Statistics brain teaser This appears to be a fairly simple question, although it stumped me:
Say there is a kitchen with 12 children, and one cookie is left. Each child with a 10 percent probability will eat the cookie today. What is the probability that the cookie is eaten today?
EDIT:
I assume the reason the answer is not $0.1^{12}$ is because that would be the Pr(all children eating the cookie), which is illogical?
 A: Each child is expected to eat $\frac1{10}$ cookies. There are $12$ children. By the linearity of expectation, the expected number of eaten cookies is $\frac{12}{10}\gt1$. Since there is only one cookie, this shows that the specified probabilities are inconsistent.
A consistent specification that may have been intended would be: Each child will try to eat the cookie today with probability $\frac1{10}$. These attempts are independent. The cookie will be eaten if and only if at least one child tries to eat it.
In this case, the probability that there is no attempt to eat the cookie is $\left(\frac9{10}\right)^{12}\approx28\%$, so the probability that the cookie is eaten is the complement, approximately $72\%$.
A: You can also model the problem more formally. Indeed, imagine that, given the information you have, a child eating a cookie follows a Bernoulli distribution with parameter $p=0.1$, i.e. $X_1$~$Bern(p)$ and the event {'Child 1 eats cookie'} equals the event {$X_1$ = 1}. 
Now, you can model all the children in the kitchen by $n=12$ independent Bernoulli random variable, $X_1,...,X_{12}$. 
Now, how would you model the fact that the cookie is eaten or not? The cookie is eaten if at least on child eats it, so we can look at the sum of our random variables, say $Y=\sum_{i=1}^{12}X_i$ which we know follows a Binomial distribution with parameters $n=12$ and $p=0.1$, i.e. $Y$~$Bin(n,p)$.
So the event {'Cookie doesn't get eaten'} equals the event {$Y=0$}, and we know that $\forall k \in [\![0, 12 ]\!]$ $\mathbb P(Y=k)=C_{k}^{n}p^{k}(1-p)^{n-k}$. So: 
$\mathbb P(${'Cookie doesn't get eaten'}$) = \mathbb P(Y=0)=C_{0}^{n}p^{0}(1-0.1)^{12-0}=(\frac{9}{10})^{12}$ but we are interested in $\mathbb P(${'Cookie gets eaten'}$)= 1-\mathbb P(${'Cookie doesn't get eaten'}$)=1-(\frac{9}{12})^{12}\approx 0.717$
To your point, that wouldn't be equivalent to computing $\mathbb P(Y=12)$ as it doesn't make any sense to say that the twelve kids ate the cookie. 
