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Math Help - non-fair coin

  1. #1
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    non-fair coin

    Suppose that a non-fair coin with  P(\text{Heads}) = 0.6 is tossed repeatedly.

    (a) What is the probability that the first time we get "Heads" happens on an even numbered toss?

    (b) What is the probability that the first "Heads" occurs on an odd-numbered toss?

    (c) Explain why the answer (b) is larger than the answer in (a). Does it depend on  P(\text{Heads}) ?

    So let  P_{n} denote the probability that the first head appears on the n th toss. Then  P_{1} = 0.6, \ P_{2} = (0.6)(0.4), \ P(3) = (0.4)^{2}(0.6) \dots

    In general,  P_{n} =  \frac{3}{5}\left(\frac{2}{5}\right)^{n-1}  . So for part (a) it would be  P_{2} + P_{4}+P_{6}+ \ldots + P_{2n} where  n \to \infty .

    Similarly, for part (b) it would be  P_{1}+P_{3} + P_{5} + \ldots + P_{2n+1} where  n \to \infty .

    So  P_{2n} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n-1} and  P_{2n+1} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n} = \frac{3}{5} \left(\frac{4}{25} \right)^{n} .

    Then  P_{2n+1} = \sum_{n=1}^{\infty} \frac{12}{125} \left(\frac{4}{25} \right)^{n-1} = \frac{\frac{12}{125}}{\frac{21}{25}} \approx 0.1142 . Then  P_{2n} = 1-0.1142 = 0.8858 ?

    It seems like  P_{2n+1} < P_{2n} .
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lord12 View Post
    Suppose that a non-fair coin with  P(\text{Heads}) = 0.6 is tossed repeatedly.

    (a) What is the probability that the first time we get "Heads" happens on an even numbered toss?

    (b) What is the probability that the first "Heads" occurs on an odd-numbered toss?

    (c) Explain why the answer (b) is larger than the answer in (a). Does it depend on  P(\text{Heads}) ?

    So let  P_{n} denote the probability that the first head appears on the n th toss. Then  P_{1} = 0.6, \ P_{2} = (0.6)(0.4), \ P(3) = (0.4)^{2}(0.6) \dots

    In general,  P_{n} = \frac{3}{5}\left(\frac{2}{5}\right)^{n-1} . So for part (a) it would be  P_{2} + P_{4}+P_{6}+ \ldots + P_{2n} where  n \to \infty .

    Similarly, for part (b) it would be  P_{1}+P_{3} + P_{5} + \ldots + P_{2n+1} where  n \to \infty .

    So  P_{2n} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n-1} and  P_{2n+1} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n} = \frac{3}{5} \left(\frac{4}{25} \right)^{n} .

    Then  P_{2n+1} = \sum_{n=1}^{\infty} \frac{12}{125} \left(\frac{4}{25} \right)^{n-1} = \frac{\frac{12}{125}}{\frac{21}{25}} \approx 0.1142 . Then  P_{2n} = 1-0.1142 = 0.8858 ?

    It seems like  P_{2n+1} < P_{2n} .
    You seem to have lost P_1 somewhere.

     <br />
P_{odd}=0.6 + 0.4^2 0.6 + ... =0.6 (1 + 0.4^2 + 0.4^4 ...)=0.6 \frac{1}{1-0.4^2} \approx 0.714..<br />


    RonL
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