a very long and wordy problem, any hint will be appreciated, Thanks.
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  2. #2
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    Quote Originally Posted by akolman View Post
    Hello, I am lost with the following problem.

    Consider two players, player A and player B, in a game that involves tossing a (not necessarily fair) coin with $\displaystyle p = P(H)$ such that A gains one dollar when a head $\displaystyle (H)$ occurs and B gains one dollar if a $\displaystyle T (tail)$ comes up. Assume that player A needs $\displaystyle n$ while player B needs $\displaystyle m$ more dollars to win the game (both $\displaystyle n$ and $\displaystyle m$ are non-negative integers)

    Let $\displaystyle A$ be the event that $\displaystyle n$ heads occur before $\displaystyle m$ tails. $\displaystyle A$ occurs if and only if there are at least $\displaystyle n$ heads in the fi…rst $\displaystyle n +m- 1$ trials. Explain why this reasoning is correct. Then use the binomial distribution to write an expression for $\displaystyle P(A) = P_{n,m}.$


    I know it is a very long and wordy problem, any hint will be appreciated, Thanks.
    The question is not correct because there are contradictions between the statements 1 and 2 ; and statements 2 and 3, as labeled below:

    Statement 1: Assume that player A needs $\displaystyle n$ while player B needs $\displaystyle m$ "more" dollars to win the game (both $\displaystyle n$ and $\displaystyle m$ are non-negative integers)

    Statement 2:$\displaystyle A$ occurs if and only if there are at least $\displaystyle n$ heads in the fi…rst $\displaystyle n +m- 1$ trials.

    Statement 3: Then use the binomial distribution to write an expression for $\displaystyle P(A) = P_{n,m}.$

    In Statement 1, the word "more" should have been struck out.
    P(A) in Statement 3 should have been $\displaystyle P(A)=\binom{n+m-1}{n}p_h^np_t^{m-1}$

    -------------------
    Assuming A needs to win $\displaystyle n$ dollars, and B, $\displaystyle m$ dollars, A will occur, if and only if there are $\displaystyle n$ heads in the first $\displaystyle n+m-1$ trials.

    Example: Say n=3 dollars, m=2 dollars. For A to win there are n+m-1 trials, which have n+m-1=4 possible sample spaces {HHH},{HHTH},{HTHH},{THHH}

    $\displaystyle P(A)=\binom{3+2-1}{3}p_h^3p_t^{3+2-1-3}=\binom{4}{3}p_h^3p_t^{1}$

    Note: This is not a fair question because it does not tell us who will go first in the game. There is a big difference as to who will go first because the first to go will have a much better probability of winning. This you can test yourself by considering the expected value of each player. A flaw is found in this question in which it expected the solution to align with binomial distribution, for for which after a string of $\displaystyle n$ number of heads appeared, B would bother to complete the remaining number trials.
    Last edited by novice; Oct 24th 2009 at 07:48 AM.
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