A probability within a probability Let's say I am doing 10 coin flips and I want to calculate the probability of getting at most 3 heads. I have calculated this out to be
$$p=\left(\binom{10}0+\binom{10}1+\binom{10}2+\binom{10}3\right)\left(\frac12\right)^{10}$$
Now, let's say 100 people are performing this experiment. Now the probability of everyone flipping at most 3 heads is $p^{100}$.
Now, what if I were to bend the question and say what is the probability of at least one person flipping at most 3 heads. Then, my intuition is to do something like: probability of exactly one person getting this outcome ($\binom{100}1$), plus the probability of exactly two people getting this outcome, and so on. However, this can quickly get out of hand. Hence, is there a more concise way to approach this? I thought of approaching it as "probability that no one gets this outcome - 1". However, I am not sure how I would represent "no one getting this outcome," since it is not the same as "everyone" getting the same output.
 A: The natural way to tackle that problem is to compute the probability of that to not happen. Let me fix some notation first: set $X$ to be the number of people that got at most three heads, and let $p$ be the probability that you computed first, of each person individually to get at most three heads.
So, you are interested in computing $\mathbb{P}[X \geq 1 ]$, try instead to compute $\mathbb{P}[X < 1 ] = \mathbb{P}[X =  0] = 1 - \mathbb{P}[X\geq 1]$.
This probability is much easier to compute, as it is simply $(1 - p )^{100}$, so the number you are looking for is $\mathbb{P}[X\geq 1] = 1 - (1 - p)^{100} $.
This corresponds to compute the "other tail" of the binomial distribution, which is significantly smaller. Note that you couldn't get away with this strategy if you have asked for the probability of at least 50 people to get at most three heads. For that you need other kind of arguments (either a powerful computer or some approximation with controled error, see Chernoff bound).
Note that this is a rather general approach. Instead of trying to compute the raw probability you can always try to compute the probability of its complement. Other similar tactics are conditioning your event on other and Bayes theorem. Once you have endowed yourself with a lot of practice with these tactics, to use them iterativelly is a very strong weapon and should be your first approach to a combinatorics/probabilistic problem, before the cmoputations explode, like what you observed.
A: Denote (as you already have) with $p$ the probability of a person flipping at most $3$ heads, (i.e. the probability of success) $$p:= \left(\binom{10}0+\binom{10}1+\binom{10}2+\binom{10}3\right)\left(\frac12\right)^{10}
$$ and denote with $n$ the total number of people that you are considering $($e.g. $n=100)$. 
The number of people $X$ among the $n$ people, that flip at most $3$ heads is a binomial random variable with parameters $p$ (probability of success) and $n$ (total number of people that flip). Hence, for any $0\le k\le n$ you have that $$P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}$$ Hence to find the probability of at least one person flipping at most $3$ heads, among $n=100$ persons, you write $$P(X\ge 1)=1-P(X=0)=1-\dbinom{100}{0}p^0(1-p)^{100-0}=1-(1-p)^{100}$$ 
