Fliping 11 coins with $0$ and $1$ on their sides, probability of the event $\text{ there are more zeros than ones }$ you flip 11 coins wich has sides of $1$ and $0$, what is the probability of the event $$\text{ there are more zeros than ones }$$
my first attempt was just write $$6\cdot \frac{1}{2}$$ but first of all this is not resonable because it is bigger than 1. then I realised 2 things:

*

*I dont understand how to determine my probability space

*How do I considers another cases with there is more zeros than ones, for example I got 7 zeros and 4 ones.

I hope some 1 can help
 A: The answer is of course $1/2$, because there are only two equally likely outcomes:  more $0$s or more $1$s.
A: By symmetry, the probability of having more heads than tails will be exactly equal to the probability of having more tails than heads, given the coin is fair.
You cannot have an equal number of heads than tails with an odd number of coins, so the required probability is simply $\frac 12$.
Slightly more formally, let the probability of more heads than tails be $p$. Then the probability of more tails than heads is also $p$. Finally, let the probability of equal heads and tails be $q$. You have that $p +p +q =1 \implies p = \frac{1-q}2$. Now note that $q=0$ for an odd number of coins (it would be non-zero for an even number). Hence, $p = \frac 12$.
A: Basically, the situation when there are more 0s than 1s are the cases where you flipped 6 times 0s + 7 times + ...  until you hit 11 times.
The probability that you flipped $n$ zeros out of $11$ is:
$$ \binom{11}{n}  \left( \frac{1}{2} \right)^{11}$$
Because we choose $n$ coins out of $11$ and then calculating each probability that it hits a zero or a one: $ \overset{\text{0s}}{\frac{1}{2^n}} \cdot \overset{\text{1s}}{\frac{1}{2^{11-n}}}$
You need to calculate the sum:
$$ \sum_{n = 6}^{11}  \binom{11}{n}  \left( \frac{1}{2} \right)^{11}$$
You can also interpret this question how David has, because this question is a dichotomy (Either you get more 0s or more 1s)
