Markov's and Tchebychev's Inequality to find Upper Bound I'm not sure how to solve this problem:
C = random variable; the number of heads in 100 independent fair coin flips
Find E(C) and V(C). 
Find the upper bounds on P(C >75), using Markov's Inequality and Tchebychev's Inequality.
I understand how to get E(C) (which I believe is 50), but I'm confused on how to complete the rest.
 A: $C$ is distributed according to the binomial distribution with probability $p$, which I assume is $1/2$ in this case.
According to
http://en.wikipedia.org/wiki/Binomial_distribution
we have $E(C)=np$ which in your case is indeed 50. Furthermore, $V(C)=np(1-p)$.
You can use the Markov inequality
$$P(X\geq a)\leq a^{-1} E(X)$$ to calculate an upper bound to $P(C>75)$ since you already know $E(C)$.
A: You should have have $E[C]=50$ and $Var(C)=25$.
Dima McGreen has shown Markov's inequality gives $P(C \gt 75) \lt \tfrac{50}{75}$ or $P(C \ge 76) \le \tfrac{50}{76}$ 
For Chebyshev's inequality, note $75 = 50 + 5 \times \sqrt{25}$, and 


*

*either use the two-sided $P\left(|X-\mu_X| \gt k\sigma_X \right) \lt \tfrac{1}{k^2}$ in this case $P\left(|C-50| \gt 25\right) \lt \tfrac{1}{5^2}=\tfrac{1}{25}$ 

*or the one sided $P\left(X-\mu_X \gt k\sigma_X \right) \lt \tfrac{1}{k^2+1}$ in this case $P\left({C-50} \gt 25\right) \lt \tfrac{1}{5^2+1}=\tfrac{1}{26}$.  
Again you could marginally improve this by looking at $P(C \ge 76)$.
So the Chebyshev inequality gives a substantially tighter bound than the Markov inequality.  But this is still very loose.  In fact, using the binomial distribution, $P(C \gt 75)$ is slightly less than $0.0000001$.
