Five people are tossing a coin ten times. What is the probability that at least 1 person gets heads 10 times? I know that the probability that one person getting heads all ten times is 
1/(2^10) or 1/1024.  

I also know the calculation when using "at least one" is
P(At least one) = 1 - P(None).  

The probability of one person getting no heads is the same:
1/1024.  

The part that is confusing is the number of people.  Is it 
P(At least one) = 1 - (1/1024)^5?  

If there were 100 people tossing a coin, the probability that at least one person gets all ten heads is 
P(At least one) = 1 - (1/1024)^100?  

Is this correct?
 A: The probability of getting $10$ heads is $\frac1{2^{10}}$.
Let $X$ be the number people getting $10$ heads, this follows binomial distribution $Bin(n, \frac1{2^{10}})$.
Hence $$P(X \ge 1) = 1-P(X=0)=1-\left( 1-\frac1{2^{10}}\right)^{n}$$
A: The probability that at least one person gets heads ten times, equals 1 minus the probability that no one gets heads ten times. As you correctly derived, the probability of getting heads for a specific individual equals $\frac{1}{2^{10}}$. Then, the probability we are looking for equals:
$$1 - \left(1 - \frac{1}{2^{10}}\right)^5 \approx 0.0049$$ 
A: Suppose we have $n$ people, each of whom tosses a coin $m$ times and that coin falls heads with probability $p$. Then the probability that the exact person chosen by us gets all heads is $p^m$. Thus the probability that he does not is $1- p^m$. So the probability that none of them had all heads is $(1 - p^m)^n$. And thus the probability, that at least one of them had all heads is $1 - (1 - p^m)^n$. That is the answer.
