Fair coin

mhitch03 · #1 $Old$ October 29th, 2009, 07:17 PM

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.

theodds · #2 $Old$ October 29th, 2009, 08:28 PM

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.

mhitch03 · #3 $Old$ October 29th, 2009, 08:51 PM

Thanks!

mhitch03 · #4 $Old$ October 30th, 2009, 06:13 AM

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

mr fantastic · #5 $Old$ October 30th, 2009, 06:37 AM

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.