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Math Help - Fair coin

  1. #1
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    Fair coin

    A fair coin is tossed 6 times. Let x = number of heads obtained. Find the value of K for which Pr(x=K+1)-Pr(x=K) is the maximum possible.
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  2. #2
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    There are only 7 possible outcomes, you can do this by exhaustion. Note in this case that P(X = x) = \binom{6}{x} \left ( \frac{1}{2} \right ) ^ 6 or for a general binomial distributed RV, P(X = x) = \binom{n}{x} p^x (1 - p) ^ {n - x}.

    You probably know this, but just in case \binom{n}{x} = \frac{n!}{(x!)(n-x)!}. This should also be enough to get you through the other problems you posted, just write out the P(X = K) for each K, and do the problems by exhaustion.
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  3. #3
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    Thanks!
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  4. #4
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    I'm still a little confused... what is the k-value in this problem? How do I solve for that? Thanks
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  5. #5
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    Quote Originally Posted by mhitch03 View Post
    I'm still a little confused... what is the k-value in this problem? How do I solve for that? Thanks
    Substitute the appropriate expressions into Pr(x=K+1)-Pr(x=K) and then test each value of k (k = 0, 1, 2 .... 5) and see which one gives the maximum value.
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