What is the probability that he counts head more that tails when tosses a coin 6 times Suppose kitty tosses a coin 6 times. What is the probability that she counts more "heads" than tails?
This is a question from the test and I got only partial marks so I am wondering what is the solution to this.
What I did was I assume the number of coin outcomes,for example: hhhhhh,tttttt,and so on and I use the formula P(E) = n(E)/n(S) 
where n(E) is the number of events n that has heads more than tails. 

Since head and tails have same probability, what you seek is half the probability of not getting an equal number of head and tails. And you get 3 head and 3 tails in ${6\choose 3}=20$ different ways out of $2^6=64$ possible outcomes.
So, the requested probability is:
$$
p = \frac{1 - \frac{20}{64}}{2} = \frac{11}{32}.
$$
added. A very intuitive way to understand what is happening. Consider the 6-th power of the binomial $T+H$:
$$
(T+H)^6 = T^6 + 6 T^5 H + 15 T^3 H^2 + 20 T^3 H^3 + 15 T^2 H^4 + 6 T H^5 + H^6.
$$
If you develop the product $(T+H)^6$ you have to choose T or H for six times. So you get all possible outcomes of a 6-tuple of coin tosses. The binomial formula tells us how we can put together the outcomes if we are not interested in the order. So, out of $2^6=64$ possible outcomes you have one of only tails $T^6$, six of $T^5H$ (i.e. 5 tails and one head) and so on. You see that to have more heads than tails you have to consider $15+6+1$ out of $64$ possibilities. Or, more quickly, you note that this number is half of the total minus 20: $(64-20)/2$ over $64$ possibilities. 
