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Math Help - coin flip

  1. #1
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    coin flip

    Two players, Jim and Tom, each are going to flip 3 fair coins. What is the probability that they will get the same number of heads?

    I am reviewing for an exam and cannot figure out how to get to this answer. The key says 5/16
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  2. #2
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    \frac{1}<br />
{{64}} + \frac{9}<br />
{{64}} + \frac{9}<br />
{{64}} + \frac{1}<br />
{{64}} = \frac{5}<br />
{{16}}
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  3. #3
    MHF Contributor Unknown008's Avatar
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    P(both 3 heads) = (\frac{1}{2})^3 \times (\frac{1}{2})^3

    P(both 2 heads) = 3(\frac{1}{2})^3 \times 3(\frac{1}{2})^3

    I used 3 because you can have HHT, HTH or THH

    P(both 1 head) = 3(\frac{1}{2})^3 \times 3(\frac{1}{2})^3

    Same thing here, TTH, THT, HTT

    P(both 0 head) = (\frac{1}{2})^3 \times (\frac{1}{2})^3

    Then add all of them to give 5/16
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  4. #4
    Super Member Random Variable's Avatar
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    Let  A be the event that they both flip the same number of heads.

    Condition on the number of heads Jim flips.

     P(A) = P(A|0)P(0) + P(A|1)P(1) + P(A|2)P(2) + P(A|3)P(3)

     = \binom{3}{0} (1/2)^{0} (1/2)^{3}*\binom{3}{0} (1/2)^{0} (1/2)^{3} + \binom{3}{1} (1/2)^{1} (1/2)^{2}*\binom{3}{1} (1/2)^{1} (1/2)^{2} + \binom{3}{2} (1/2)^{2} (1/2)^{1}*\binom{3}{2} (1/2)^{2} (1/2)^{1}  + \binom{3}{3} (1/2)^{3} (1/2)^{0}*\binom{3}{3} (1/2)^{3} (1/2)^{0}

     = \frac{1}{8}*\frac{1}{8} + \frac{3}{8}*\frac{3}{8} + \frac{3}{8}*\frac{3}{8} + \frac{1}{8}*\frac{1}{8} = \frac{5}{16}
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