1. ## 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.

2. There are only 7 possible outcomes, you can do this by exhaustion. Note in this case that $\displaystyle P(X = x) = \binom{6}{x} \left ( \frac{1}{2} \right ) ^ 6$ or for a general binomial distributed RV, $\displaystyle P(X = x) = \binom{n}{x} p^x (1 - p) ^ {n - x}$.

You probably know this, but just in case $\displaystyle \binom{n}{x} = \frac{n!}{(x!)(n-x)!}$. This should also be enough to get you through the other problems you posted, just write out the $\displaystyle P(X = K)$ for each K, and do the problems by exhaustion.

3. Thanks!

4. I'm still a little confused... what is the k-value in this problem? How do I solve for that? Thanks

5. Originally Posted by mhitch03
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.